{"id":626502,"date":"2023-09-12T04:17:45","date_gmt":"2023-09-12T04:17:45","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626502"},"modified":"2023-09-12T04:17:52","modified_gmt":"2023-09-12T04:17:52","slug":"kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/","title":{"rendered":"KC Sinha: Exercise 22.1- Mathematics Solution Class 12 Chapter 22 \u0938\u092e\u093e\u0915\u0932 \u0915\u0947 \u0905\u0928\u0941\u092a\u094d\u0930\u092f\u094b\u0917"},"content":{"rendered":"\n

<\/span><\/p>\n\n\n\n

Question 1<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930 y=x2<\/sup> ,\u0930\u0947\u0916\u093e\u0913 x=1, x=2 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964
[Find the area bounded by the curve y=x2<\/sup> , lines x=1, x=2 and x-axis]
Sol :
ar(ABCDA)=$\\int_{1}^{2}ydx$<\/p>\n\n\n\n

$=\\int_{1}^{2}x^2dx$<\/p>\n\n\n\n

$=\\left[\\frac{x^3}{3}\\right]_{1}^{2}$<\/p>\n\n\n\n

$=\\frac{2^3}{3}-\\frac{1^3}{3}$<\/p>\n\n\n\n

$=\\frac{8}{3}-\\frac{1}{3}$<\/p>\n\n\n\n

$=\\frac{7}{3}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 2<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930 y=x4<\/sup> ,\u0930\u0947\u0916\u093e\u0913 x=1, x=5 \u090f\u0935\u0902 x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964
[Find the area bounded by the curve y=x4<\/sup> , lines x=1, x=5 and x-axis]
Sol :
ar(ABCDA)$=\\int_{1}^{5}x^4dx$<\/p>\n\n\n\n

$=\\left[\\frac{x^5}{5}\\right]_{1}^5$<\/p>\n\n\n\n

$=\\frac{5^5}{5}-\\frac{1^5}{5}$<\/p>\n\n\n\n

=625-0.2
=624.8 \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

\u0930\u0947\u0916\u093e y=x, x-\u0905\u0915\u094d\u0937 \u0924\u0925\u093e \u0915\u094b\u091f\u093f\u092f\u094b\u0902 x=-1 \u0914\u0930 x=2 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964
[Find the area bounded by the line y=x, the x-axis and the ordinates x=-1 and x=2]
Sol :
\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(\u0394OAB)+ar(\u0394COD)<\/p>\n\n\n\n

$=-\\int_{-1}^{0}xdx+\\int_{0}^{2}xdx$<\/p>\n\n\n\n

$=-\\left[\\frac{x^2}{2}\\right]_{-1}^{0}+\\left[\\frac{x^2}{2}\\right]_{0}^{2}$<\/p>\n\n\n\n

$=-\\left[\\frac{0^2}{2}-\\frac{(-1)^2}{2}\\right]+\\left[\\frac{2^2}{2}-\\frac{0}{2}\\right]$<\/p>\n\n\n\n

$=\\frac{1}{2}+2$<\/p>\n\n\n\n

$=\\frac{1+4}{2}=\\frac{5}{2}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 4<\/h4>\n\n\n\n

\u0930\u0947\u0916\u093e y=3x+2, x-\u0905\u0915\u094d\u0937 \u090f\u0935\u0902 \u0915\u094b\u091f\u093f\u092f\u094b x=-1 \u090f\u0902\u0935 x=1 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964
[Find the area of the region bounded by the line y=3x+2, the x-axis and the ordinates x=-1 and x=1]
Sol :
ar(ABCDA)$=\\int_{-1}^{1}(3x+2)dx$<\/p>\n\n\n\n

$=\\left[\\frac{3x^2}{2}+2x\\right]_{-1}^{1}$<\/p>\n\n\n\n

$=\\left[\\frac{3(1)^2}{2}+2(1)\\right]-\\left[\\frac{3(-1)^2}{2}+2(-1)\\right]$<\/p>\n\n\n\n

$=\\frac{3}{2}+2-\\frac{3}{2}+2$<\/p>\n\n\n\n

=4 \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 5<\/h4>\n\n\n\n

\u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930 2y=5x+7, x-\u0905\u0915\u094d\u0937 \u090f\u0902\u0935 \u0930\u0947\u0916\u093e\u0913 x=2 \u0924\u0925\u093e x=8  \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964
Sol :
ar(ABCDA)$=\\int_{2}^{8}(\\frac{5}{2}x+\\frac{7}{2})dx$<\/p>\n\n\n\n

$=\\left[\\frac{5}{4}x^2+\\frac{7}{2}x\\right]_{2}^{8}$<\/p>\n\n\n\n

$=\\left[\\frac{5}{4}(8)^2+\\frac{7}{2}(8)\\right]-\\left[\\frac{5}{4}(2)^2+\\frac{7}{2}(2)\\right]$<\/p>\n\n\n\n

$=\\left[\\frac{5}{4}(64)+28\\right]-\\left[\\frac{5}{4}(4)+7\\right]$<\/p>\n\n\n\n

=[80+28]-12<\/p>\n\n\n\n

=108-12=96 \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 9<\/h4>\n\n\n\n

\u092a\u0930\u0935\u0932\u092f y2<\/sup>=4ax, \u0907\u0938\u0915\u093e \u0905\u0915\u094d\u0937 \u0924\u0925\u093e \u0926\u094b \u0915\u094b\u091f\u093f\u092f\u094b x=a \u090f\u0935\u0902 x=2a \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964 
Sol :
y2<\/sup>=4ax<\/p>\n\n\n\n

$y=2\\sqrt{a}x^\\frac{1}{2}$<\/p>\n\n\n\n

ar(ABCDA)$=\\int_{a}^{2a}2\\sqrt{a}.x^{\\frac{1}{2}}dx$<\/p>\n\n\n\n

$=2\\sqrt{a}\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]^{2a}$<\/p>\n\n\n\n

$=\\frac{4}{a}\\sqrt{a}\\left[x^{\\frac{3}{2}}\\right]_{a}^{2a}$<\/p>\n\n\n\n

$=\\frac{4}{3}\\sqrt{a}\\left[(2a)^{\\frac{3}{2}}-a^{\\frac{3}{2}}\\right]$<\/p>\n\n\n\n

$=\\frac{4}{3}\\sqrt{a}\\left[2\\sqrt{2}.a^{\\frac{3}{2}}-a^{\\frac{3}{2}}\\right]$<\/p>\n\n\n\n

$=\\frac{4}{3}a^{\\frac{1}{2}}.a^{\\frac{3}{2}}(2\\sqrt{2}-1)$<\/p>\n\n\n\n

$=\\frac{4}{3}a^2(2\\sqrt{2}-1)$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 10<\/strong><\/p>\n\n\n\n

\u0935\u0915\u094d\u0930 y2<\/sup>=x \u0938\u0930\u0932 \u0930\u0947\u0916\u093e\u0913 x=1, x=4 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964
Sol :
ar(ABCDA)$=\\int_{1}^{4}x^{\\frac{1}{2}}dx$<\/p>\n\n\n\n

$=\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]^{4}$<\/p>\n\n\n\n

$=\\frac{2}{3}(4)^{\\frac{3}{2}}-\\frac{2}{3}(1)^{\\frac{3}{2}}$<\/p>\n\n\n\n

$=\\frac{2}{3}(2)^{2\\times \\frac{3}{2}}-\\frac{2}{3}$<\/p>\n\n\n\n

$=\\frac{16}{3}-\\frac{2}{3}$<\/p>\n\n\n\n

$=\\frac{14}{3}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 11<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930 y2<\/sup>=4x \u0924\u0925\u093e \u0938\u0930\u0932 \u0930\u0947\u0916\u093e x=3 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964<\/p>\n\n\n\n

[Find the area of the region bounded by the curve y2<\/sup>=4x and the line x=3] <\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y2<\/sup>=4x \u21d2$y=2x^{\\frac{1}{2}}$<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 =2\u00d7ar(OABO)<\/p>\n\n\n\n

$2\\int_{0}^{3}2x^{\\frac{1}{2}}$<\/p>\n\n\n\n

$=4\\int_{0}^{3}x^{\\frac{1}{2}}dx$<\/p>\n\n\n\n

$=4\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{0}^3$<\/p>\n\n\n\n

$=4\\left[\\frac{2}{3}(3)^{\\frac{3}{2}}-\\frac{2}{3}(0)^{\\frac{3}{2}}\\right]$<\/p>\n\n\n\n

$=4\\left[\\frac{2}{3}\\times 3\\sqrt{3}-0\\right]$<\/p>\n\n\n\n

=8\u221a3<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 12<\/h4>\n\n\n\n

\u092a\u094d\u0930\u0925\u092e \u091a\u0924\u0941\u0930\u094d\u0925\u093e\u0936 \u092e\u0947 y2<\/sup>=4x ,x=1 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y2<\/sup>=4x \u21d2y=2\u221ax=$2x^{\\frac{1}{2}}$<\/p>\n\n\n\n

ar(ABCDA)$=\\int_{1}^{4}2x^{\\frac{1}{2}}dx$<\/p>\n\n\n\n

$=2\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{1}^{4}$<\/p>\n\n\n\n

$=2\\left[\\frac{2}{3}(4)^{\\frac{3}{2}}-\\frac{2}{3}(1)^{\\frac{3}{2}}\\right]$<\/p>\n\n\n\n

$=2\\left[\\frac{16}{3}-\\frac{2}{3}\\right]$<\/p>\n\n\n\n

$=2\\times \\frac{14}{3}$<\/p>\n\n\n\n

$=\\frac{28}{3}$ \u0935\u0930\u094d\u0917  \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 14<\/h4>\n\n\n\n

\u092a\u0930\u0935\u0932\u092f x2<\/sup>=y, y-\u0905\u0915\u094d\u0937 \u0924\u0925\u093e \u0930\u0947\u0916\u093e y=1 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

x=\u221ay \u21d2$x=y^{\\frac{1}{2}}$<\/p>\n\n\n\n

ar(OABO)$=\\int_{0}^{1}y^{\\frac{1}{2}}dy$<\/p>\n\n\n\n

$=\\left[\\frac{2}{3}y^{\\frac{3}{2}}\\right]_{0}^{1}$<\/p>\n\n\n\n

$=\\frac{2}{3}(1)^{\\frac{3}{2}}-\\frac{2}{3}(0)^{\\frac{3}{2}}$<\/p>\n\n\n\n

$=\\frac{2}{3}$ \u0935\u0930\u094d\u0917  \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 15<\/h4>\n\n\n\n

\u092a\u094d\u0930\u0925\u092e \u091a\u0924\u0941\u0930\u094d\u0925\u093e\u0936 \u092e\u0947 x2<\/sup>=4x, y=2, y=4 \u0924\u0925\u093e y-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

x2<\/sup>=4y \u21d2x=\u221a4y \u21d2$x=2y^{\\frac{1}{2}}$<\/p>\n\n\n\n

ar(ABCDA)$=\\int_{2}^{4}2y^{\\frac{1}{2}}dy$<\/p>\n\n\n\n

$=2\\int_{2}^{4}y^{\\frac{1}{2}}dy$<\/p>\n\n\n\n

$=2\\left[\\frac{2}{3}y^{\\frac{3}{2}}\\right]_{2}^{4}$<\/p>\n\n\n\n

$=\\frac{4}{3}\\left[y^{\\frac{3}{2}}\\right]_{2}^{4}$<\/p>\n\n\n\n

$=\\frac{4}{3}\\left[(4)^{\\frac{3}{2}}-(2)^{\\frac{3}{2}}\\right]$<\/p>\n\n\n\n

$=\\frac{4}{3}\\left[2^{2\\times \\frac{3}{2}-2\\sqrt{2}}\\right]$<\/p>\n\n\n\n

$=\\frac{4}{8}\\left[8-2\\sqrt{2}\\right]$<\/p>\n\n\n\n

$=\\frac{8}{3}\\left[4-\\sqrt{2}\\right]$ \u0935\u0930\u094d\u0917  \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 16<\/h4>\n\n\n\n

\u092a\u0930\u0935\u0932\u092f y2<\/sup>=4ax \u0914\u0930 \u0909\u0938\u0915\u0947 \u0928\u092d\u093f\u0932\u0902\u092c \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

[Find the area bounded by the parabola y2<\/sup>=4ax and its latus rectum]<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y2<\/sup>=4ax \u21d2$y=2\\sqrt{a}x^{\\frac{1}{2}}$<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=2\u00d7ar(OABO)<\/p>\n\n\n\n

$=2\\times \\int_{0}^{a}2\\sqrt{a}x^{\\frac{1}{2}}dx$<\/p>\n\n\n\n

$=4\\sqrt{a}\\int_{0}^{a}x^{\\frac{1}{2}}dx$<\/p>\n\n\n\n

$=4\\sqrt{a}\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{0}^{a}$<\/p>\n\n\n\n

$=\\frac{8}{3}a^{\\frac{1}{2}}\\left[x^{\\frac{3}{2}}\\right]_{0}^{a}$<\/p>\n\n\n\n

$=\\frac{8}{3}a^{\\frac{1}{2}}\\left(a^{\\frac{3}{2}-0^{\\frac{3}{2}}}\\right)$<\/p>\n\n\n\n

$=\\frac{8}{3}a^2$ \u0935\u0930\u094d\u0917  \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 17<\/h4>\n\n\n\n

\u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u093e \u0909\u092a\u092f\u094b\u0917 \u0915\u0930\u0924\u0947 \u0939\u0941\u090f \u092a\u0930\u0935\u0932\u0930 y2<\/sup>=16x \u0924\u0925\u093e \u0930\u0947\u0916\u093e x=4 \u0915\u093e \u092e\u0927\u094d\u092f\u0935\u0930\u094d\u0924\u0940 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092a\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y2<\/sup>=16x \u21d2y=4\u221ax \u21d2$y=4x^{\\frac{1}{2}}$<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 <\/p>\n\n\n\n

=2\u00d7ar(OABO)<\/p>\n\n\n\n

$=2\\times \\int_{0}^{4}4x^{x\\frac{1}{2}}dx$<\/p>\n\n\n\n

$=8\\int_{0}^{4}x^{\\frac{1}{2}}dx$<\/p>\n\n\n\n

$=8\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{0}^{4}$<\/p>\n\n\n\n

$=\\frac{16}{3}\\left[x^{\\frac{3}{2}}\\right]_{0}^{4}$<\/p>\n\n\n\n

$=\\frac{16}{3}\\left[(4)^{\\frac{3}{2}}-0^{\\frac{3}{2}}\\right]$<\/p>\n\n\n\n

$=\\frac{16}{3}(2)^{2\\times \\frac{3}{2}}$<\/p>\n\n\n\n

$=\\frac{128}{3}$ \u0935\u0930\u094d\u0917  \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 19<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930 y2<\/sup>=4ax, \u0930\u0947\u0916\u093e y=2a \u0924\u0925\u093e y-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

[Find the area bounded by the curve y2<\/sup>=4ax,the line y=2a and y-axis ]<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$x=\\frac{1}{4a}y^2$<\/p>\n\n\n\n

ar(OABO)$=\\frac{1}{4a}\\int_{0}^{2a}y^2 dy$<\/p>\n\n\n\n

$=\\frac{1}{4a}\\left[\\frac{y^3}{3}\\right]_{0}^{2a}$<\/p>\n\n\n\n

$=\\frac{1}{4a}\\left[\\frac{(2a)^3}{3}-\\frac{0^3}{3}\\right]$<\/p>\n\n\n\n

$=\\frac{1}{4a}\\left[\\frac{8a^3}{3}\\right]$<\/p>\n\n\n\n

$=\\frac{2}{3}a^2$ \u0935\u0930\u094d\u0917  \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 20<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930\u094b x=y2 <\/sup>\u0924\u0925\u093e x=4 \u0915\u0947 \u092c\u0940\u091a \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0930\u0947\u0916\u093e x=a \u0926\u094d\u0935\u093e\u0930\u093e \u0926\u094b \u092c\u0930\u093e\u092c\u0930 \u092d\u093e\u0917\u094b \u092e\u0947 \u0935\u093f\u092d\u093e\u091c\u093f\u0924 \u0939\u094b\u0924\u093e \u0939\u0948, \u0924\u094b a \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y2<\/sup>=x \u21d2$y=x^{\\frac{1}{2}}$<\/p>\n\n\n\n

ar(OCDO)=ar(ABDCA)<\/p>\n\n\n\n

$\\int_{0}^{a}x^{\\frac{1}{2}}dx$<\/p>\n\n\n\n

$=\\int_{a}^{4}x^{\\frac{1}{2}}dx$<\/p>\n\n\n\n

$\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{0}^{a}$<\/p>\n\n\n\n

$=\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{a}^{4}$<\/p>\n\n\n\n

$=\\frac{2}{3}(a)^{\\frac{3}{2}}-\\frac{2}{3}(0)^{\\frac{3}{2}}$<\/p>\n\n\n\n

$=\\frac{2}{3}(4)^{\\frac{3}{2}}-\\frac{2}{3}(a)^{\\frac{3}{2}}$<\/p>\n\n\n\n

$a^{\\frac{2}{3}}=\\frac{2}{3}\\times 8 \\times \\frac{3}{4}$<\/p>\n\n\n\n

$a^{\\frac{3}{2}}=4$<\/p>\n\n\n\n

$a=(4)^{\\frac{2}{3}}$<\/p>\n\n\n\n

$a=[(4)^2]^{\\frac{1}{3}}$<\/p>\n\n\n\n

$a=(16)^{\\frac{1}{3}}$<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 21<\/h4>\n\n\n\n

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924\u094b \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964<\/p>\n\n\n\n

[Find the area of the region bounded by the following ellipses]<\/p>\n\n\n\n

(i)<\/strong> $\\frac{x^2}{4}+\\frac{y^2}{9}=1$ <\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$\\frac{x^2}{2^2}+\\frac{y^2}{3^2}=1$<\/p>\n\n\n\n

$\\frac{y^2}{3^2}=1-\\frac{x^2}{2^2}$<\/p>\n\n\n\n

$\\frac{y^2}{3^2}=\\frac{2^2-x^2}{2^2}$<\/p>\n\n\n\n

$y^2=\\frac{3^2}{2^2}(2^2-x^2)$<\/p>\n\n\n\n

$y=\\sqrt{\\frac{3^2}{2^2}(2^2-x^2)}$<\/p>\n\n\n\n

$=\\frac{3}{2}\\sqrt{2^2-x^2}$<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924\u094d\u0924 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=4\u00d7ar(OABO)<\/p>\n\n\n\n

$=4\\times \\int_{0}^{2}\\frac{3}{2}\\sqrt{2^2-x^2}dx$<\/p>\n\n\n\n

$=6\\int_{0}^{2}\\sqrt{2^2-x^2}$<\/p>\n\n\n\n

$=6\\left[\\frac{x}{2}\\sqrt{2^2-x^2}+\\frac{2^2}{2}\\sin^{-1}\\frac{x}{2}\\right]_{0}^{2}$<\/p>\n\n\n\n

$=6\\left[\\left(0+2\\times \\frac{\\pi}{2}\\right)-(0+0)\\right]$<\/p>\n\n\n\n

=6\u03c0 \u0935\u0930\u094d\u0917  \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

(ii)<\/strong> $\\frac{x^2}{16}+\\frac{y^2}{9}=1$<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 22<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930 $\\frac{x^2}{4}+\\frac{y^2}{9}=1$ \u0915\u0947 \u0906\u0932\u0947\u0916 \u0915\u093e \u091a\u093f\u0924\u094d\u0930 \u0916\u0940\u091a\u0947 \u0924\u0925\u093e \u0935\u0915\u094d\u0930 \u0915\u0947 \u0928\u0940\u091a\u0947 \u0914\u0930 x-\u0905\u0915\u094d\u0937 \u0915\u0947 \u090a\u092a\u0930 \u0915\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$\\frac{x^2}{2^2}+\\frac{y^2}{3^2}=1$<\/p>\n\n\n\n

$\\frac{y^2}{3^2}=1-\\frac{x^2}{2^2}$<\/p>\n\n\n\n

$\\frac{y^2}{3^2}=\\frac{2^2-x^2}{2^2}$<\/p>\n\n\n\n

$y^2=\\frac{3^2}{2^2}(2^2-x^2)$<\/p>\n\n\n\n

$y=\\sqrt{\\frac{3^2}{2^2}(2^2-x^2)}$<\/p>\n\n\n\n

$=\\frac{3}{2}\\sqrt{2^2-x^2}$<\/p>\n\n\n\n

\u0918\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=2\u00d7ar(OABO)<\/p>\n\n\n\n

$=2\\times \\int_{0}^{2}\\frac{3}{2}\\sqrt{2^2-x^2}dx$<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 23<\/h4>\n\n\n\n

\u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u093e \u0909\u092a\u092f\u094b\u0917 \u0915\u0930 \u0394ABC \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u0915\u0947 \u0936\u0940\u0930\u094d\u0937 A(2,3),B(4,7) \u0924\u0925\u093e C(6,2) \u0939\u0948\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092d\u093e\u0917 \u0394ABC \u0915\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0936\u0940\u0930\u094d\u0937 \u0938\u0947 x-\u0905\u0915\u094d\u0937 \u092a\u0930 \u0932\u0902\u092c AP, BQ \u0924\u0925\u093e CR \u0916\u093f\u091a\u093e \u0939\u0948\u0964<\/p>\n\n\n\n

\u0926\u094b \u092c\u093f\u0902\u0926\u0941\u0913 (x1<\/sub>,y1<\/sub>) \u0924\u0925\u093e (x2<\/sub>,y2<\/sub>) \u0938\u0947 \u0939\u094b\u0915\u0930 \u091c\u093e\u0928\u0947\u0935\u093e\u0932\u0940 \u0930\u0947\u0916\u093e \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

$y-y_1=\\frac{y_2-y_1}{x_2-x_1}(x-x_1)$<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e AB \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 <\/p>\n\n\n\n

$y-3=\\frac{7-3}{4-2}(x-2)$<\/p>\n\n\n\n

$y-3=\\frac{4}{2}(x-2)$<\/p>\n\n\n\n

y-3=2x-4<\/p>\n\n\n\n

y=2x-1<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e BC \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 <\/p>\n\n\n\n

$y-7=\\frac{2-7}{6-4}(x-4)$<\/p>\n\n\n\n

$y-7=\\frac{-5}{2}(x-4)$<\/p>\n\n\n\n

$y=\\frac{-5}{2}x+10+7$<\/p>\n\n\n\n

$y=17-\\frac{5}{2}x$<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e AC \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

 $y-3=\\frac{2-3}{6-2}(x-2)$<\/p>\n\n\n\n

$y=3=\\frac{-1}{4}(x-2)$<\/p>\n\n\n\n

$y-3=\\frac{-1}{4}x+\\frac{1}{2}$<\/p>\n\n\n\n

$y=\\frac{7}{2}-\\frac{1}{4}x$<\/p>\n\n\n\n

\u0394ABC \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(ABQP)+ar(BCRQ)-ar(ACRP)<\/p>\n\n\n\n

$=\\int_{2}^{4}(2x-1)dx+\\int_{4}^{6}\\left(17-\\frac{5}{2}x\\right)dx-\\int_{2}^{6}\\left(\\frac{7}{2}-\\frac{1}{4}x\\right)dx$<\/p>\n\n\n\n

$=\\left[x^2-x\\right]_{2}^{4}+\\left[17x-\\frac{5}{4}x\\right]_{4}^{6}-\\left[\\frac{7}{2}x-\\frac{1}{8}x^2\\right]_{2}^{6}$<\/p>\n\n\n\n

=12-2+(102-45)-(68-20)-$\\left[\\left(21-\\frac{9}{2}\\right)-\\left(7-\\frac{1}{2}\\right)\\right]$<\/p>\n\n\n\n

=10+57-48-$\\left[21-\\frac{9}{20}-7+\\frac{1}{2}\\right]$<\/p>\n\n\n\n

=10+9-14+4<\/p>\n\n\n\n

=23-14<\/p>\n\n\n\n

=9 \u0935\u0930\u094d\u0917  \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 25<\/h4>\n\n\n\n

(i) \u0938\u092e\u093e\u0915\u0932\u0928 \u0935\u093f\u0927\u093f \u0915\u093e \u0909\u092a\u092f\u094b\u0917 \u0915\u0930\u0924\u0947 \u0939\u0941\u090f, \u0930\u0947\u0916\u093e\u0913 2x+y=4, 3x-2y=6 \u090f\u0935\u0902 x-3y+5=0 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u0924\u0940\u0928\u094b \u0930\u0947\u0916\u093e\u0913 \u0915\u0947 \u0906\u092a\u0938 \u092e\u0947 \u092a\u094d\u0930\u0924\u093f\u091a\u094d\u091b\u0947\u0926 \u0915\u0930\u0928\u0947 \u0938\u0947 \u0394ABC \u092c\u0928\u093e \u0939\u0948\u0964<\/p>\n\n\n\n

A \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 (2,0)<\/p>\n\n\n\n

B \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 (4,3)<\/p>\n\n\n\n

C \u0915\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 (1,2)<\/p>\n\n\n\n

\u0394ABC \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(PQBC)-ar(\u0394PAC)-ar(\u0394AQB)<\/p>\n\n\n\n

$=\\int_{1}^{4}\\left(\\frac{1}{3}x+\\frac{5}{3}\\right)dx-\\int_{1}^{2}(4-2x)dx-\\int_{2}^{4}\\left(\\frac{3}{2}x-3\\right)dx$<\/p>\n\n\n\n

$=\\left[\\frac{x^2}{6}+\\frac{5}{3}x\\right]_{1}^{4}-\\left[4x-x^2\\right]_{1}^{2}-\\left[\\frac{3}{4}x^2-3x\\right]_{2}^{4}$<\/p>\n\n\n\n

$=\\left(\\frac{8}{3}+\\frac{20}{3}\\right)-\\left(\\frac{1}{6}+\\frac{5}{3}\\right)-[4-3]-[0-(-3)]$<\/p>\n\n\n\n

$=\\frac{8}{3}+\\frac{20}{3}-\\frac{1}{6}-\\frac{5}{3}-1-3$<\/p>\n\n\n\n

$=\\frac{23}{3}-\\frac{1}{6}-4$<\/p>\n\n\n\n

$=\\frac{46-1-24}{6}$<\/p>\n\n\n\n

$=\\frac{46-25}{6}=\\frac{21}{6}=\\frac{7}{2}$ \u0935\u0930\u094d\u0917  \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 26<\/h4>\n\n\n\n

\u092a\u094d\u0930\u0925\u092e \u091a\u0924\u0941\u0930\u094d\u0925\u093e\u0902\u0936 \u092e\u0947 \u0935\u0943\u0924 4x2<\/sup>+9y2<\/sup>=36 \u0924\u0925\u093e \u0928\u093f\u0930\u094d\u0926\u0947\u0936\u093e\u0902\u0915 \u0905\u0915\u094d\u0937\u094b \u0938\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

4x2<\/sup>+9y2<\/sup>=36<\/p>\n\n\n\n

\u0926\u094b\u0928\u094b \u0924\u0930\u092b 36 \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930<\/p>\n\n\n\n

$\\frac{4x^2}{36}+\\frac{9y^2}{36}=\\frac{36}{36}$<\/p>\n\n\n\n

$\\frac{x^2}{3^2}+\\frac{y^2}{2^2}=1$<\/p>\n\n\n\n

$y=\\frac{2}{3}\\sqrt{3^2-x^2}$<\/p>\n\n\n\n

ar(OABO)$=\\frac{2}{3}\\int_{0}^{3}\\sqrt{3^2-x^2}dx$<\/p>\n\n\n\n

$=\\frac{2}{3}\\left[\\frac{x}{2}\\sqrt{3^2-x^2}+\\frac{3^2}{2}\\sin^{-1}\\frac{x}{3}\\right]_{0}^{3}$<\/p>\n\n\n\n

$=\\frac{2}{3}\\left[\\frac{9}{2}\\times \\frac{\\pi}{2}-0\\right]$<\/p>\n\n\n\n

$=\\frac{2}{3}\\times \\frac{9\\pi}{4}$ \u0935\u0930\u094d\u0917  \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 27<\/h4>\n\n\n\n

\u0938\u0930\u0932 \u0930\u0947\u0916\u093e $x=\\frac{a}{2}$ \u0926\u094d\u0935\u093e\u0930\u093e \u0935\u093f\u092d\u093e\u091c\u093f\u0924 \u0935\u0943\u0924 x2<\/sup>+y2<\/sup>=a2<\/sup> \u0915\u0947 \u092d\u093e\u0917\u094b \u092e\u0947 \u0938\u0947 \u091b\u094b\u091f\u0947 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u092b\u092b\u0932 \u0928\u093f\u0915\u093e\u0932\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u0935\u0943\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 <\/p>\n\n\n\n

x2<\/sup>+y2<\/sup>=a2<\/sup><\/p>\n\n\n\n

$y=\\sqrt{a^2-x^2}$<\/p>\n\n\n\n

ABCA \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 $=\\int_{\\frac{a}{2}}^{a}\\sqrt{a^2-x^2}dx$<\/p>\n\n\n\n

$=\\left[\\frac{x}{2}\\sqrt{a^2-x^2}+\\frac{a^2}{2}\\sin^{-1}\\frac{x}{a}\\right]_{\\frac{a}{2}}^{a}$<\/p>\n\n\n\n

$=\\frac{a^2}{2}\\times \\frac{\\pi}{2}-\\left(\\frac{a}{4}\\times \\frac{\\sqrt{3}a}{2}+\\frac{a^2}{2}\\times \\frac{\\pi}{6}\\right)$<\/p>\n\n\n\n

$=\\frac{a^2 \\pi}{4}-\\frac{\\sqrt{3}}{8}a^2-\\frac{a^2}{12}\\pi$<\/p>\n\n\n\n

$=\\frac{3a^2\\pi-a^2 \\pi}{12}-\\frac{\\sqrt{3}}{8}a^2$<\/p>\n\n\n\n

$=\\frac{2a^2\\pi}{12}-\\frac{\\sqrt{3}}{8}a^2$<\/p>\n\n\n\n

$=\\frac{a^2}{2}\\left(\\frac{\\pi}{3}-\\frac{\\sqrt{3}}{4}\\right)$<\/p>\n\n\n\n

\u091b\u094b\u091f\u0947 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=2\u00d7ar(ABCA)<\/p>\n\n\n\n

$=2\\times \\frac{a^2}{2}\\left(\\frac{\\pi}{3}-\\frac{\\sqrt{3}}{4}\\right)$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

$=a^2\\left(\\frac{\\pi}{3}-\\frac{\\sqrt{3}}{4}\\right)$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 28<\/h4>\n\n\n\n

\u0938\u0930\u0932 \u0930\u0947\u0916\u093e $x=\\frac{a}{\\sqrt{2}}$ \u0926\u094d\u0935\u093e\u0930\u093e \u0935\u093f\u092d\u093e\u091c\u093f\u0924 \u0935\u0943\u0924 x2<\/sup>+y2<\/sup>=a2<\/sup> \u0915\u0947 \u092d\u093e\u0917\u094b \u092e\u0947 \u0938\u0947 \u091b\u094b\u091f\u0947 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0928\u093f\u0915\u093e\u0932\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 29<\/h4>\n\n\n\n

\u0938\u0930\u0932 \u0930\u0947\u0916\u093e y=2x, x=0 \u0924\u0925\u093e y=2 \u0938\u0947 \u092c\u0928\u0947 \u0924\u094d\u0930\u093f\u092d\u0941\u091c \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0938\u092e\u093e\u0915\u0932\u0928 \u0926\u094d\u0935\u093e\u0930\u093e \u0928\u093f\u0915\u093e\u0932\u0947\u0964<\/p>\n\n\n\n

[Find the area of the triangle formed by the straight lines y=2x, x=0 and y=2 by integration]<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u0394OAC \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 $=\\frac{1}{2}\\int_{0}^{2}ydy$<\/p>\n\n\n\n

$=\\frac{1}{2}\\left[\\frac{y^2}{2}\\right]_{0}^{2}$<\/p>\n\n\n\n

$=\\frac{1}{2}(2-0)=1$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 30<\/h4>\n\n\n\n

x-\u0905\u0915\u094d\u0937 \u0914\u0930 \u0935\u0915\u094d\u0930 y=sin x \u0915\u0947 \u092c\u0940\u091a x=0 \u0938\u0947 x=\u03c0 \u0924\u0915 \u0915\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0928\u093f\u0915\u093e\u0932\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

OABO \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932$=\\int_{0}^{\\pi} \\sin xdx$<\/p>\n\n\n\n

$=\\left[-\\cos x\\right]_{0}^{\\pi}$<\/p>\n\n\n\n

=-cos \u03c0-(-cos 0)<\/p>\n\n\n\n

=-(-1)-(-1)<\/p>\n\n\n\n

=1+1=2  \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 31<\/h4>\n\n\n\n

x-\u0905\u0915\u094d\u0937 \u0914\u0930 \u0935\u0915\u094d\u0930 y=cos x \u0915\u0947 \u092c\u0940\u091a x=0 \u0938\u0947 x=2\u03c0 \u0924\u0915 \u0915\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0928\u093f\u0915\u093e\u0932\u0947\u0964<\/p>\n\n\n\n

[Find the area between x-axis and the curve y=cos x from x=0 and x=2\u03c0]<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

$=\\int_{0}^{\\frac{\\pi}{2}}\\cos xdx-\\int_{\\frac{\\pi}{2}}^{\\frac{3\\pi}{2}}\\cos xdx+\\int_{\\frac{3\\pi}{2}}^{2\\pi}\\cos xdx$<\/p>\n\n\n\n

$=\\left[\\sin x \\right]_{0}^{\\frac{\\pi}{2}}-\\left[\\sin x\\right]_{\\frac{\\pi}{2}}^{\\frac{3\\pi}{2}}+\\left[\\sin x\\right]_{\\frac{3\\pi}{2}}^{2\\pi}$<\/p>\n\n\n\n

=1-0-(-1-1)+(0+1)<\/p>\n\n\n\n

=1+2+1=4 \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 32<\/h4>\n\n\n\n

$y=2\\sqrt{1-x^2}$ , x\u2208<\/strong>[0,1] \u0915\u093e \u0930\u092b \u0906\u0932\u0947\u0916 \u0916\u0940\u091a\u0947 \u0924\u0925\u093e \u0935\u0915\u094d\u0930 \u0914\u0930 x-\u0905\u0915\u094d\u0937 \u0915\u0947 \u092c\u0940\u091a\u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

OABO \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

$=2\\int_{0}^{1}\\sqrt{1^2-x^2}dx$<\/p>\n\n\n\n

$=2\\left[\\frac{x}{2}\\sqrt{1^2-x^2}+\\frac{1^2}{2}\\sin ^{-1}\\frac{x}{1}\\right]_{0}^{1}$<\/p>\n\n\n\n

$=2\\left[\\frac{1}{2}\\times \\frac{\\pi}{2}-0\\right]$<\/p>\n\n\n\n

$=2\\times \\frac{\\pi}{4}=\\frac{\\pi}{2}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 33<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930 y=x|x| , x-\u0905\u0915\u094d\u0937 \u090f\u0935\u0902 \u0915\u094b\u091f\u093f\u092f\u094b x=-1 \u0924\u0925\u093e x=1 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y=x|x| $=\\left\\{\\begin{array}-x^2,& \\text{if }x<0\\\\x^2,&\\text{if } x\\geq 0\\end{array}\\right.$<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

$=-\\int_{-1}^{0}(-x^2)dx+\\int_{0}^{1}x^2dx$<\/p>\n\n\n\n

$=\\left[\\frac{x^3}{3}\\right]_{-1}^{0}+\\left[\\frac{x^3}{3}\\right]_{0}^{1}$<\/p>\n\n\n\n

$=0-\\left(\\frac{-1}{3}\\right)+\\frac{1}{3}-0$<\/p>\n\n\n\n

$=\\frac{1}{3}+\\frac{1}{3}=\\frac{2}{3}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 34<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930 y=4x-x2<\/sup>, x-\u0905\u0915\u094d\u0937 \u0924\u0925\u093e \u0915\u094b\u091f\u093f\u092f\u094b x=1 \u0924\u0925\u093e x=3 \u0915\u0947 \u092c\u0940\u091a \u0918\u093f\u0930\u0947 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0928\u093f\u0915\u093e\u0932\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Question 35<\/h4>\n\n\n\n

\u092a\u0930\u0935\u0932\u092f y2<\/sup>=x , \u0930\u0947\u0916\u093e y+x=2 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y2<\/sup>=x \u0930\u0947\u0916\u093e y+x=2 or x=2y<\/p>\n\n\n\n

y2<\/sup>=2-y<\/p>\n\n\n\n

y2<\/sup>+y-2=0<\/p>\n\n\n\n

y2<\/sup>+2y-y-2=0<\/p>\n\n\n\n

y(y+2)-1(y+2)=0<\/p>\n\n\n\n

(y+2)(y-1)=0<\/p>\n\n\n\n

y=-2,1<\/p>\n\n\n\n

\u2235y=1 \u21d2x=1<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(OCBO)+ar(ACBA)<\/p>\n\n\n\n

$=\\int_{0}^{1}x^{\\frac{1}{2}}dx+\\int_{1}^{2}(2-x)dx$<\/p>\n\n\n\n

$=\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{0}^{1}+\\left[2x-\\frac{x^2}{2}\\right]_{1}^{2}$<\/p>\n\n\n\n

$=\\frac{2}{3}-0+2-\\frac{3}{2}$<\/p>\n\n\n\n

$=\\frac{4+12-9}{6}=\\frac{7}{6}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 36<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930\u094b y=x2<\/sup>+2, y=x, x=0 \u090f\u0935\u0902 x=3 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y=x2<\/sup>+2..(i) ,y=x..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947,<\/p>\n\n\n\n

x2<\/sup>+2=x<\/p>\n\n\n\n

$x=\\frac{-(-1)\\pm \\sqrt{(-1)^2-4\\times 1\\times 2}}{2\\times 1}$<\/p>\n\n\n\n

$x=\\frac{1\\pm \\sqrt{-7}}{2}\\notin R$<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(OADEO)-ar(OAC)<\/p>\n\n\n\n

$\\int_{0}^{3}(x^2+2)dx-\\int_{0}^{3}xdx$<\/p>\n\n\n\n

$=\\left[\\frac{x^2}{3}+2x\\right]_{0}^{3}-\\left[\\frac{x^2}{2}\\right]_{0}^{3}$<\/p>\n\n\n\n

$=9+6-0-\\frac{9}{2}+0$<\/p>\n\n\n\n

$=15-\\frac{9}{2}=\\frac{30-9}{2}=\\frac{21}{2}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

Question 37<\/h4>\n\n\n\n

\u092a\u0930\u0935\u0932\u092f x2<\/sup>=y ,\u0930\u0947\u0916\u093e y=x+2 \u090f\u0935\u0902 x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

x2<\/sup>=y..(i) ,y=x+2..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947 ,<\/p>\n\n\n\n

x2<\/sup>=x+2<\/p>\n\n\n\n

x2<\/sup>-x-2=0<\/p>\n\n\n\n

x2<\/sup>-2x+x-2=0<\/p>\n\n\n\n

x(x-2)+1(x-2)=0<\/p>\n\n\n\n

(x-2)(x+1)=0<\/p>\n\n\n\n

x=-1,2<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

x=-1,y=1<\/p>\n\n\n\n

x=2,y=4<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(ABCDA)-ar(AOBCOD)<\/p>\n\n\n\n

$=\\int_{-1}^{2}(x+2)dx-\\int_{-1}^{2}x^2dx$<\/p>\n\n\n\n

$=\\left[\\frac{x^2}{2}+2x\\right]_{-1}^{2}-\\left[\\frac{x^3}{3}\\right]_{-1}^{2}$<\/p>\n\n\n\n

$=\\left(6+\\frac{3}{2}\\right)-\\left(\\frac{8}{3}+\\frac{1}{3}\\right)$<\/p>\n\n\n\n

$=6+\\frac{3}{2}-3$<\/p>\n\n\n\n

$=3+\\frac{3}{2}$<\/p>\n\n\n\n

$=\\frac{6+3}{2}=\\frac{9}{2}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 38<\/h4>\n\n\n\n

\u092a\u0930\u0935\u0932\u092f y=x2 <\/sup>\u0924\u0925\u093e y=|x| \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

CASE-I<\/p>\n\n\n\n

y=x2 <\/sup>\u0924\u0925\u093e y=x<\/p>\n\n\n\n

\u2234x2<\/sup>=x<\/p>\n\n\n\n

x(x-1)=0<\/p>\n\n\n\n

x=0,1<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

x=1,y=1<\/p>\n\n\n\n

x=0,y=0<\/p>\n\n\n\n

ODAO \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(OCAO)-ar(OCADO)<\/p>\n\n\n\n

$=\\int_{0}^{1}xdx-\\int_{0}^{1}x^2dx$<\/p>\n\n\n\n

$=\\left[\\frac{x^2}{2}\\right]_{0}^{1}-\\left[\\frac{x^3}{3}\\right]_{0}^{1}$<\/p>\n\n\n\n

$=\\frac{1}{2}-\\frac{1}{3}=\\frac{3-2}{6}$<\/p>\n\n\n\n

$=\\frac{1}{6}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=2\u00d7(ODAO) \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

$=2\\times \\frac{1}{6}$<\/p>\n\n\n\n

$=\\frac{1}{3}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 39<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930 $\\frac{x^2}{16}+\\frac{y^2}{9}=1$ \u0924\u0925\u093e \u0938\u0930\u0932 \u0930\u0947\u0916\u093e $\\frac{x}{4}+\\frac{y}{3}=1$ \u0938\u0947 \u0918\u093f\u0930\u0947 \u0932\u0918\u0941 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

$\\frac{x^2}{16}+\\frac{y^2}{9}=1$ \u0924\u0925\u093e \u0930\u0947\u0916\u093e \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

$\\frac{x}{4}+\\frac{y}{3}=1$<\/p>\n\n\n\n

$\\frac{3x+4y}{12}=1$<\/p>\n\n\n\n

3x+4y=12<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

\u0935\u0915\u094d\u0930 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 <\/p>\n\n\n\n

$=\\frac{x^2}{16}+\\frac{y^2}{9}=1$<\/p>\n\n\n\n

$y=\\frac{3}{4}\\sqrt{4^2-x^2}$<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 <\/p>\n\n\n\n

3x+4y=12<\/p>\n\n\n\n

4y=12-3x<\/p>\n\n\n\n

$y=3-\\frac{3}{4}x$ \u092f\u093e $y=\\frac{3}{4}(4-x)$<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(OADBO)-ar(OABO)<\/p>\n\n\n\n

$=\\frac{3}{4}\\int_{0}^{4}\\sqrt{4^2-x^2}dx-\\frac{3}{4}\\int_{0}^{4}(4-x)dx$<\/p>\n\n\n\n

$=\\frac{3}{4}\\left[\\frac{x}{2}\\sqrt{4^2-x^2}+\\frac{4^2}{x}\\sin^{-1}\\frac{x}{4}\\right]_{0}^{4}-\\frac{3}{4}\\left[4x-\\frac{x^2}{2}\\right]_{0}^{4}$<\/p>\n\n\n\n

$=\\frac{3}{4}\\left[8\\times \\frac{\\pi}{2}-0\\right]-\\frac{3}{4}[8-0]$<\/p>\n\n\n\n

$=\\frac{3}{4}(4\\pi)-\\frac{3}{4}(8)$<\/p>\n\n\n\n

=3\u03c0-6<\/p>\n\n\n\n

=3(\u03c0-2) \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 40<\/h4>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924 $\\frac{x^2}{9}+\\frac{y^2}{4}=1$ \u090f\u0935\u0902 $\\frac{x}{3}+\\frac{y}{2}=1$ \u0938\u0947 \u0918\u093f\u0930\u0947 \u0932\u0918\u0941 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u0926\u0940\u0930\u094d\u0918\u0935\u0943\u0924 $\\frac{x^2}{9}+\\frac{y^2}{4}=1$<\/p>\n\n\n\n

$\\frac{x^2}{3^2}+\\frac{y^2}{2^2}=1$<\/p>\n\n\n\n

$y=\\frac{2}{3}\\sqrt{3^2-x^2}$<\/p>\n\n\n\n

\u0930\u0947\u0916\u093e \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 $\\frac{x}{3}+\\frac{y}{2}=1$<\/p>\n\n\n\n

$\\frac{2x+3y}{6}=1$<\/p>\n\n\n\n

2x+3y=6<\/p>\n\n\n\n

$y=\\frac{2}{3}(3-x)$<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(OABCO)-ar(\u0394OAC)<\/p>\n\n\n\n

$=\\frac{2}{3}\\int_{0}^{3}\\sqrt{3^2-x^2}dx-\\frac{2}{3}\\int_{0}^{3}(3-x)dx$<\/p>\n\n\n\n

$=\\frac{2}{3}\\left[\\frac{x}{2}\\sqrt{3^2-x^2}+\\frac{3^2}{2}\\sin^{-1}\\frac{x}{3}\\right]_{0}^{3}-\\frac{2}{3}\\left[3x-\\frac{x^2}{2}\\right]_{0}^{3}$<\/p>\n\n\n\n

$=\\frac{2}{3}\\left[\\frac{9}{2}\\times \\frac{\\pi}{2}-0\\right]-\\frac{2}{3}\\left[9-\\frac{9}{2}-0\\right]$<\/p>\n\n\n\n

$=\\frac{2}{3}\\left(\\frac{9\\pi}{4}\\right)-\\frac{2}{3}\\left(\\frac{9}{2}\\right)$<\/p>\n\n\n\n

$=\\frac{3\\pi}{2}-3$<\/p>\n\n\n\n

$=\\frac{3\\pi-6}{2}=\\frac{3}{2}(\\pi-2)$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 41<\/h4>\n\n\n\n

\u092a\u0930\u0935\u0932\u092f y2<\/sup>=4x \u0915\u0947 \u0909\u0938 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u094b \u0930\u0947\u0916\u093e y=x \u0938\u0947 \u0915\u091f\u0924\u093e \u0939\u0948\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y2<\/sup>=4x..(i), y=x..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947,<\/p>\n\n\n\n

x2<\/sup>=4x<\/p>\n\n\n\n

x2<\/sup>-4x=0<\/p>\n\n\n\n

x(x-4)=0<\/p>\n\n\n\n

x=0,4<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

x=0,y=0<\/p>\n\n\n\n

x=4,y=4<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(OABCO)-ar(\u0394OAB)<\/p>\n\n\n\n

$=2\\int_{0}^{4}x^{\\frac{1}{2}}dx-\\int_{0}^{4}xdx$<\/p>\n\n\n\n

$=2\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{0}^{4}-\\left[\\frac{x^2}{2}\\right]_{0}^{4}$<\/p>\n\n\n\n

$=2\\left[\\frac{2}{3}(4)^{\\frac{3}{2}}-0\\right]-\\left[\\frac{4^2}{2}-0\\right]$<\/p>\n\n\n\n

$=2\\left[\\frac{2}{3}\\times (2)^{2\\times \\frac{3}{2}}\\right]-\\frac{16}{2}$<\/p>\n\n\n\n

$=2\\times \\frac{2}{3}\\times 8-8$<\/p>\n\n\n\n

$=\\frac{32}{3}-8$<\/p>\n\n\n\n

$=\\frac{32-24}{3}=\\frac{8}{3}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 42<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930\u094b y=x \u0924\u0925\u093e y2<\/sup>=4x \u0915\u0947 \u092c\u0940\u091a \u0915\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y=x..(i) ,y=x2<\/sup>..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947,<\/p>\n\n\n\n

x2<\/sup>=x<\/p>\n\n\n\n

x2<\/sup>-x=0<\/p>\n\n\n\n

x(x-1)=0<\/p>\n\n\n\n

x=0,1<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

x=0,y=0<\/p>\n\n\n\n

x=1,y=1<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=ar(OABO)-ar(\u0394ABCO)<\/p>\n\n\n\n

$=\\int_{0}^{1}xdx-\\int_{0}^{1}x^2dx$<\/p>\n\n\n\n

$=\\left[\\frac{x^2}{2}\\right]_{0}^{1}-\\left[\\frac{x^3}{3}\\right]_{0}^{1}$<\/p>\n\n\n\n

$=\\left(\\frac{1}{2}-0\\right)-\\left(\\frac{1}{3}-0\\right)$<\/p>\n\n\n\n

$=\\frac{1}{2}-\\frac{1}{3}$<\/p>\n\n\n\n

$=\\frac{3-2}{6}=\\frac{1}{6}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 43<\/h4>\n\n\n\n

\u0935\u0943\u0924 x2<\/sup>+y2<\/sup>=25 \u0924\u0925\u093e \u0938\u0930\u0932 \u0930\u0947\u0916\u093e x+y=5 \u0915\u0947 \u092e\u0927\u094d\u092f\u0935\u0930\u094d\u0924\u0940 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u0935\u0943\u0924 \u0903 x2<\/sup>+y2<\/sup>=25 <\/p>\n\n\n\n

x2<\/sup>+y2<\/sup>=52<\/sup> <\/p>\n\n\n\n

$y=\\sqrt{5^2-x^2}$<\/p>\n\n\n\n

\u0938\u0930\u0932 \u0930\u0947\u0916\u093e <\/p>\n\n\n\n

x+y=5<\/p>\n\n\n\n

y=5-x<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

=ar(OABCO)-ar(\u0394OAC)<\/p>\n\n\n\n

$=\\int_{0}^{5}\\sqrt{5^2-x^2}dx-\\int_{0}^{5}(5-x)dx$<\/p>\n\n\n\n

$=\\left[\\frac{x}{2}\\sqrt{5^2-x^2}+\\frac{5^2}{2}\\sin^{-1}\\frac{x}{5}\\right]_{0}^{5}-\\left[5x-\\frac{x^2}{2}\\right]_{0}^{5}$<\/p>\n\n\n\n

$=\\frac{25}{2}\\times \\frac{\\pi}{2}-0-\\left[25-\\frac{25}{2}-0\\right]$<\/p>\n\n\n\n

$=\\frac{25}{4}\\pi-\\frac{25}{2}$<\/p>\n\n\n\n

$=\\frac{25\\pi-50}{4}=\\frac{25}{4}(\\pi-2)$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 44<\/h4>\n\n\n\n

y2<\/sup>=4ax \u0924\u0925\u093e x2<\/sup>=y \u0915\u0947 \u092e\u0927\u094d\u092f\u0935\u0930\u094d\u0924\u0940 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y2<\/sup>=4ax..(i) \u0924\u0925\u093e x2<\/sup>=y..(ii)<\/p>\n\n\n\n

(x2<\/sup>)2<\/sup>=4ax (\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u0938\u0947)<\/p>\n\n\n\n

x4<\/sup>-4ax=0<\/p>\n\n\n\n

x(x3<\/sup>-4a)=0<\/p>\n\n\n\n

x=0, $(4a)^{\\frac{1}{3}}$<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

x=0,y=02<\/sup>=0<\/p>\n\n\n\n

$x=(4a)^{\\frac{1}{3}},y=(4a)^{\\frac{2}{3}}$<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

=ar(OABCO)-ar(OABDO)<\/p>\n\n\n\n

$=2\\sqrt{a}\\int_{0}^{(4a)^{\\frac{1}{3}}}-\\int_{0}^{(4a)^{\\frac{1}{3}}}$<\/p>\n\n\n\n

$=2\\sqrt{a}\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{0}^{(4a)^{\\frac{1}{3}}}-\\left[\\frac{x^3}{3}\\right]_{0}^{(4a)^{\\frac{1}{3}}}$<\/p>\n\n\n\n

$=2\\sqrt{a}\\left[\\frac{2}{3}(4a)^{\\frac{1}{3}\\times \\frac{3}{2}}-0\\right]-\\left[\\frac{(4a)^{\\frac{1}{3}\\times 3}}{3}-0\\right]$<\/p>\n\n\n\n

$=\\frac{4}{3}\\sqrt{a}\\times \\sqrt{4a}-\\frac{4a}{3}$<\/p>\n\n\n\n

$=\\frac{8}{3}a-\\frac{4a}{3}$<\/p>\n\n\n\n

$=\\frac{4a}{3}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 45<\/h4>\n\n\n\n

\u0935\u0915\u094d\u0930\u094b y=4x2 <\/sup>\u0924\u0925\u093e y2<\/sup>=2x \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0928\u093f\u0915\u093e\u0932\u0947\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u0935\u0915\u094d\u0930\u0903 y=4x2<\/sup>..(i) \u0924\u0925\u093e y2<\/sup>=2x..(ii)<\/p>\n\n\n\n

y=4x2 <\/sup>\u0938\u092e\u0940\u0915\u0930\u0923 (ii) \u092e\u0947 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

(4x2<\/sup>)2<\/sup>=2x<\/p>\n\n\n\n

16x4<\/sup>=2x<\/p>\n\n\n\n

$x^3=\\frac{2}{16}$<\/p>\n\n\n\n

$x^3=\\left(\\frac{1}{2}\\right)^3$<\/p>\n\n\n\n

$x=\\frac{1}{2}$<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0938\u0947 , y=4x2<\/sup><\/p>\n\n\n\n

$y=4\\left(\\frac{1}{2}\\right)^2$<\/p>\n\n\n\n

y=1<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

=OBADO \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 -OBACO \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 <\/p>\n\n\n\n

$=\\sqrt{2}\\int_{0}^{\\frac{1}{2}}x^{\\frac{1}{2}}dx$<\/p>\n\n\n\n

$=\\sqrt{2}\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{0}^{\\frac{1}{2}}-4\\left[\\frac{x^3}{3}\\right]_{0}^{\\frac{1}{2}}$<\/p>\n\n\n\n

$=\\sqrt{2}\\left[\\frac{2}{3}\\left(\\frac{1}{2}\\right)^{\\frac{3}{2}}-0\\right]-4\\left[\\dfrac{\\left(\\frac{1}{2}\\right)^3}{3}-0\\right]$<\/p>\n\n\n\n

$=\\sqrt{2}\\times \\frac{2}{3}\\times \\frac{1}{2\\sqrt{2}}-4\\times \\frac{1}{24}$<\/p>\n\n\n\n

$=\\frac{1}{3}-\\frac{1}{6}=\\frac{2-1}{6}=\\frac{1}{6}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 46<\/h4>\n\n\n\n

\u092a\u094d\u0930\u0925\u092e \u091a\u0924\u0941\u0930\u094d\u0925\u093e\u0936 \u092e\u0947 \u0935\u0943\u0924 x2<\/sup>+y2<\/sup>=32 ,\u0930\u0947\u0916\u093e y=x \u090f\u0935\u0902 x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u0935\u0943\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

x2<\/sup>+y2<\/sup>=(4\u221a2)2<\/sup><\/p>\n\n\n\n

$y=\\sqrt{(4\\sqrt{2})^2-x^2}$<\/p>\n\n\n\n

\u2235y=x \u0930\u0947\u0916\u093e \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

\u2234x2<\/sup>+x2<\/sup>=32<\/p>\n\n\n\n

2x2<\/sup>=32<\/p>\n\n\n\n

x2<\/sup>=16<\/p>\n\n\n\n

x=\u00b14<\/p>\n\n\n\n

x=4 \u092a\u0930,y=4 ;<\/p>\n\n\n\n

x=-4, y=-4<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

=ar(OACO)+ar(ABCA)<\/p>\n\n\n\n

$=\\int_{0}^{4}xdx+\\int_{4}^{4\\sqrt{2}}\\sqrt{(4\\sqrt{2})^2-x^2}dx$<\/p>\n\n\n\n

$=\\left[\\frac{x^2}{2}\\right]_{0}^{4}+\\left[\\frac{x}{2}\\sqrt{(4\\sqrt{2})^2-x^2}+\\frac{(4\\sqrt{2})}{2}\\sin^{-1}\\frac{x}{4\\sqrt{2}}\\right]_{4}^{4\\sqrt{2}}$<\/p>\n\n\n\n

$=\\frac{4^2}{2}-\\frac{0^2}{2}+16\\times \\frac{\\pi}{2}-\\left(2\\times 4+16\\times \\frac{\\pi}{4}\\right)$<\/p>\n\n\n\n

=8+8\u03c0-8-4\u03c0<\/p>\n\n\n\n

=4\u03c0 \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 47<\/h4>\n\n\n\n

\u0935\u0943\u0924 4x2<\/sup>+4y2<\/sup>=9 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u094b \u092a\u0930\u0935\u0932\u092f y2<\/sup>=4x \u0915\u0947 \u0905\u0902\u0926\u0930 \u0939\u0948\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u0935\u0943\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 <\/p>\n\n\n\n

4x2<\/sup>+4y2<\/sup>=9…(i)<\/p>\n\n\n\n

$x^2+y^2=\\left(\\frac{3}{2}\\right)^2$<\/p>\n\n\n\n

$y=\\sqrt{\\left(\\frac{3}{2}\\right)^2-x^2}$<\/p>\n\n\n\n

\u092a\u0930\u0935\u0932\u092f \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 y2<\/sup>=4x<\/p>\n\n\n\n

\u22344x2<\/sup>+4(4x)=9 (\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0938\u0947)<\/p>\n\n\n\n

4x2<\/sup>+16x-9=0<\/p>\n\n\n\n

4x2<\/sup>+18x-2x-9=0<\/p>\n\n\n\n

2x(2x+9)-1(2x+9)=0<\/p>\n\n\n\n

(2x+9)(2x-1)=0<\/p>\n\n\n\n

$x=\\frac{-9}{2},\\frac{1}{2}$<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

$x=-\\frac{9}{2}$ \u092a\u0930, <\/p>\n\n\n\n

$y^2=4\\left(-\\frac{5}{2}\\right)$<\/p>\n\n\n\n

$y=\\sqrt{-18}\\notin R$<\/p>\n\n\n\n

$x=\\frac{1}{2}$\u092a\u0930, <\/p>\n\n\n\n

$y^2=4\\left(\\frac{1}{2}\\right)$<\/p>\n\n\n\n

y=\u00b1\u221a2<\/p>\n\n\n\n

ar(OABO)=ar(OCBO)+ar(CABC)<\/p>\n\n\n\n

$=2\\int_{0}^{\\frac{1}{2}}x^{\\frac{1}{2}}dx+\\int_{\\frac{1}{2}}^{\\frac{3}{2}}\\sqrt{\\left(\\frac{3}{2}\\right)^2-x^2}dx$<\/p>\n\n\n\n

$=2\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{0}^{\\frac{1}{2}}+\\left[\\frac{x}{2}\\sqrt{\\left(\\frac{3}{2}\\right)^2-x^2}+\\dfrac{\\left(\\frac{3}{2}\\right)^2}{2}\\sin^{-1}\\dfrac{x}{\\frac{3}{2}}\\right]_{\\frac{1}{2}}^{\\frac{3}{2}}$<\/p>\n\n\n\n

$=2\\left[\\frac{2}{3}\\left(\\frac{1}{2}\\right)^{\\frac{3}{2}}-0\\right]+\\left[\\frac{9}{8}\\times \\frac{\\pi}{2}-\\left(\\frac{1}{4}\\sqrt{2}+\\frac{9}{8}\\sin^{-1}\\frac{1}{3}\\right)\\right]$<\/p>\n\n\n\n

$=\\frac{4}{3}\\times \\frac{1}{2\\sqrt{2}}+\\frac{9\\pi}{16}-\\frac{\\sqrt{2}}{4}-\\frac{9}{8}\\sin^{-1}\\frac{1}{3}$<\/p>\n\n\n\n

$=\\frac{\\sqrt{2}}{3}-\\frac{\\sqrt{2}}{4}+\\frac{9}{8}\\left(\\frac{\\pi}{2}-\\sin^{-1}\\frac{1}{3}\\right)$<\/p>\n\n\n\n

$=\\frac{4\\sqrt{2}-3\\sqrt{2}}{12}+\\frac{9}{8}\\left(\\frac{\\pi}{2}-\\sin^{-1}\\frac{1}{3}\\right)$<\/p>\n\n\n\n

$=\\frac{\\sqrt{2}}{12}+\\frac{9}{8}\\left(\\frac{\\pi}{2}-\\sin^{-1}\\frac{1}{3}\\right)$<\/p>\n\n\n\n

\u0935\u0943\u0924\u094d\u0924 \u0924\u0925\u093e \u092a\u0930\u0935\u0932\u092f \u0926\u094d\u0935\u093e\u0930\u093e \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

=2\u00d7ar(OABO)<\/p>\n\n\n\n

$=2\\left[\\frac{\\sqrt{2}}{12}+\\frac{9}{8}\\left(\\frac{\\pi}{2}-\\sin^{-1}\\frac{1}{3}\\right)\\right]$<\/p>\n\n\n\n

$=\\frac{\\sqrt{2}}{6}+\\frac{9}{4}\\left(\\frac{\\pi}{2}-sin^{-1}\\frac{1}{3}\\right)$<\/p>\n\n\n\n

\u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 48<\/h4>\n\n\n\n

\u092a\u0939\u0932\u0947 \u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923 x2<\/sup>+y2<\/sup>+=4 \u090f\u0935\u0902 (x+2)2<\/sup>+y2<\/sup>=4 \u0915\u0947 \u092e\u0927\u094d\u092f\u0935\u0930\u094d\u0924\u0940 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092a\u0939\u0932\u0947 \u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

x2<\/sup>+y2<\/sup>+=22 <\/sup>\u21d2y2<\/sup>=22<\/sup>-x2<\/sup>..(i)<\/p>\n\n\n\n

$y=\\sqrt{2^2-y^2}$<\/p>\n\n\n\n

\u0926\u0942\u0938\u0930\u0947 \u0935\u0943\u0924\u094d\u0924 \u0915\u093e \u0938\u092e\u0940\u0915\u0930\u0923<\/p>\n\n\n\n

(x-2)2<\/sup>+y2<\/sup>=22<\/sup><\/p>\n\n\n\n

y2<\/sup>=22<\/sup>-(x-2)2<\/sup>..(ii)<\/p>\n\n\n\n

$y=\\sqrt{2^2-(x-2)^2}$<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947,<\/p>\n\n\n\n

22<\/sup>-x2<\/sup>=22<\/sup>-(x-2)2<\/sup><\/p>\n\n\n\n

x2<\/sup>=x2<\/sup>-2.x.1+22<\/sup><\/p>\n\n\n\n

4x=4<\/p>\n\n\n\n

x=1 \u21d2y=\u00b1\u221a3<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

ar(OCBAO)=ar(OCAO)+ar(BACB)<\/p>\n\n\n\n

$=\\int_{0}^{1}\\sqrt{2^2-(x-2)^2}dx+\\int_{1}^{2}\\sqrt{2^2-x^2}dx$<\/p>\n\n\n\n

$=\\left[\\frac{x-2}{2}\\sqrt{2^2-(x-2)^2}+\\frac{2^2}{2}\\sin^{-1}\\frac{x-2}{2}\\right]_{0}^{1}+\\left[\\frac{x}{2}\\sqrt{2^2-x^2}+\\frac{2^2}{2}\\sin^{-1}\\frac{x}{2}\\right]_{1}^{2}$<\/p>\n\n\n\n

$=-\\frac{1}{2}\\times \\sqrt{3}+2\\left(-\\frac{\\pi}{6}\\right)-\\left[-1\\times 0+2\\times \\left(-\\frac{\\pi}{2}\\right)\\right]+2\\times \\frac{\\pi}{2}-\\left(\\frac{1}{2}\\times \\sqrt{3}+2\\times \\frac{\\pi}{6}\\right)$<\/p>\n\n\n\n

$=-\\frac{\\sqrt{3}}{2}-\\frac{\\pi}{3}+\\pi+\\pi-\\frac{\\sqrt{3}}{2}-\\frac{\\pi}{3}$<\/p>\n\n\n\n

$=2\\pi-\\frac{2\\pi}{3}-\\sqrt{3}$<\/p>\n\n\n\n

$=\\frac{6\\pi-2\\pi}{3}-\\sqrt{3}=\\frac{4\\pi}{3}-\\sqrt{3}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

\u0926\u094b\u0928\u094b \u0935\u0943\u0924\u094d\u0924\u094b \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932=2\u00d7ar(OCBAO)<\/p>\n\n\n\n

$=2\\left(\\frac{4\\pi}{3}-\\sqrt{3}\\right)$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 49<\/h4>\n\n\n\n

y=|x-5| \u0915\u093e \u0906\u0932\u0947\u0916 \u0916\u0940\u0902\u091a\u093f\u090f $\\int_{0}^{1}|x-5|dx$ \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964 \u0906\u0932\u0947\u0916 \u092f\u0939 \u0938\u092e\u093e\u0915\u0932 \u0915\u094d\u092f\u093e \u0928\u093f\u0930\u0942\u092a\u093f\u0924 \u0915\u0930\u0924\u093e \u0939\u0948\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y=|x-5|$=\\left\\{\\begin{array}{l}x-5,\\text{if }x\\geq 5\\\\ -(x-5), \\text{if }x<5\\end{array}\\right.$<\/p>\n\n\n\n

CASE-I<\/p>\n\n\n\n

y=x-5, if x\u22655<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

$=\\int_{0}^{1}|x-5|dx=-\\int_{0}^{1}(x-5)dx$<\/p>\n\n\n\n

$=-\\left[\\frac{x^2}{2}-5x\\right]_{0}^{1}$<\/p>\n\n\n\n

$=-\\left[\\frac{1}{2}-5-0\\right]$<\/p>\n\n\n\n

$=-\\left(-\\frac{9}{2}\\right)=\\frac{9}{2}$<\/p>\n\n\n\n

\u0938\u092e\u093e\u0915\u0932 \u0915\u093e \u092e\u093e\u0928 \u0935\u0915\u094d\u0930 y=|x-5|, x-\u0905\u0915\u094d\u0937 \u0924\u0925\u093e x=0,x=1 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0915\u094b \u0928\u093f\u0930\u0941\u092a\u093f\u0924 \u0915\u0930\u0924\u093e \u0939\u0948\u0964<\/p>\n\n\n\n

Question 50<\/h4>\n\n\n\n

(i)<\/strong> y=|x+1| \u0915\u093e \u0906\u0932\u0947\u0916 \u0916\u0940\u0902\u091a\u0947 $\\int_{0}^{1}|x+1|dx$ \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964 \u0906\u0932\u0947\u0916 \u0938\u092e\u093e\u0915\u0932 \u0915\u094d\u092f\u093e \u0928\u093f\u0930\u0941\u092a\u093f\u0924 \u0915\u0930\u0924\u093e \u0939\u0948\u0964<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

y=|x+| $=\\left\\{\\begin{array}{l}x+1, \\text{if } x\\leq -1\\\\ -(x+1),\\text{if } x<-1\\end{array}\\right.$<\/p>\n\n\n\n

CASE-I<\/p>\n\n\n\n

y=x+1, x\u2265-1<\/p>\n\n\n\n

(ii)<\/strong> \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928 \u0915\u093e \u0906\u0932\u0947\u0916 \u0916\u0940\u0902\u091a\u0947\u0964<\/p>\n\n\n\n

(Sketch the graph of the following function)<\/p>\n\n\n\n

$f(x)=\\left\\{\\begin{array}{l}|x-2|+2, x\\leq 2\\\\x^2-2, x>2\\end{array}\\right.$<\/p>\n\n\n\n

$\\int_{0}^{4} f(x)dx$ \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0964 \u0906\u0932\u0947\u0916 \u092a\u0930 \u092f\u0939 \u0938\u092e\u093e\u0915\u0932 \u0915\u094d\u092f\u093e \u0928\u093f\u0930\u0942\u092a\u093f\u0924 \u0915\u0930\u0924\u093e \u0939\u0948?<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

f<\/em>(x)=|x-2|+2,x\u22642<\/p>\n\n\n\n

Let y=f<\/em>(x)<\/p>\n\n\n\n

y=-(x-2)+2<\/p>\n\n\n\n

y2<\/sup>=x2<\/sup>-2, x>2<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

$\\int_{0}^{4}f(x)dx=\\int_{0}^{2}(4-x)dx+\\int_{0}^{4}(x^2-2)dx$<\/p>\n\n\n\n

$=\\left[4x-\\frac{x^2}{2}\\right]_{0}^{2}+\\left[\\frac{x^3}{3}-2x\\right]_{2}^{4}$<\/p>\n\n\n\n

=8-2-0+$\\frac{64}{3}$-8$\\left(\\frac{8}{3}-4\\right)$<\/p>\n\n\n\n

$=\\frac{64}{3}-\\frac{8}{3}+2$<\/p>\n\n\n\n

$=\\frac{64-8+6}{3}=\\frac{62}{3}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

\u0938\u092e\u093e\u0915\u0932 \u0903<\/p>\n\n\n\n

y=f<\/em>(x): x=0,x=4 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0915\u094b \u0928\u093f\u0930\u0942\u092a\u093f\u0924 \u0915\u0930\u0928\u093e \u0939\u0948<\/p>\n\n\n\n

Question 51<\/h4>\n\n\n\n

\u0938\u092e\u093e\u0915\u0932\u0928 \u0935\u093f\u0927\u093f \u0915\u093e \u0909\u092a\u092f\u094b\u0917 \u0915\u0930\u0924\u0947 \u0939\u0941\u090f \u0935\u0915\u094d\u0930 |x|+|y|=1 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

[Using integration find , the area bounded by the curve |x|+|y|=1]<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

|x|+|y|=1<\/p>\n\n\n\n

CASE-I<\/p>\n\n\n\n

x>0, y>0<\/p>\n\n\n\n

x+y=1<\/p>\n\n\n\n

CASE-II<\/p>\n\n\n\n

x>0, y<0<\/p>\n\n\n\n

x-y=1<\/p>\n\n\n\n

CASE-III<\/p>\n\n\n\n

x<0, y>0<\/p>\n\n\n\n

-x+y=1<\/p>\n\n\n\n

CASE-IV<\/p>\n\n\n\n

x<0, y<0<\/p>\n\n\n\n

-x-y=1<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 <\/p>\n\n\n\n

=4\u00d7ar(OAB)<\/p>\n\n\n\n

$=4\\times \\int_{0}^{1}(1-x)dx$<\/p>\n\n\n\n

$=4\\times \\left[x-\\frac{x^2}{2}\\right]_{0}^{1}$<\/p>\n\n\n\n

$=4\\left[1-\\frac{1}{2}-0\\right]$<\/p>\n\n\n\n

$=4\\times \\frac{1}{2}=2$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 52<\/h4>\n\n\n\n

\u0915\u094d\u0937\u0947\u0924\u094d\u0930 {(x,y): x2<\/sup>\u2264y\u2264x} \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

[Find the area of the region {(x,y): x2<\/sup>\u2264y\u2264x}]<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u092a\u0930\u0935\u0932\u092f \u0903 x2<\/sup>=y..(i) \u0924\u0925\u093e \u0930\u0947\u0916\u093e y=x..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947,<\/p>\n\n\n\n

x2<\/sup>=x<\/p>\n\n\n\n

x2<\/sup>-x=0<\/p>\n\n\n\n

x(x-1)=0<\/p>\n\n\n\n

x=0,1<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

x=0,y=0<\/p>\n\n\n\n

x=1,y=1<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

=ar(OBAO)-ar(OBACO)<\/p>\n\n\n\n

$=\\int_{0}^{1}xdx-\\int_{0}^{1}x^2dx$<\/p>\n\n\n\n

$=\\left[\\frac{x^2}{2}\\right]_{0}^{1}-\\left[\\frac{x^3}{3}\\right]_{0}^{1}$<\/p>\n\n\n\n

$=\\frac{1}{2}-0-\\frac{1}{3}+0$<\/p>\n\n\n\n

$=\\frac{1}{6}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 53<\/h4>\n\n\n\n

\u0915\u094d\u0937\u0947\u0924\u094d\u0930 {(x,y): x2<\/sup>+y2<\/sup>}\u22641\u2264x+y} \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

[Find the area of the region {(x,y): x2<\/sup>+y2<\/sup>}\u22641\u2264x+y}]<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u0926\u093f\u090f \u0917\u090f \u0905\u0938\u092e\u093f\u0915\u093e \u0915\u094b \u0938\u092e\u0940\u0915\u0930\u0923 \u092e\u0947 \u0932\u0947\u0928\u0947 \u092a\u0930<\/p>\n\n\n\n

x2<\/sup>+y2<\/sup>=1..(i) , <\/p>\n\n\n\n

x+y=1..(ii) <\/p>\n\n\n\n

y=1-x<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0938\u0947,<\/p>\n\n\n\n

x2<\/sup>+(1-x2<\/sup>+)=1<\/p>\n\n\n\n

x2<\/sup>+1-2x+x2<\/sup>=1<\/p>\n\n\n\n

2x(x-1)=0<\/p>\n\n\n\n

x=0,1<\/p>\n\n\n\n

x=0,y=1<\/p>\n\n\n\n

x=1,y=0<\/p>\n\n\n\n

[Diagram to be added]<\/p>\n\n\n\n

1\u2264x+y<\/p>\n\n\n\n

1\u22640+0<\/p>\n\n\n\n

1\u22640 \u0905\u0938\u0924\u094d\u092f<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u0940 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

=ar(OABCO)-ar(OAOC)<\/p>\n\n\n\n

$=\\int_{0}^{1}\\sqrt{1^2-x^2}dx-\\int_{0}^{1}(1-x)dx$<\/p>\n\n\n\n

$=\\left[\\frac{x}{2}\\sqrt{1^2-x^2}+\\frac{1^2}{2}\\sin^{-1}\\frac{x}{1}\\right]_{0}^{1}-\\left[x-\\frac{x^2}{2}\\right]_{0}^{1}$<\/p>\n\n\n\n

$=\\left[\\frac{1}{2}\\times \\frac{\\pi}{2}-0\\right]-\\left[\\left(1-\\frac{1}{2}\\right)-0\\right]$<\/p>\n\n\n\n

$=\\left(\\frac{\\pi}{4}-\\frac{1}{2}\\right)$ \u0935\u0930\u094d\u0917 \u0908\u0915\u093e\u0907<\/p>\n\n\n\n

Question 54<\/h4>\n\n\n\n

\u0915\u094d\u0937\u0947\u0924\u094d\u0930 {(x,y):x2<\/sup>+y2<\/sup>\u22642ax ,y2<\/sup>\u2265ax, x\u22650, y\u22650} \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964<\/p>\n\n\n\n

[Find the area of the region {(x,y):x2<\/sup>+y2<\/sup>\u22642ax ,y2<\/sup>\u2265ax, x\u22650, y\u22650}]<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

Let R={(x,y): x2<\/sup>+y2<\/sup>\u22642ax ,y2<\/sup>\u2265ax, x\u22650, y\u22650}<\/p>\n\n\n\n

R={(x,y): x2<\/sup>+y2<\/sup>\u22642ax}\u2229{(x,y): y2<\/sup>\u2265ax, x\u22650, y\u22650}\u2229{(x,y): x\u22650, y\u22650}<\/p>\n\n\n\n

R=R1<\/sub>\u2229R2<\/sub>\u2229R3<\/sub><\/p>\n\n\n\n

R1 <\/sub>\u0915\u0947 \u0932\u093f\u090f , x2<\/sup>+y2<\/sup>=2ax (\u0938\u092e\u0940\u0915\u0930\u0923 \u0930\u0942\u092a \u092e\u0947)<\/p>\n\n\n\n

x2<\/sup>-2ax+a2<\/sup>+y2<\/sup>=a2<\/sup><\/p>\n\n\n\n

(x-a)2<\/sup>+y2<\/sup>=a2<\/sup>\u21d2y=$\\sqrt{a^2-(x-a)^2}$<\/p>\n\n\n\n

R2 <\/sub>\u0915\u0947 \u0932\u093f\u090f , y2<\/sup>=ax..(ii) (\u0938\u092e\u0940\u0915\u0930\u0923 \u0930\u0942\u092a \u092e\u0947)<\/p>\n\n\n\n

$y=\\sqrt{a}x^{\\frac{1}{2}}$<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u092e\u0947 y2 <\/sup>\u0915\u093e \u092e\u093e\u0928 \u0930\u0916\u0928\u0947 \u092a\u0930,<\/p>\n\n\n\n

x2<\/sup>+ax=2ax<\/p>\n\n\n\n

x2<\/sup>-ax=0<\/p>\n\n\n\n

x(x-a)=0<\/p>\n\n\n\n

x=0, a<\/p>\n\n\n\n

x=0,y=0<\/p>\n\n\n\n

x=a,y=\u00b1a<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932<\/p>\n\n\n\n

=ar(OABCO)-ar(OABDO)<\/p>\n\n\n\n

$=\\int_{0}^{a}\\sqrt{a^2-(x-a)^2}dx-\\sqrt{a}\\int_{0}^{a}{x^\\frac{1}{2}}$<\/p>\n\n\n\n

$=\\left[\\frac{x-a}{2}\\sqrt{a^2-(x-a)^2}+\\frac{a^2}{2}\\sin^{-1}\\frac{x-a}{a}\\right]_{0}^{a}-\\sqrt{a}\\left[\\frac{2}{3}x^{\\frac{3}{2}}\\right]_{0}^{a}$<\/p>\n\n\n\n

$=0-\\left[\\frac{-a}{2}\\times 0+\\frac{a^2}{2}\\times \\left(-\\frac{\\pi}{2}\\right)\\right]-\\sqrt{a}\\times \\frac{2}{3}(a)^{\\frac{3}{2}}-0$<\/p>\n\n\n\n

$=\\frac{\\pi}{4}a^2-\\sqrt{a}\\times \\frac{2}{3}\\times (\\sqrt{a})^3$<\/p>\n\n\n\n

$=\\frac{\\pi}{4}a^2-\\frac{2}{3}a^2$<\/p>\n\n\n\n

$=\\left(\\frac{\\pi}{4}-\\frac{2}{3}\\right)a^2$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

Question 55<\/h4>\n\n\n\n

{(x,y): y\u2265x2<\/sup>}\u0924\u0925\u093e y=|x| \u0938\u0947 \u0927\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0964<\/p>\n\n\n\n

[Find the area bounded by {(x,y): y\u2265x2<\/sup>}]<\/p>\n\n\n\n

Sol :<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 \u0930\u0942\u092a \u092e\u0947 \u0905\u0938\u092e\u093f\u0915\u093e<\/p>\n\n\n\n

y=x2<\/sup>..(i) ,y=x..(ii)<\/p>\n\n\n\n

\u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0924\u0925\u093e (ii) \u0938\u0947<\/p>\n\n\n\n

x2<\/sup>=x<\/p>\n\n\n\n

x2<\/sup>-x=0<\/p>\n\n\n\n

x(x-1)=0<\/p>\n\n\n\n

\u21d2x=0,1<\/p>\n\n\n\n

x=0,y=0<\/p>\n\n\n\n

x=1,y=1<\/p>\n\n\n\n

ar(OCBO)=ar(OABO)-ar(OABCO)<\/p>\n\n\n\n

$=\\int_{0}^{1}xdx-\\int_{0}^{1}x^2dx$<\/p>\n\n\n\n

$=\\frac{1}{6}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

\u091b\u093e\u092f\u093e\u0902\u0915\u093f\u0924 \u092d\u093e\u0917 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 <\/p>\n\n\n\n

=2\u00d7ar(OCBO)<\/p>\n\n\n\n

$=2\\times \\frac{1}{6}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

$=\\frac{1}{3}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 12 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Question 1 \u0935\u0915\u094d\u0930 y=x2 ,\u0930\u0947\u0916\u093e\u0913 x=1, x=2 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964[Find the area bounded by the curve y=x2 , lines x=1, x=2 and x-axis]Sol :ar(ABCDA)=$\\int_{1}^{2}ydx$ $=\\int_{1}^{2}x^2dx$ $=\\left[\\frac{x^3}{3}\\right]_{1}^{2}$ $=\\frac{2^3}{3}-\\frac{1^3}{3}$ $=\\frac{8}{3}-\\frac{1}{3}$ $=\\frac{7}{3}$ \u0935\u0930\u094d\u0917 \u0907\u0915\u093e\u0908 [Diagram to be added] Question 2 \u0935\u0915\u094d\u0930 y=x4 ,\u0930\u0947\u0916\u093e\u0913 x=1, x=5 \u090f\u0935\u0902 x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964[Find the […]<\/p>\n","protected":false},"author":302,"featured_media":626514,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[25],"tags":[],"boards":[],"class_list":["post-626502","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-12","entry"],"yoast_head":"\nKC Sinha: Exercise 22.1- Mathematics Solution Class 12 Chapter 22 \u0938\u092e\u093e\u0915\u0932 \u0915\u0947 \u0905\u0928\u0941\u092a\u094d\u0930\u092f\u094b\u0917 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1 \u0935\u0915\u094d\u0930 y=x2 ,\u0930\u0947\u0916\u093e\u0913 x=1, x=2 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964Sol :ar(ABCDA)=$int_{1}^{2}ydx$ $=int_{1}^{2}x^2dx$\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 22.1- Mathematics Solution Class 12 Chapter 22 \u0938\u092e\u093e\u0915\u0932 \u0915\u0947 \u0905\u0928\u0941\u092a\u094d\u0930\u092f\u094b\u0917\" \/>\n<meta property=\"og:description\" content=\"Question 1 \u0935\u0915\u094d\u0930 y=x2 ,\u0930\u0947\u0916\u093e\u0913 x=1, x=2 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964Sol :ar(ABCDA)=$int_{1}^{2}ydx$ $=int_{1}^{2}x^2dx$\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/\" \/>\n<meta property=\"og:site_name\" content=\"IndCareer Schools\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/indcareer\" \/>\n<meta property=\"article:published_time\" content=\"2023-09-12T04:17:45+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-12T04:17:52+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-49-scaled.jpg\" \/>\n\t<meta property=\"og:image:width\" content=\"1600\" \/>\n\t<meta property=\"og:image:height\" content=\"901\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/jpeg\" \/>\n<meta name=\"author\" content=\"Pooja\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@indcareer\" \/>\n<meta name=\"twitter:site\" content=\"@indcareer\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Pooja\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"12 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/\"},\"author\":{\"name\":\"Pooja\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e\"},\"headline\":\"KC Sinha: Exercise 22.1- Mathematics Solution Class 12 Chapter 22 \u0938\u092e\u093e\u0915\u0932 \u0915\u0947 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\u0915\u0940\u091c\u093f\u090f\u0964Sol :ar(ABCDA)=$\\\\int_{1}^{2}ydx$ $=\\\\int_{1}^{2}x^2dx$\",\"breadcrumb\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/#primaryimage\",\"url\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-49-scaled.jpg\",\"contentUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-49-scaled.jpg\",\"width\":1600,\"height\":901,\"caption\":\"KC Sinha: Exercise 22.1- Mathematics 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\/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"KC Sinha: Exercise 22.1- Mathematics Solution Class 12 Chapter 22 \u0938\u092e\u093e\u0915\u0932 \u0915\u0947 \u0905\u0928\u0941\u092a\u094d\u0930\u092f\u094b\u0917 - IndCareer Schools","description":"Question 1 \u0935\u0915\u094d\u0930 y=x2 ,\u0930\u0947\u0916\u093e\u0913 x=1, x=2 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964Sol :ar(ABCDA)=$int_{1}^{2}ydx$ $=int_{1}^{2}x^2dx$","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/","og_locale":"en_US","og_type":"article","og_title":"KC Sinha: Exercise 22.1- Mathematics Solution Class 12 Chapter 22 \u0938\u092e\u093e\u0915\u0932 \u0915\u0947 \u0905\u0928\u0941\u092a\u094d\u0930\u092f\u094b\u0917","og_description":"Question 1 \u0935\u0915\u094d\u0930 y=x2 ,\u0930\u0947\u0916\u093e\u0913 x=1, x=2 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964Sol :ar(ABCDA)=$int_{1}^{2}ydx$ $=int_{1}^{2}x^2dx$","og_url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/","og_site_name":"IndCareer Schools","article_publisher":"https:\/\/www.facebook.com\/indcareer","article_published_time":"2023-09-12T04:17:45+00:00","article_modified_time":"2023-09-12T04:17:52+00:00","og_image":[{"width":1600,"height":901,"url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-49-scaled.jpg","type":"image\/jpeg"}],"author":"Pooja","twitter_card":"summary_large_image","twitter_creator":"@indcareer","twitter_site":"@indcareer","twitter_misc":{"Written by":"Pooja","Est. reading time":"12 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/#article","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/"},"author":{"name":"Pooja","@id":"https:\/\/www.indcareer.com\/schools\/#\/schema\/person\/d6945cf059726f162259ba738092301e"},"headline":"KC Sinha: Exercise 22.1- Mathematics Solution Class 12 Chapter 22 \u0938\u092e\u093e\u0915\u0932 \u0915\u0947 \u0905\u0928\u0941\u092a\u094d\u0930\u092f\u094b\u0917","datePublished":"2023-09-12T04:17:45+00:00","dateModified":"2023-09-12T04:17:52+00:00","mainEntityOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/"},"wordCount":3609,"publisher":{"@id":"https:\/\/www.indcareer.com\/schools\/#organization"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-49-scaled.jpg","articleSection":["Class 12"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/","url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/","name":"KC Sinha: Exercise 22.1- Mathematics Solution Class 12 Chapter 22 \u0938\u092e\u093e\u0915\u0932 \u0915\u0947 \u0905\u0928\u0941\u092a\u094d\u0930\u092f\u094b\u0917 - IndCareer Schools","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/#website"},"primaryImageOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/#primaryimage"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-49-scaled.jpg","datePublished":"2023-09-12T04:17:45+00:00","dateModified":"2023-09-12T04:17:52+00:00","description":"Question 1 \u0935\u0915\u094d\u0930 y=x2 ,\u0930\u0947\u0916\u093e\u0913 x=1, x=2 \u0924\u0925\u093e x-\u0905\u0915\u094d\u0937 \u0938\u0947 \u0918\u093f\u0930\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930 \u0915\u093e \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f\u0964Sol :ar(ABCDA)=$\\int_{1}^{2}ydx$ $=\\int_{1}^{2}x^2dx$","breadcrumb":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-22-1-mathematics-solution-class-12-chapter-22-samakal-ke-anuprayog\/#primaryimage","url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-49-scaled.jpg","contentUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-49-scaled.jpg","width":1600,"height":901,"caption":"KC Sinha: Exercise 22.1- Mathematics Solution Class 12 Chapter 22 \u0938\u092e\u093e\u0915\u0932 \u0915\u0947 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