{"id":626457,"date":"2023-09-12T02:22:38","date_gmt":"2023-09-12T02:22:38","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626457"},"modified":"2023-09-12T02:22:50","modified_gmt":"2023-09-12T02:22:50","slug":"kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/","title":{"rendered":"KC Sinha: Exercise 17.1- Mathematics Solution Class 12 Chapter 17 \u0938\u0928\u094d\u0928\u093f\u0915\u091f\u0928"},"content":{"rendered":"\n

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

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

\u0905\u0935\u0915\u0932 \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0924\u0947 \u0939\u0941\u090f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u093e \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092e\u093e\u0928 \u0926\u0936\u092e\u0932\u0935 \u0915\u0947 3 \u0938\u094d\u0925\u093e\u0928\u094b \u0924\u0915 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
[Using differentials, find the approximate value of each of the following upto 3 places of decimal]
(i)<\/strong> $\\sqrt{36.6}$
Sol :
 \u092e\u093e\u0928\u093e f<\/em>(x)=\u221ax<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{2\\sqrt{x}}$<\/p>\n\n\n\n

36 \u0915\u093e \u0935\u0930\u094d\u0917\u092e\u0942\u0932 6 \u0939\u094b\u0924\u093e \u0939\u0948 \u0924\u0925\u093e<\/p>\n\n\n\n

36.6 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0964<\/p>\n\n\n\n

x=36 ,<\/p>\n\n\n\n

x+\u1e9fx=36.6
36+\u1e9fx=36.6
\u1e9fx=0.6<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f<\/em> ‘(x).\u1e9fx<\/p>\n\n\n\n

f<\/em>(36.6)$=\\sqrt{x}+\\frac{1}{2\\sqrt{36}} \\times 0.6$<\/p>\n\n\n\n

$\\sqrt{36.6}=\\sqrt{36.6}+\\frac{1}{2\\sqrt{36}}\\times 0.6$<\/p>\n\n\n\n

$=6+\\frac{1}{2\\times 6} \\times 0.6$<\/p>\n\n\n\n

$=6+\\frac{0.1}{2}$
=6+0.050
=6.050<\/p>\n\n\n\n

(ii)<\/strong> $\\sqrt{0.6}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)=\u221ax<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{2\\sqrt{x}}$<\/p>\n\n\n\n

0.64 , 0.6 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 ,<\/p>\n\n\n\n

\u0924\u0925\u093e 0.64 \u0915\u093e \u0935\u0930\u094d\u0917\u092e\u0942\u0932 0.8 \u0939\u094b\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

x=0.64 , x+\u1e9fx=0.6
0.64+\u1e9fx=0.6
\u1e9fx=-0.04<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f<\/em> ‘(x).\u1e9fx
f<\/em>(0.6)$=\\sqrt{x}+\\frac{1}{2\\sqrt{x}}\\times (-0.04)$<\/p>\n\n\n\n

$\\sqrt{0.6}=\\sqrt{0.64}+\\frac{1}{2\\sqrt{0.64}}\\times (-0.04)$<\/p>\n\n\n\n

$=0.8-\\frac{0.04}{2\\times 0.8}$<\/p>\n\n\n\n

$=0.8-\\frac{2\\times 10}{4 \\times 10}$<\/p>\n\n\n\n

$=0.8-\\frac{1}{40}$
=0.8-0.025
=0.775<\/p>\n\n\n\n

(iii)<\/strong> $\\sqrt{49.5}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)=\u221ax<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{2\\sqrt{x}}$<\/p>\n\n\n\n

49 , 49.5 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0964<\/p>\n\n\n\n

49 \u0915\u093e \u0935\u0930\u094d\u0917\u092e\u0942\u0932 7 \u0939\u094b\u0924\u093e \u0939\u0948\u0964<\/p>\n\n\n\n

x=49 ,<\/p>\n\n\n\n

x+\u1e9fx=49.5
49+\u1e9fx=49.5
\u1e9fx=0.5<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f<\/em> ‘(x).\u1e9fx
f<\/em>(49.5)=\u221ax+$\\frac{1}{2\\sqrt{x}}.(0.5)$
$\\sqrt{49.5}=\\sqrt{49}+\\frac{1}{2\\sqrt{49}}\\times (0.5)$
=7+$\\frac{1}{2\\times 7} \\times 0.5$
=7+$\\frac{0.5}{14}$
=7+0.035
=7.035<\/p>\n\n\n\n

(iv)<\/strong> $(401)^{\\frac{1}{2}}$
Sol :
=\u221a401<\/p>\n\n\n\n

x=400 ,<\/p>\n\n\n\n

x+\u1e9fx=401<\/p>\n\n\n\n

\u1e9fx=1<\/p>\n\n\n\n

(v)<\/strong> $\\sqrt{0.0037}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)=\u221ax<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{2\\sqrt{x}}$<\/p>\n\n\n\n

0.0036 , 0.0037 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0964 0.0036 \u0915\u093e \u0935\u0930\u094d\u0917\u092e\u0942\u0932 0.06 \u0939\u094b\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

x=0.0036 ,<\/p>\n\n\n\n

x+\u1e9fx=0.0037
0.0036+\u1e9fx=0.0037
\u1e9fx=0.001<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f<\/em> ‘(x).\u1e9fx
f<\/em>(0.0037)$=\\sqrt{x}+\\frac{1}{2\\sqrt{x}} \\times (0.001)$
$\\sqrt{0.0037}=\\sqrt{0.0036}+\\frac{1}{2\\times \\sqrt{0.0036}}\\times 0.0001$
$=0.006+\\frac{0.0001}{2 \\times 0.06}$
$=0.06+\\frac{0.0001}{0.12}$
=0.06+0.00083
=0.06083<\/p>\n\n\n\n

(vi)<\/strong> $\\sqrt{25.3}$
Sol :<\/p>\n\n\n\n

(vii)<\/strong> $(0.009)^{\\frac{1}{3}}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)=\u221ax<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{2\\sqrt{x}}$<\/p>\n\n\n\n

0.008 , 0.009 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0924\u0925\u093e 0.008 \u0915\u093e \u0918\u0928\u092e\u0942\u0932 0.2 \u0939\u094b\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

x=0.008 ,<\/p>\n\n\n\n

x+\u1e9fx=0.009
0.008+\u1e9fx=0.009
\u1e9fx=0.001<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f<\/em> ‘(x).\u1e9fx<\/p>\n\n\n\n

f<\/em>(0.009)$=x^{\\frac{1}{3}}+\\frac{1}{3x^{2\/3}}\\times (0.001)$<\/p>\n\n\n\n

$(0.009)^{\\frac{1}{3}}=(0.008)^{\\frac{1}{3}}+\\frac{1}{3(0.008)^{\\frac{2}{3}}}\\times (0.001)$
$(0.009)^{\\frac{1}{3}}=(0.2)^{3\\times \\frac{1}{3}}+\\frac{1}{3(0.2)^{3 \\times \\frac{2}{3}}}\\times (0.001)$
$=0.2+\\frac{0.001}{0.12}$
=0.2+0.0083
=0.2083<\/p>\n\n\n\n

(viii)<\/strong> $(26.57)^{\\frac{1}{3}}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)=\u221ax<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{2\\sqrt{x}}$<\/p>\n\n\n\n

27, 26.57 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0924\u0925\u093e 27 \u0915\u093e \u0918\u0928\u092e\u0942\u0932 3 \u0939\u094b\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

x=27 ,<\/p>\n\n\n\n

x+\u1e9fx=26.57<\/p>\n\n\n\n

27+\u1e9fx=26.57<\/p>\n\n\n\n

\u1e9fx=-0.43<\/p>\n\n\n\n

f<\/em>(26.57)$=x^{\\frac{1}{3}}+\\frac{1}{3x^{\\frac{2}{3}}} \\times (-0.43)$<\/p>\n\n\n\n

$(26.57)^{\\frac{1}{3}}=(27)^{\\frac{1}{3}}+\\frac{1}{3 \\times (27)^{\\frac{2}{3}}} \\times (-0.43)$
$=(3^3)^{\\frac{1}{3}}+\\frac{1}{3(3)^{3\\times \\frac{2}{3}}} \\times (-0.43)$
$=3-\\frac{0.43}{27}$
=3-0.016
=2.984<\/p>\n\n\n\n

(ix)<\/strong> $(3.968)^{\\frac{3}{2}}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)$=x^{\\frac{3}{2}}$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{3}{2}x^{\\frac{1}{2}}$ or $\\frac{3}{2}^{\\sqrt{x}}$<\/p>\n\n\n\n

4, 3.968 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0924\u0925\u093e 4 \u0915\u093e $\\frac{3}{2}$ \u0918\u093e\u0924 \u0932\u0947\u0928\u0947 \u092a\u0930 8 \u092a\u094d\u0930\u093e\u092a\u094d\u0924 \u0939\u094b\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

\u2234x=4 ,<\/p>\n\n\n\n

x+\u1e9fx=3.968<\/p>\n\n\n\n

4+\u1e9fx=3.968<\/p>\n\n\n\n

\u1e9fx=-0.032<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f <\/em>‘(x)<\/p>\n\n\n\n

f<\/em>(3.968)$x^{\\frac{3}{2}}+\\frac{3}{2}\\sqrt{x} \\times (-0.032)$
$=(4)^{\\frac{3}{2}}+\\frac{3}{2}\\sqrt{4} \\times (-0.032)$
$=(2^2)^{\\frac{3}{2}}+\\frac{3}{2}\\times 2 \\times (-0.032)$
=8-0.096
=7.904<\/p>\n\n\n\n

(x)<\/strong> $(0.999)^{\\frac{1}{10}}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)$=x^{\\frac{1}{10}}$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{10x^{\\frac{9}{10}}}$<\/p>\n\n\n\n

1 , 0.999 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0924\u0925\u093e 1 \u0915\u093e \u0918\u093e\u0924 $\\frac{1}{10}$ \u0932\u0947\u0928\u0947 \u092a\u0930 1 \u0939\u0940 \u092a\u094d\u0930\u093e\u092a\u094d\u0924 \u0939\u094b\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

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

x+\u1e9fx=0.999<\/p>\n\n\n\n

1+\u1e9fx=0.999<\/p>\n\n\n\n

\u1e9fx=-0.001<\/p>\n\n\n\n

f<\/em>(0.999)$=x^{\\frac{1}{10}}+\\frac{1}{10x^{\\frac{9}{10}}} \\times (-0.001)$<\/p>\n\n\n\n

$(0.999)^{\\frac{1}{10}}=(1)^{\\frac{1}{10}}+\\frac{1}{10(1)^{\\frac{9}{10}}}\\times (-0.001)$<\/p>\n\n\n\n

$=1-\\frac{0.001}{10}$
=1-0.0001
=0.9999<\/p>\n\n\n\n

(xi)<\/strong> $(32.15)^{\\frac{1}{5}}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)$=x^{\\frac{1}{5}}$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{5x^{\\frac{4}{5}}}$<\/p>\n\n\n\n

32 , 32.15 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0924\u0925\u093e $(32)^{\\frac{1}{5}}$ , 2 \u0939\u094b\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

\u2234x=32 ,<\/p>\n\n\n\n

x+\u1e9fx=32.15
32+\u1e9fx=32.15
\u1e9fx=0.15<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f <\/em>‘(x)\u1e9fx
f<\/em>(32.15)=$x^{\\frac{1}{5}}+\\frac{1}{5x^{\\frac{4}{5}}}.(0.15)$
$(32.15)^{\\frac{1}{5}}=(32)^{\\frac{1}{5}}+\\frac{1}{5\\times 32^{\\frac{4}{5}}} (0.15)$
$=(2^5)^{\\frac{1}{5}}+\\frac{1}{5\\times (2)^{5 \\times \\frac{4}{5}}}\\times 0.15$
$=2+\\frac{0.15}{80}$
=2+0.00187
=2.00187<\/p>\n\n\n\n

(xii)<\/strong> $(81.5)^{\\frac{1}{4}}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)$=x^{\\frac{1}{4}}$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{4x^{\\frac{3}{4}}}$<\/p>\n\n\n\n

81 , 81.5 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0924\u0925\u093e $(81)^{\\frac{1}{4}}$ , 3 \u0939\u094b\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

x=81 ,<\/p>\n\n\n\n

x+\u1e9fx=81.5
81+\u1e9fx=81.5
\u1e9fx=0.5<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f <\/em>‘(x)\u1e9fx<\/p>\n\n\n\n

f<\/em>(81.5)$=x^{\\frac{1}{4}}+\\frac{1}{4x^{\\frac{3}{4}}} (0.5)$<\/p>\n\n\n\n

$(81.5)^{\\frac{1}{4}}=(81)^{\\frac{1}{4}}+\\frac{1}{4 \\times (81)^{\\frac{3}{4}}} \\times 0.5$
$=(3)^{4\\times \\frac{1}{4}}+\\frac{1}{4(3)^{4\\times \\frac{3}{4}}}\\times 0.5$
$=3+\\frac{0.5}{108}$
=3+0.0046
=3.0046<\/p>\n\n\n\n

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

\u0905\u0935\u0915\u0932 \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u093e \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u0915\u093e \u092e\u093e\u0928 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964
[Using differentials find the approximate value of each of the following]
(i)<\/strong> $(25)^{\\frac{1}{3}}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)$=x^{\\frac{1}{3}}$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

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

27 , 25 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0924\u0925\u093e $(27)^{\\frac{1}{3}}$=3 \u0939\u094b\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

x=27 ,<\/p>\n\n\n\n

x+\u1e9fx=25
27+\u1e9fx=25
\u1e9fx=-2<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f <\/em>‘(x)\u1e9fx<\/p>\n\n\n\n

f<\/em>(25)$=x^{\\frac{1}{3}}+\\frac{1}{3x^{\\frac{2}{3}}} \\times (-2)$<\/p>\n\n\n\n

$(25)^{\\frac{1}{3}}=(27)^{\\frac{1}{3}}-\\frac{2}{3(27)^{\\frac{2}{3}}}$
$=(3^3)^{\\frac{1}{3}}-\\frac{2}{3(3)^{3\\times \\frac{2}{3}}}$
$=3-\\frac{2}{27}$
=3-0.074
=2.926<\/p>\n\n\n\n

(ii)<\/strong> $(15)^{\\frac{1}{4}}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)$=x^{\\frac{1}{4}}$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{4x^{\\frac{3}{4}}}$<\/p>\n\n\n\n

16 , 15 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0924\u0925\u093e $(16)^{\\frac{1}{4}}=2$ \u0939\u094b\u0924\u093e \u0939\u0948 \u0964
x=16 ,<\/p>\n\n\n\n

x+\u1e9fx=15
16+\u1e9fx=15
\u1e9fx=-1<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f <\/em>‘(x)\u1e9fx<\/p>\n\n\n\n

f<\/em>(15)$=x^{\\frac{1}{4}}+\\frac{1}{4x^{\\frac{3}{4}}} \\times (-1)$
$(15)^{\\frac{1}{4}}=(16)^{\\frac{1}{4}}+\\frac{1}{4(16)^{\\frac{3}{4}}} \\times (-1)$
$=(2)^{4 \\times \\frac{1}{4}}+\\frac{1}{4(2)^{4\\times \\frac{3}{4}}} (-1)$
$=2-\\frac{1}{32}$
=2-0.03125
=1.96875<\/p>\n\n\n\n

(iv)<\/strong> $(33)^{-\\frac{1}{5}}$
Sol :
\u092e\u093e\u0928\u093e f<\/em>(x)=$x^{-\\frac{1}{5}}$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=-\\frac{1}{5x^{\\frac{6}{5}}}$<\/p>\n\n\n\n

32 , 33 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0924\u0925\u093e $(32)^{-\\frac{1}{5}}=\\frac{1}{2}$ \u0915\u0947 \u092c\u0930\u093e\u092c\u0930 \u0939\u0948\u0964<\/p>\n\n\n\n

x=32 ,<\/p>\n\n\n\n

x+\u1e9fx=33
32+\u1e9fx=33
\u1e9fx=1<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f<\/em> ‘(\u1e9fx)
f<\/em>(33)=$x^{-\\frac{1}{5}}+\\frac{-1}{5x^{\\frac{6}{5}}}\\times (1)$<\/p>\n\n\n\n

$(33)^{-\\frac{1}{5}}=(32)^{-\\frac{1}{5}}-\\frac{1}{5(32)^{\\frac{6}{5}}}$<\/p>\n\n\n\n

$=(2^5)^{-\\frac{1}{5}}-\\frac{1}{5(2)^{5\\times \\frac{6}{5}}}$<\/p>\n\n\n\n

$=\\frac{1}{2}-\\frac{1}{5\\times 64}$<\/p>\n\n\n\n

$=0.5-\\frac{1}{320}$
=0.5-0.003125
=0.496875<\/p>\n\n\n\n

(v)<\/strong> $\\left(\\frac{17}{81}\\right)^{\\frac{1}{4}}$
Sol :
\u092e\u093e\u0928\u093e $f(x)=\\frac{1}{3}x^{\\frac{1}{4}}$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$f'(x)=\\frac{1}{12x^{\\frac{3}{4}}}$<\/p>\n\n\n\n

16 , 17 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0924\u0925\u093e $(16)^{\\frac{1}{4}}=2$ \u0939\u094b\u0924\u093e \u0939\u0948 \u0964<\/p>\n\n\n\n

x=16 ,<\/p>\n\n\n\n

x+\u1e9fx=17
16+\u1e9fx=17
\u1e9fx=1<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f<\/em> ‘(\u1e9fx)<\/p>\n\n\n\n

f<\/em>(17)$=\\frac{1}{3}x^{\\frac{1}{4}}+\\frac{1}{12x^{\\frac{3}{4}}}\\times (1)$<\/p>\n\n\n\n

$\\frac{1}{3}(17)^{\\frac{1}{4}}=\\frac{1}{3}(16)^{\\frac{1}{4}}+\\frac{1}{12(16)^{\\frac{3}{4}}}$<\/p>\n\n\n\n

$=\\frac{1}{3}(2)^{4 \\times \\frac{1}{4}}+\\frac{1}{12(2)^{4 \\times \\frac{3}{4}}}$<\/p>\n\n\n\n

$=\\frac{2}{3}+\\frac{1}{96}$
=0.666+0.0104
0.6764<\/p>\n\n\n\n

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

f<\/em>(3.02) \u0915\u093e \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u0939\u093e\u0901 f<\/em>(x)=3x2<\/sup>+5x+3
[Find the approximate value of f<\/em>(3.02) , where f<\/em>(x)=3x2<\/sup>+5x+3]
Sol :
f<\/em>(x)=3x2<\/sup>+5x+3<\/p>\n\n\n\n

Differentiate w.r.t x<\/p>\n\n\n\n

f <\/em>‘(x)=6x+5<\/p>\n\n\n\n

Let x=3 ,<\/p>\n\n\n\n

x+\u1e9fx=3.02
3+\u1e9fx=3.02
\u1e9fx=0.02<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f<\/em> ‘(\u1e9fx)<\/p>\n\n\n\n

f<\/em>(3.02)=(3x2<\/sup>+5x+3)+(6x+5)\u00d7(0.02)<\/p>\n\n\n\n

=(3(3)2<\/sup>+5(3)+3)+(6\u00d73+5)\u00d7(0.02)<\/p>\n\n\n\n

=27+15+3+23\u00d70.02
=45+0.46
=45.46<\/p>\n\n\n\n

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

f<\/em>(2.01) \u0915\u093e \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u091c\u0939\u093e\u0901 f<\/em>(x)=4x2<\/sup>+5x+2
[Find the approximate value of f<\/em>(2.01) , where f<\/em>(x)=4x2<\/sup>+5x+2]
Sol :
f<\/em>(x)=4x2<\/sup>+5x+2<\/p>\n\n\n\n

Differentiate w.r.t x<\/p>\n\n\n\n

f <\/em>‘(x)=8x+5<\/p>\n\n\n\n

Let x=2 ,<\/p>\n\n\n\n

x+\u1e9fx=2.01
2+\u1e9fx=2.01
\u1e9fx=0.01<\/p>\n\n\n\n

f<\/em>(x+\u1e9fx)=f<\/em>(x)+f<\/em> ‘(\u1e9fx)<\/p>\n\n\n\n

f<\/em>(2.01)=4x2<\/sup>+5x+2+(8x+5)\u00d7<\/em>(0.01)<\/p>\n\n\n\n

=4(2)2<\/sup>+5(2)+2+(8(2)+5)\u00d7<\/em>(0.01)
=16+10+2+21\u00d7<\/em>0.01
=28+0.21
=28.21<\/p>\n\n\n\n

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

\u090f\u0915 \u0917\u094b\u0932\u0947 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e 9m \u092e\u093e\u092a\u0940 \u091c\u093e\u0924\u0940 \u0939\u0948 \u091c\u093f\u0938\u092e\u0947 0.03 m \u0924\u094d\u0930\u0941\u091f\u093f \u0939\u0948 \u0964 \u0907\u0938\u0915\u0947 \u092a\u0943\u0937\u094d\u0920 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0915\u0947 \u092a\u0930\u093f\u0915\u0932\u0928 \u092e\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u0924\u094d\u0930\u0941\u091f\u093f \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u0964
Sol :
\u092e\u093e\u0928\u093e \u0917\u094b\u0932\u0947 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e r \u0939\u0948 \u0964<\/p>\n\n\n\n

\u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e \u092e\u0947 \u0924\u094d\u0930\u0941\u091f\u093f , \u1e9fr=0.03m ,r=9m<\/p>\n\n\n\n

\u0917\u094b\u0932\u0947 \u0915\u093e \u092a\u0943\u0937\u094d\u0920 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 A=4\u03c0r2<\/sup><\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{dA}{dr}=8\\pi r$<\/p>\n\n\n\n

\u092a\u0943\u0937\u094d\u0920 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u092e\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u0924\u094d\u0930\u0941\u091f\u093f \u1e9fA$=\\frac{dA}{dr} \\times \\delta r$<\/p>\n\n\n\n

=8\u03c0r\u00d70.03m2<\/sup>
=8\u03c0r\u00d79\u00d70.03m2<\/sup>
=2.16\u03c0m2<\/sup><\/p>\n\n\n\n

Question 6<\/h4>\n\n\n\n

\u090f\u0915 \u0917\u094b\u0932\u0947 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e 9m \u092e\u093e\u092a\u0940 \u091c\u093e\u0924\u0940 \u0939\u0948 \u091c\u093f\u0938\u092e\u0947 0.03 m \u0924\u094d\u0930\u0941\u091f\u093f \u0939\u0948 \u0964 \u0907\u0938\u0915\u0947 \u0906\u092f\u0924\u0928 \u0915\u0947 \u092a\u0930\u093f\u0915\u0932\u0928 \u092e\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u0924\u094d\u0930\u0941\u091f\u093f \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u0964
Sol :<\/p>\n\n\n\n

\u092e\u093e\u0928\u093e \u0917\u094b\u0932\u0947 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e r \u0939\u0948 \u0964<\/p>\n\n\n\n

\u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e \u092e\u0947 \u0924\u094d\u0930\u0941\u091f\u093f , \u1e9fr=0.03cm ,r=9cm<\/p>\n\n\n\n

\u0917\u094b\u0932\u0947 \u0915\u093e \u0906\u092f\u0924\u0928 , $v=\\frac{4}{3}\\pi r^3$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{dv}{dr}=4\\pi r^2$<\/p>\n\n\n\n

\u0906\u092f\u0924\u0928 \u092e\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u0924\u094d\u0930\u0941\u091f\u093f , $\\delta v=\\frac{dv}{dr}\\times \\delta r$<\/p>\n\n\n\n

=4\u03c0r2<\/sup>\u00d70.03cm3<\/sup>
=4\u03c0\u00d792<\/sup>\u00d70.03cm3<\/sup>
=9.72\u03c0cm3<\/sup><\/p>\n\n\n\n

Question 7<\/h4>\n\n\n\n

\u090f\u0915 \u0917\u094b\u0932\u0947 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e 7m \u092e\u093e\u092a\u0940 \u091c\u093e\u0924\u0940 \u0939\u0948 \u091c\u093f\u0938\u092e\u0947 0.02 m \u0924\u094d\u0930\u0941\u091f\u093f \u0939\u0948 \u0964 \u0907\u0938\u0915\u0947 \u0906\u092f\u0924\u0928 \u0915\u0947 \u092a\u0930\u093f\u0915\u0932\u0928 \u092e\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u0924\u094d\u0930\u0941\u091f\u093f \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 \u0917\u094b\u0932\u0947 \u0915\u0940 \u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e r \u0939\u0948 \u0964<\/p>\n\n\n\n

\u0924\u094d\u0930\u093f\u091c\u094d\u092f\u093e \u092e\u0947 \u0924\u094d\u0930\u0941\u091f\u093f , \u1e9fr=0.02m ,r=7m<\/p>\n\n\n\n

\u0917\u094b\u0932\u0947 \u0915\u093e \u0906\u092f\u0924\u0928 , $v=\\frac{4}{3}\\pi r^3$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{dv}{dr}=4\\pi r^2$<\/p>\n\n\n\n

\u0906\u092f\u0924\u0928 \u092e\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u0924\u094d\u0930\u0941\u091f\u093f , $\\delta v=\\frac{dv}{dr}\\times \\delta r$<\/p>\n\n\n\n

=4\u03c0r2<\/sup>\u00d70.02m3<\/sup>
=4\u03c0\u00d772<\/sup>\u00d70.02m3<\/sup>
=3.92\u03c0m3<\/sup><\/p>\n\n\n\n

Question 8<\/h4>\n\n\n\n

x \u092e\u0940\u091f\u0930 \u092d\u0941\u091c\u093e \u0935\u093e\u0932\u0947 \u0918\u0928 \u0915\u0940 \u092d\u0941\u091c\u093e \u092e\u0947 2% \u0915\u0940 \u0935\u0943\u0926\u094d\u0927\u093f \u0915\u0947 \u0915\u093e\u0930\u0923 \u0938\u0947 \u0918\u0928 \u0915\u0947 \u0906\u092f\u0924\u0928 \u092e\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092a\u0930\u093f\u0935\u0930\u094d\u0924\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u0964
Sol :
\u0918\u0928 \u0915\u0940 \u092d\u0941\u091c\u093e=x<\/p>\n\n\n\n

\u0918\u0928 \u0915\u0947 \u092d\u0941\u091c\u093e \u092e\u0947 \u092a\u0930\u093f\u0935\u0930\u094d\u0924\u0928 , \u1e9fx=x \u0915\u093e 2%
$=x\\times \\frac{2}{100}$
\u1e9fx=0.02x<\/p>\n\n\n\n

\u0918\u0928 \u0915\u093e \u0906\u092f\u0924\u0928, v=x2<\/sup><\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

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

\u0906\u092f\u0924\u0928 \u092e\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u0930 \u092a\u0930\u093f\u0935\u0930\u094d\u0924\u0928 , $\\delta v=\\frac{dv}{dx} \\times \\delta x$
=3x2<\/sup>\u00d70.02x m3<\/sup>
=0.06x3<\/sup>m3<\/sup><\/p>\n\n\n\n

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

x \u092e\u0940\u091f\u0930 \u092d\u0941\u091c\u093e \u0935\u093e\u0932\u0947 \u0918\u0928 \u0915\u0940 \u092d\u0941\u091c\u093e \u092e\u0947 1% \u0915\u0940 \u0935\u0943\u0926\u094d\u0927\u093f \u0915\u0947 \u0915\u093e\u0930\u0923 \u0938\u0947 \u0918\u0928 \u0915\u0947 \u092a\u0943\u0937\u094d\u091f \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u092e\u0947 \u0939\u094b\u0928\u0947 \u0935\u093e\u0932\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092a\u0930\u093f\u0935\u0930\u094d\u0924\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u0964
Sol :<\/p>\n\n\n\n

\u0918\u0928 \u0915\u093e \u092d\u0941\u091c\u093e=x<\/p>\n\n\n\n

\u0918\u0928 \u0915\u0940 \u092d\u0941\u091c\u093e \u092e\u0947 \u0924\u094d\u0930\u0941\u091f\u093f , \u1e9fx=x \u0915\u093e 1%<\/p>\n\n\n\n

$=x \\times \\frac{1}{100}$m<\/p>\n\n\n\n

\u1e9fx=0.01 x m<\/p>\n\n\n\n

\u0918\u0928 \u0915\u093e \u092a\u0943\u0937\u094d\u0920 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932, A=6x2<\/sup><\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{dA}{dx}=12x$<\/p>\n\n\n\n

\u092a\u0943\u0937\u094d\u0920 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u092e\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u0924\u094d\u0930\u0941\u091f\u093f, $\\delta A=\\frac{dA}{dx} \\delta x$
=12x\u00d7 0.01x m2<\/sup><\/p>\n\n\n\n

=0.12x2 <\/sup>m2<\/sup><\/p>\n\n\n\n

Question 10<\/h4>\n\n\n\n

\u090f\u0915 \u0936\u0902\u0915\u0941 \u0915\u0940 \u090a\u0901\u091a\u093e\u0908 \u092e\u0947 \u0935\u0943\u0926\u094d\u0927\u093f 2% \u0939\u0948 \u0924\u0925\u093e \u0907\u0938\u0915\u093e \u0905\u0930\u094d\u0927 \u0936\u0940\u0930\u094d\u0937 \u0915\u094b\u0923 \u0905\u092a\u0930\u093f\u0935\u0930\u094d\u0924\u093f\u0924 \u0939\u0948 \u0964 \u0907\u0938\u0915\u0947
(i) \u0915\u0941\u0932 \u092a\u0943\u0937\u094d\u091f \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932
(ii) \u0906\u092f\u0924\u0928 \u092e\u0947
\u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092a\u094d\u0930\u0924\u093f\u0936\u0924 \u092c\u0922\u093c\u094b\u0924\u0930\u0940 \u0935\u0943\u0926\u094d\u0927\u093f \u0915\u094d\u092f\u093e \u0939\u0948 ?
Sol :
(i)
\u092e\u093e\u0928\u093e \u0936\u0902\u0915\u0941 \u0915\u0940 \u090a\u0901\u091a\u093e\u0908 x \u0939\u0948 \u0964 \u0936\u0902\u0915\u0941 \u0915\u093e \u0905\u0930\u094d\u0927 \u0936\u0940\u0930\u094d\u0937 \u0915\u094b\u0923 \u03b8 \u0939\u0948 \u0964<\/p>\n\n\n\n

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

$\\tan \\theta =\\frac{AB}{OA}$<\/p>\n\n\n\n

$\\tan \\theta =\\frac{AB}{x}$<\/p>\n\n\n\n

$\\cos \\theta =\\frac{OA}{OB}$<\/p>\n\n\n\n

$\\cos \\theta =\\frac{x}{OB}$<\/p>\n\n\n\n

$OB=\\frac{x}{\\cos \\theta}$<\/p>\n\n\n\n

OB=xsec\u03b8<\/p>\n\n\n\n

\u090a\u0901\u091a\u093e\u0908 \u092e\u0947 \u0935\u0943\u0926\u094d\u0927\u093f , \u1e9fx=x \u0915\u093e 2%$=x \\times \\frac{2}{100}$=0.02%<\/p>\n\n\n\n

\u0936\u0902\u0915\u0941 \u0915\u0941\u0932 \u092a\u0943\u0937\u094d\u091f \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 , A=\u03c0.xtan\u03b8.xsec\u03b8+\u03c0.(xtan\u03b8)2<\/sup>
A=\u03c0.x2<\/sup>tan\u03b8.xsec\u03b8+\u03c0x2<\/sup>.tan2<\/sup>\u03b8
A=(tan\u03b8sec\u03b8+tan2<\/sup>\u03b8)\u03c0x2<\/sup><\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{dA}{dx}=(\\tan \\theta \\sec \\theta + \\tan ^2 \\theta)\\times 2 \\pi x$<\/p>\n\n\n\n

\u092a\u0943\u0937\u094d\u091f\u0940\u092f \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u092e\u0947 \u0935\u0943\u0926\u094d\u0927\u093f , $\\delta A=\\frac{dA}{dx} \\times \\delta x$<\/p>\n\n\n\n

=(tan\u03b8sec\u03b8+tan2<\/sup>\u03b8)2\u03c0x\u00d70.02x<\/p>\n\n\n\n

=(tan\u03b8sec\u03b8+tan2<\/sup>\u03b8)0.04\u03c0x2<\/sup><\/p>\n\n\n\n

\u092a\u0943\u0937\u094d\u091f\u0940\u092f \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u092e\u0947 \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u0935\u0943\u0926\u094d\u0927\u093f \u092a\u094d\u0930\u0924\u093f\u0936\u0924$=\\frac{\\delta A}{A}\\times 100$<\/p>\n\n\n\n

$=\\frac{(\\tan \\theta \\sec \\theta+\\tan ^2 \\theta)\\times 0.04 \\pi x^2}{(\\tan \\theta \\sec \\theta +\\tan^2 \\theta).\\pi x^2}\\times 100$<\/p>\n\n\n\n

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

(ii)
\u092e\u093e\u0928\u093e \u0936\u0902\u0915\u0941 \u0915\u0940 \u090a\u0901\u091a\u093e\u0908 x \u0939\u0948 \u0964 \u0936\u0902\u0915\u0941 \u0915\u093e \u0905\u0930\u094d\u0927 \u0936\u0940\u0930\u094d\u0937 \u0915\u094b\u0923 \u03b8 \u0939\u0948 \u0964<\/p>\n\n\n\n

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

$\\tan \\theta =\\frac{AB}{OA}$<\/p>\n\n\n\n

$\\tan \\theta =\\frac{AB}{x}$<\/p>\n\n\n\n

$\\cos \\theta =\\frac{OA}{OB}$<\/p>\n\n\n\n

$\\cos \\theta =\\frac{x}{OB}$<\/p>\n\n\n\n

$OB=\\frac{x}{\\cos \\theta}$<\/p>\n\n\n\n

OB=xsec\u03b8<\/p>\n\n\n\n

\u090a\u0901\u091a\u093e\u0908 \u092e\u0947 \u0935\u0943\u0926\u094d\u0927\u093f , \u1e9fx=x \u0915\u093e 2%$=x \\times \\frac{2}{100}$=0.02%<\/p>\n\n\n\n

\u0936\u0902\u0915\u0941 \u0915\u093e \u0906\u092f\u0924\u0928 , $v=\\frac{1}{3} \\pi (x\\tan \\theta )^2.x$
$=\\frac{1}{3} \\pi x^3 \\tan^2 \\theta$<\/p>\n\n\n\n

Differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{dv}{dx}=\\pi x^2. \\tan ^2 \\theta$<\/p>\n\n\n\n

\u0906\u092f\u0924\u0928 \u092e\u0947 \u0935\u0943\u0926\u094d\u0927\u093f , $\\delta v=\\frac{dv}{dx}\\times \\delta x$
=\u03c0x2<\/sup>tan2<\/sup>\u03b8\u00d70.02x<\/p>\n\n\n\n

=(tan\u03b8sec\u03b8+tan2<\/sup>\u03b8)0.04\u03c0x2<\/sup>
=0.02\u03c0x3<\/sup>tan2<\/sup>\u03b8<\/p>\n\n\n\n

\u0906\u092f\u0924\u0928 \u0915\u093e \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092a\u094d\u0930\u0924\u093f\u0936\u0924 \u0935\u0943\u0926\u094d\u0927\u093f $=\\frac{\\delta v}{v} \\times 100$<\/p>\n\n\n\n

$=\\dfrac{0.02 \\pi x^3 \\tan^2 \\theta}{ \\frac{1}{3} \\pi x^3 \\tan ^2 \\theta} \\times 100$<\/p>\n\n\n\n

=0.06\u00d7100%
=6%<\/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 \u0905\u0935\u0915\u0932 \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0924\u0947 \u0939\u0941\u090f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u093e \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092e\u093e\u0928 \u0926\u0936\u092e\u0932\u0935 \u0915\u0947 3 \u0938\u094d\u0925\u093e\u0928\u094b \u0924\u0915 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964[Using differentials, find the approximate value of each of the following upto 3 places of decimal](i) $\\sqrt{36.6}$Sol : \u092e\u093e\u0928\u093e f(x)=\u221ax Differentiating w.r.t x $f'(x)=\\frac{1}{2\\sqrt{x}}$ 36 \u0915\u093e \u0935\u0930\u094d\u0917\u092e\u0942\u0932 6 \u0939\u094b\u0924\u093e \u0939\u0948 \u0924\u0925\u093e 36.6 \u0915\u0947 \u0928\u093f\u0915\u091f\u0924\u092e \u0939\u0948 \u0964 x=36 […]<\/p>\n","protected":false},"author":302,"featured_media":626466,"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-626457","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-12","entry"],"yoast_head":"\nKC Sinha: Exercise 17.1- Mathematics Solution Class 12 Chapter 17 \u0938\u0928\u094d\u0928\u093f\u0915\u091f\u0928 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1 \u0905\u0935\u0915\u0932 \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0924\u0947 \u0939\u0941\u090f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u093e \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092e\u093e\u0928 \u0926\u0936\u092e\u0932\u0935 \u0915\u0947 3 \u0938\u094d\u0925\u093e\u0928\u094b \u0924\u0915 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964(i) $sqrt{36.6}$Sol : \u092e\u093e\u0928\u093e f(x)=\u221ax\" \/>\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-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 17.1- Mathematics Solution Class 12 Chapter 17 \u0938\u0928\u094d\u0928\u093f\u0915\u091f\u0928\" \/>\n<meta property=\"og:description\" content=\"Question 1 \u0905\u0935\u0915\u0932 \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0924\u0947 \u0939\u0941\u090f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u093e \u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092e\u093e\u0928 \u0926\u0936\u092e\u0932\u0935 \u0915\u0947 3 \u0938\u094d\u0925\u093e\u0928\u094b \u0924\u0915 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964(i) $sqrt{36.6}$Sol : \u092e\u093e\u0928\u093e f(x)=\u221ax\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/\" \/>\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=\"2023-09-12T02:22:38+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2023-09-12T02:22:50+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-44-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=\"7 minutes\" \/>\n<script type=\"application\/ld+json\" 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\u0938\u0928\u094d\u0928\u093f\u0915\u091f\u0928\",\"datePublished\":\"2023-09-12T02:22:38+00:00\",\"dateModified\":\"2023-09-12T02:22:50+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/\"},\"wordCount\":1275,\"publisher\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/#organization\"},\"image\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-44-scaled.jpg\",\"articleSection\":[\"Class 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\u0915\u0947 3 \u0938\u094d\u0925\u093e\u0928\u094b \u0924\u0915 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964(i) $\\\\sqrt{36.6}$Sol : \u092e\u093e\u0928\u093e f(x)=\u221ax\",\"breadcrumb\":{\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/#primaryimage\",\"url\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-44-scaled.jpg\",\"contentUrl\":\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-44-scaled.jpg\",\"width\":1600,\"height\":901,\"caption\":\"KC Sinha: Exercise 17.1- Mathematics Solution Class 12 Chapter 17 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: \u092e\u093e\u0928\u093e f(x)=\u221ax","og_url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/","og_site_name":"IndCareer Schools","article_publisher":"https:\/\/www.facebook.com\/indcareer","article_published_time":"2023-09-12T02:22:38+00:00","article_modified_time":"2023-09-12T02:22:50+00:00","og_image":[{"width":1600,"height":901,"url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-44-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":"7 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\u0938\u0928\u094d\u0928\u093f\u0915\u091f \u092e\u093e\u0928 \u0926\u0936\u092e\u0932\u0935 \u0915\u0947 3 \u0938\u094d\u0925\u093e\u0928\u094b \u0924\u0915 \u0928\u093f\u0915\u093e\u0932\u0947 \u0964(i) $\\sqrt{36.6}$Sol : \u092e\u093e\u0928\u093e f(x)=\u221ax","breadcrumb":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-17-1-mathematics-solution-class-12-chapter-17-sannikatan\/#primaryimage","url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-44-scaled.jpg","contentUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-44-scaled.jpg","width":1600,"height":901,"caption":"KC Sinha: Exercise 17.1- Mathematics Solution Class 12 Chapter 17 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