{"id":626453,"date":"2023-09-12T02:16:32","date_gmt":"2023-09-12T02:16:32","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626453"},"modified":"2023-09-12T02:16:53","modified_gmt":"2023-09-12T02:16:53","slug":"kc-sinha-exercise-15-1-mathematics-solution-class-12-chapter-15-vardhaman-aur-hasman","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-15-1-mathematics-solution-class-12-chapter-15-vardhaman-aur-hasman\/","title":{"rendered":"KC Sinha: Exercise 15.1- Mathematics Solution Class 12 Chapter 15 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0914\u0930 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928"},"content":{"rendered":"\n

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

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

\u0905\u0928\u094d\u0924\u0930\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u092e\u0947 \u0928\u093f\u092e\u094d\u0930\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092f\u093e \u0939\u093e\u0938\u092e\u093e\u0928 \u0939\u0947
[Find the intervals in which the following functions are increasing or decreasing]
(i)<\/strong> $f(x)=10-6 x-2 x^{2}$
(ii)<\/strong> $f(x)=2 x^{3}-12 x^{2}+18 x+15$
(iii)<\/strong> f<\/em>(x)=5+36x+3x2<\/sup>-2x3<\/sup>
(iv) <\/strong>f<\/em>(x)=8+36+3x2<\/sup>-2x3<\/sup>
(v)  <\/strong>f<\/em>(x)=5x3<\/sup>-15x2<\/sup>-120x+3
(vi) <\/strong>f<\/em>(x)=x3<\/sup>-6x2<\/sup>-36x+2
(xi)<\/strong> f<\/em>(x)=2x3<\/sup>-24x+107
(xii)<\/strong> f<\/em>(x)=(x-1)(x-2)2<\/sup>
(xv)<\/strong> f<\/em>(x)=[x(x-2)]2<\/sup>
Sol :
(i) <\/strong>$f(x)=10-6 x-2 x^{2}$
Sol :<\/p>\n\n\n\n

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

$f^{\\prime}(x)=-6-4 x$<\/p>\n\n\n\n

=-2(3+2 x)<\/p>\n\n\n\n

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

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

$f^{\\prime}(x)$ \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e:<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 $\\left(-\\infty,-\\frac{3}{2}\\right)$ , \u092e\u0947 f <\/em>‘(x)>0<\/p>\n\n\n\n

\u2234f<\/em>(x) , \u0905\u0902\u0924\u0930\u093e\u0932 $\\left(-\\infty,-\\frac{3}{2}\\right)$ \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 $\\left(-\\frac{3}{2}, \\infty\\right)$ ,\u092e\u0947 , f <\/em>‘(x)<0<\/p>\n\n\n\n

\u2234f<\/em>(x) , \u0905\u0902\u0924\u0930\u093e\u0932 $\\left(-\\frac{3}{2}, \\infty\\right)$ \u092e\u0947 \u0939\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii)<\/strong> $f(x)=2 x^{3}-12 x^{2}+18 x+15$<\/p>\n\n\n\n

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

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

f<\/em> ‘(x)=6x2<\/sup>-24x+18<\/p>\n\n\n\n

=6[x2<\/sup>-4x+3]<\/p>\n\n\n\n

=6[x2<\/sup>-3x-x+3]<\/p>\n\n\n\n

=6[x(x-3)-1(x-3)]<\/p>\n\n\n\n

f<\/em> ‘(x)=6(x-1)(x-3)<\/p>\n\n\n\n

f<\/em> ‘(x)=0
6(x-1)(x-3)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}x-1=0&x-3=0\\\\x=1&x=3\\end{array}$<\/p>\n\n\n\n

$f^{\\prime}(x)$ \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e:<\/p>\n\n\n\n

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

(i) \u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,1) \u0914\u0930 (3,\u221e) \u092e\u0947 f<\/em>‘(c)>0<\/p>\n\n\n\n

\u2234f<\/em>(x) , (-\u221e,1)\u222a(3,\u221e) \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (1,3) \u092e\u0947  f<\/em>‘(x)<0<\/p>\n\n\n\n

\u2234f<\/em>(x) , (1,3) \u092e\u0947 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(iv)<\/strong> f<\/em>(x)=8+36+3x2<\/sup>-2x3<\/sup>
Sol :<\/p>\n\n\n\n

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

f <\/em>‘(x)=36+6x-6x2<\/sup><\/p>\n\n\n\n

=-6x2<\/sup>+6x+36<\/p>\n\n\n\n

=-6[x2<\/sup>-x-6]<\/p>\n\n\n\n

=-6[x2<\/sup>-3x+2x-6]<\/p>\n\n\n\n

=-6[x(x-3)+2(x-3)]<\/p>\n\n\n\n

f<\/em> ‘(x)=-6(x+2)(x-3)<\/p>\n\n\n\n

\u2234f<\/em>‘(x)=0<\/p>\n\n\n\n

-6(x+2)(x-3)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}x+2=0&x-3=0 \\\\ x=-2&x=3\\end{array}$<\/p>\n\n\n\n

f <\/em>‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 (-2,3) \u092e\u0947 f<\/em>‘(x)>0
\u2234f<\/em>(x) , (-2,3) \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,-2) \u0914\u0930 (3,\u221e) \u092e\u0947 f<\/em> ‘(x)<0
\u2234f<\/em>(x) , (-\u221e,-2)\u222a(3,\u221e) \u092e\u0947 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(xii)<\/strong> f<\/em>(x)=(x-1)(x-2)2<\/sup>
Sol :<\/p>\n\n\n\n

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

f<\/em> ‘(x)=1.(x-2)2<\/sup>+(x-1).2(x-2).1<\/p>\n\n\n\n

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

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

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

\u2234f<\/em> ‘(x)=0
(x-2)(3x-4)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}x-2=0 & 3 x-4=0\\\\x=2 & 3 x=4\\\\&x=\\frac{4}{3}\\end{array}$<\/p>\n\n\n\n

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 $\\left(-\\infty , \\frac{4}{3}\\right)$ \u0914\u0930 (2,\u221e) \u092e\u0947 , f<\/em>‘(x)>0
\u2234 f<\/em>(x) , $\\left(-\\infty , \\frac{4}{3}\\right)$\u22c3(2,\u221e) \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932  $\\left(\\frac{4}{3},2\\right)$ \u092e\u0947 ,f <\/em>‘(x)<0
\u2234 f<\/em>(x) , $\\left(\\frac{4}{3},2\\right)$ \u092e\u0947 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(xv)<\/strong> f<\/em>(x)=[x(x-2)]2<\/sup>
Sol :
f<\/em>(x)=x2<\/sup>(x-2)2<\/sup><\/p>\n\n\n\n

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

f<\/em> ‘(x)=2x(x-2)2<\/sup>+x2<\/sup>.2(x-2).1
=2x(x-2)(x-2+x)
=2x(x-2)(2x-2)
=4x(x-2)(x-1)<\/p>\n\n\n\n

\u2234f<\/em> ‘(x)=0
4x(x-2)(x-1)=0<\/p>\n\n\n\n

$\\begin{array}{l|l|l}4x=0&x-2=0&x-1=0\\\\x=0&x=2&x=1\\end{array}$
f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 (0,1) \u0914\u0930 (2,\u221e) \u092e\u0947 f<\/em> ‘(x)>0
\u2234 f<\/em>(x) (0,1)\u22c3(2,\u221e) \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,0) \u0914\u0930 (1,2) \u092e\u0947 f<\/em> ‘(x)<0
\u2234 f<\/em>(x) , (-\u221e,0)\u22c3(1,2) \u092e\u0947 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

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

\u0905\u0928\u094d\u0924\u0930\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0932\u092e\u0947\u0902 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928 \u0928\u093f\u0930\u0902\u0924\u0930 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092f\u093e \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u0939\u0948 \u0964
[Find the intervals in which the following functions are strictly increasing or decreasing]
(i)<\/strong> f<\/em>(x)=2x3<\/sup>-3x2<\/sup>-36x+7
(ii)<\/strong> f<\/em>(x)=2x3<\/sup>-9x2<\/sup>-24x-5
(xii)<\/strong> f<\/em>(x)=(x+1)3<\/sup>(x-3)3<\/sup>
Sol :
(i)<\/strong>
 f<\/em>(x)=2x3<\/sup>-3x2<\/sup>-36x+7<\/p>\n\n\n\n

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

 f <\/em>‘(x)=6x2<\/sup>-6x-36
=6[x2<\/sup>-x-6]
=6[x2<\/sup>-3x+2x-6]
=6[x(x-3)+2(x-3)]<\/p>\n\n\n\n

f <\/em>‘(x)=6(x+2)(x-3)<\/p>\n\n\n\n

\u2234f <\/em>‘(x)=0
6(x+2)(x-3)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}x+2=0&x-3=0\\\\x=-2&x=3\\end{array}$<\/p>\n\n\n\n

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932, (-\u221e,-2) \u0924\u0925\u093e (3,\u221e) , f<\/em> ‘(x)>0
\u2234f<\/em>(x) , (-\u221e,-2)\u22c3(3,\u221e) \u092e\u0947 \u0928\u093f\u0930\u0902\u0924\u0930 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (-2,3) \u092e\u0947 , f<\/em> ‘(x)<0
\u2234f<\/em>(x) , (-2,3) \u092e\u0947 \u0928\u093f\u0930\u0902\u0924\u0930 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(xii)<\/strong>
f<\/em>(x)=(x+1)3<\/sup>(x-3)3<\/sup><\/p>\n\n\n\n

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

f <\/em>‘(x)=3(x+1)2<\/sup>.1(x-3)3<\/sup>+(x+1)3<\/sup>.3(x-3)2<\/sup>.1
=3(x+1)2<\/sup>.(x-3)2<\/sup>(x-3+x+1)
=3(x+1)2<\/sup>.(x-3)2<\/sup>(2x-2)
=6(x+1)2<\/sup>.(x-3)2<\/sup>(x-1)<\/p>\n\n\n\n

\u2234f <\/em>‘(x)=0
6(x+1)2<\/sup>(x-3)2<\/sup>(x-1)=0<\/p>\n\n\n\n

[(x+1)2<\/sup>\u22650
(x-3)2<\/sup>\u22650]<\/p>\n\n\n\n

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

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

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 (1,\u221e) \u092e\u0947 , f <\/em>‘(x)>0
\u2234 f<\/em>(x) , (1,\u221e) \u092e\u0947 , f <\/em>‘(x)<0<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,1) \u092e\u0947 \u0928\u093f\u0930\u0902\u0924\u0930 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

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

\u0905\u0928\u094d\u0924\u0930\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u092e\u0947 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092f\u093e \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u0939\u0948 \u0964
[Find the intervals in which the following functions are increasing or decreasing]
(i)<\/strong> f<\/em>(x)=2x3<\/sup>-24x+7
(ii)<\/strong> f<\/em>(x)=x4<\/sup>-2x2<\/sup>
(iii)<\/strong> f<\/em>(x)=x4<\/sup>-8x3<\/sup>+22x2<\/sup>-24x+21
(iv)<\/strong> $f=x^4-\\frac{x^3}{3}$
(v)<\/strong> $f(x)=\\frac{x}{2}+\\frac{2}{x},$x\u22600
(vi)<\/strong> $f(x)=\\frac{x}{x^2+1}$
(vii)<\/strong> $f(x)=\\frac{x-2}{x+1}$,x\u2260-1
Sol :
(i)<\/strong> f<\/em>(i)=2x3<\/sup>-24x+7
Differentiating w.r.t x<\/p>\n\n\n\n

f<\/em> ‘(x)=6x2<\/sup>-24
=6(x2<\/sup>-4)
=6(x2<\/sup>-22<\/sup>)
f<\/em> ‘(x)=6(x-2)(x+2)<\/p>\n\n\n\n

\u2234f<\/em> ‘(x)=0
6(x-2)(x+2)=0<\/p>\n\n\n\n

$\\begin{array}{l|l}x-2=0&x+2=0\\\\x=2&x=-2\\end{array}$<\/p>\n\n\n\n

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,-2) \u0914\u0930 (2,\u221e) \u092e\u0947 f <\/em>‘(x)>0
\u2234 f<\/em>(x) , (-\u221e,-2)\u22c3(2,\u221e) \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (-2,2) \u092e\u0947\u0902 , f<\/em> ‘(x)<0
\u2234 f<\/em>(x) , (-2,2) \u092e\u0947 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii)<\/strong> f<\/em>(x)=x4<\/sup>-2x2<\/sup>
Sol :<\/p>\n\n\n\n

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

f<\/em> ‘(x)=4x3<\/sup>-4x
=4x(x2<\/sup>-1)
=4x(x2<\/sup>-12<\/sup>)<\/p>\n\n\n\n

f<\/em> ‘(x)=4x(x+1)(x-1)
\u2234f<\/em> ‘(x)=0<\/p>\n\n\n\n

4x(x+1)(x-1)=0<\/p>\n\n\n\n

$\\begin{array}{l|l|l}4x=0&x+1&x-1=0\\\\x=0&x=-1&x=1\\end{array}$<\/p>\n\n\n\n

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 (-1,0) \u0924\u0925\u093e (1,\u221e) \u092e\u0947 , f<\/em> ‘(x)>0
\u2234f<\/em>(x) , (-1,0) \u0924\u0925\u093e (0,1) \u092e\u0947 , f<\/em> ‘(x)<0<\/p>\n\n\n\n

(ii)\u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,-1) \u0924\u0925\u093e (0,1) \u092e\u0947 , f<\/em> ‘(x)<0
\u2234f<\/em>(x), (-\u221e,-1)\u22c3(0,1) \u092e\u0947 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(iii)<\/strong> f<\/em>(x)=x4<\/sup>-8x3<\/sup>+22x2<\/sup>-24x+21<\/p>\n\n\n\n

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

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

f <\/em>‘(x)=4x3<\/sup>-24x2<\/sup>+44x-24<\/p>\n\n\n\n

=4[x3<\/sup>-6x2<\/sup>+11x-6]<\/p>\n\n\n\n

=4[x3<\/sup>-x2<\/sup>-5x2<\/sup>+5x+6x-6]<\/p>\n\n\n\n

=4[x2<\/sup>(x-1)-5x(x-1)+6(x-1)]<\/p>\n\n\n\n

=4(x-1)(x2<\/sup>-5x+6)
=4(x-1)(x2<\/sup>-3x-2x+6)<\/p>\n\n\n\n

=4(x-1)(x(x-3)-2(x-3))<\/p>\n\n\n\n

f <\/em>‘(x)=4(x-1)(x-2)(x-3)<\/p>\n\n\n\n

\u2234f <\/em>‘(x)=0
4(x-1)(x-2)(x-3)=0<\/p>\n\n\n\n

$\\begin{array}{l|l|l}x-1=0&x-2=0&x-3=0\\\\x=1&x=2&x=3\\end{array}$<\/p>\n\n\n\n

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 (1,2) \u0914\u0931 (3,\u221e) \u092e\u0947 , f<\/em> ‘(x)>0
\u2234f<\/em>(x) , (1,2)\u22c3(3,\u221e) \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,1) \u0914\u0931 (2,3) \u092e\u0947 , f<\/em> ‘(x)<0
\u2234f<\/em>(x) , (-\u221e,1)\u22c3(2,3) \u092e\u0947 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(iv)<\/strong> $f=x^4-\\frac{x^3}{3}$
Sol :
Differentiating w.r.t x<\/p>\n\n\n\n

f<\/em> ‘(x)=4x3<\/sup>$-\\frac{3x^2}{3}$<\/p>\n\n\n\n

f<\/em> ‘(x)=x2<\/sup>(4x-1)<\/p>\n\n\n\n

\u2234f<\/em> ‘(x)=0<\/p>\n\n\n\n

x2<\/sup>(4x-1)=0 [\u2235x2<\/sup>\u22650]<\/p>\n\n\n\n

4x-1=0<\/p>\n\n\n\n

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

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 ($\\frac{1}{4}$,\u221e) \u092e\u0947 ,  f<\/em> ‘(x)>0
\u2234f<\/em>(x) , ($\\frac{1}{4}$,\u221e) \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,$\\frac{1}{4}$) \u092e\u0947 , f<\/em> ‘(x)<0
\u2234f<\/em>(x) , (-\u221e,$\\frac{1}{4}$) \u092e\u0947 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(v)<\/strong> $f(x)=\\frac{x}{2}+\\frac{2}{x},$x\u22600
Sol :
Differentiating w.r.t x<\/p>\n\n\n\n

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

f<\/em> ‘(x) $=\\frac{x^2-4}{2x^2}$<\/p>\n\n\n\n

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

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

\u2234f<\/em> ‘(x)<\/p>\n\n\n\n

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

$\\begin{array}{l|l|l}x\\neq 0&x-2=0&x+2=0\\\\&x=2&x=-2\\end{array}$<\/p>\n\n\n\n

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,-2) \u0914\u0930 (2,\u221e) \u092e\u0947 , f <\/em>‘(x)>0
\u2234 f<\/em>(x) , (-\u221e,-2)\u22c3(2,\u221e) \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (-2,0)\u22c3(0,2) , \u092e\u0947 f <\/em>‘(x)<0
\u2234f<\/em>(x) , (-2,0)\u22c3(0,2) \u092e\u0947 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(vi)<\/strong> $f(x)=\\frac{x}{x^2+1}$
Sol :
Differentiating w.r.t x<\/p>\n\n\n\n

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

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

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

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

\u2234f<\/em> ‘(x)=0<\/p>\n\n\n\n

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

[$(x^2+1^2)^2 \\geq 0$]<\/p>\n\n\n\n

-x2<\/sup>+1=0<\/p>\n\n\n\n

x=\u00b1\u221a1<\/p>\n\n\n\n

x=\u00b11<\/p>\n\n\n\n

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 (-1,1) \u092e\u0947\u0902 f<\/em> ‘(x)>0
\u2234 f<\/em>(x) , (-1,1) \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,1) \u0924\u0925\u093e (1,\u221e) \u092e\u0947\u0902 f<\/em> ‘(x)<0
\u2234 f<\/em>(x) , (-\u221e,1)\u22c3(1,\u221e) \u092e\u0947\u0902 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(vii)<\/strong> $f(x)=\\frac{x-2}{x+1}$,x\u2260-1
Sol :
Differentiating w.r.t x<\/p>\n\n\n\n

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

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

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

\u2235x\u2260-1<\/p>\n\n\n\n

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,-1) \u0914\u0930 (-1,\u221e) \u092e\u0947\u0902 f<\/em> ‘(x)>0
\u2234 f<\/em>(x) , (-\u221e,-1)\u22c3(-1,\u221e) \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

(ii) \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0915\u0947 \u0932\u093f\u090f \u0915\u094b\u0908 \u0905\u0902\u0924\u0930\u093e\u0932 \u0928\u0939\u0940 \u0939\u0948 \u0964<\/p>\n\n\n\n

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

\u092e\u093e\u0928\u093e \u0915\u093f 1 , \u0905\u0928\u094d\u0924\u0930\u093e\u0932 \u0939\u0948 \u091c\u094b (-1,1) \u0938\u0947 \u0905\u0938\u0902\u092f\u0941\u0915\u094d\u0924 \u0939\u0948 \u0964 \u0926\u093f\u0916\u093e\u0901\u090f \u0915\u093f \u092b\u0932\u0928 f<\/em> \u091c\u094b $f(x)=x+\\frac{1}{x}$ \u0938\u0947 \u092a\u094d\u0930\u0926\u0924\u094d\u0924 \u0939\u0948 , I \u092e\u0947 \u0928\u093f\u0930\u0902\u0924\u0930 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0939\u0948 \u0964
Sol :
$f(x)=x+\\frac{1}{x}$<\/p>\n\n\n\n

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

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

\u2234f<\/em> ‘(x)=0<\/p>\n\n\n\n

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

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

x2<\/sup>=1<\/p>\n\n\n\n

x=\u00b1\u221a1<\/p>\n\n\n\n

x=\u00b11<\/p>\n\n\n\n

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

\u0905\u0902\u0924\u0930\u093e\u0932 (-\u221e,-1) \u0914\u0931 (1,\u221e) \u092e\u0947\u0902 f<\/em> ‘(x)>0<\/p>\n\n\n\n

\u2234f<\/em>(x) \u0905\u0902\u0924\u0930\u093e\u0932 I\u2208R-(-1,1) \u092e\u0947\u0902 \u0928\u093f\u0930\u0902\u0924\u0930 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

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

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f f<\/em>(x)=7x-3 \u0926\u094d\u0935\u093e\u0930\u093e \u092a\u094d\u0930\u0926\u0924\u094d\u0924 \u092b\u0932\u0928, R \u092e\u0947 \u0928\u093f\u0930\u0902\u0924\u0930 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0939\u0948 \u0964
[Show that the function given by f<\/em>(x)=7x-3 is strictly increasing on R]
Sol :
f<\/em>(x)=7x-3<\/p>\n\n\n\n

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

f<\/em> ‘(x)=7<\/p>\n\n\n\n

x\u2208R , f<\/em> ‘(x)>0<\/p>\n\n\n\n

\u2234f<\/em>(x) , R \u092e\u0947\u0902 \u0928\u093f\u0930\u0902\u0924\u0930 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

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

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f f<\/em>(x)=x3<\/sup>-6x2<\/sup>+12x-18 , R \u092e\u0947\u0902 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964
[Show that f<\/em>(x)=x3<\/sup>-6x2<\/sup>+12x-18 is an increasing function on R]
Sol :
f<\/em>(x)=x3<\/sup>-6x2<\/sup>+12x-18<\/p>\n\n\n\n

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

f<\/em> ‘(x)=3x2<\/sup>-12x+12
=3(x2<\/sup>-4x+4)
=3(x2<\/sup>-2.x.2+22<\/sup>)
=3(x-2)2<\/sup><\/p>\n\n\n\n

\u0905\u092c , x\u2208R\u21d2(x-2)2<\/sup>\u22650
\u21d23(x-2)2<\/sup>\u22650
\u21d2f<\/em> ‘(x)\u22650<\/p>\n\n\n\n

\u2234f<\/em>(x) is an increasing function on R<\/p>\n\n\n\n

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

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f f<\/em>(x)=x3<\/sup>-15x2<\/sup>+75x-50 , R \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964
[Show that f<\/em>(x)=x3<\/sup>-15x2<\/sup>+75x-50 is an increasing function in R]
Sol :
f<\/em>(x)=x3<\/sup>-15x2<\/sup>+75x-50<\/p>\n\n\n\n

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

f<\/em> ‘(x)=3x2<\/sup>-30x+75
=3(x2<\/sup>-10x+25)
=3(x2<\/sup>-2.x.5+52<\/sup>)
=3(x-5)2<\/sup><\/p>\n\n\n\n

\u0905\u092c , x\u2208R\u21d2(x-5)2<\/sup>\u22650<\/p>\n\n\n\n

\u21d23(x-5)2<\/sup>\u22650
\u21d2f<\/em> ‘(x)\u22650<\/p>\n\n\n\n

\u2234f<\/em>(x) is an increasing function on R<\/p>\n\n\n\n

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

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f \u092b\u0932\u0928 f<\/em> \u091c\u094b f<\/em>(x)=x3<\/sup>-3x2<\/sup>+3x , x\u2208R \u0938\u0947 \u092a\u094d\u0930\u0926\u0924\u094d\u0924 \u0939\u0948 , R \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964
[Show that the function f<\/em> given by f<\/em>(x)=x3<\/sup>-3x2<\/sup>+3x , x\u2208R is increasing on R]
Sol :
f<\/em>(x)=x3<\/sup>-3x2<\/sup>+3x<\/p>\n\n\n\n

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

f<\/em> ‘(x)=3x2<\/sup>-6x+3
=3(x2<\/sup>-2x+1)
=3(x2<\/sup>-2.x.1+12<\/sup>)
=3(x-1)2<\/sup><\/p>\n\n\n\n

\u0905\u092c , x\u2208R\u21d2(x-1)2<\/sup>\u22650<\/p>\n\n\n\n

\u21d23(x-1)2<\/sup>\u22650<\/p>\n\n\n\n

\u21d2f<\/em> ‘(x)\u22650<\/p>\n\n\n\n

\u2234f<\/em>(x) is an increasing function on R<\/p>\n\n\n\n

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

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928 R \u092e\u0947 \u0928\u093f\u0930\u0902\u0924\u0930 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0939\u0948 \u0964
[Show that the following functions are strictly increasing on R]
(i)<\/strong> f<\/em>(x)=3x+17
Sol :
Differentiating w.r.t x<\/p>\n\n\n\n

f<\/em> ‘(x)=3>0 \u2200x\u2208R and f<\/em>(x) is one-one function<\/p>\n\n\n\n

\u2234 f<\/em>(x) is strictly increasing function on R<\/p>\n\n\n\n

(ii)<\/strong> f<\/em>(x)=e2x<\/sup>
Sol :
Differentiating w.r.t x<\/p>\n\n\n\n

f<\/em> ‘(x)=2e2x<\/sup>>0 , \u2200x\u2208R<\/p>\n\n\n\n

and f<\/em>(x) is one-one function.<\/p>\n\n\n\n

\u2234f<\/em>(x) is strictly increasing function on R<\/p>\n\n\n\n

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

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f f<\/em>(x)=cosx(0,\u03c0) \u092e\u0947\u0902 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0938\u093e\u0925 \u0939\u0940 \u092f\u0939 \u092d\u0940 \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f
[Show that f<\/em>(x)=cosx is decreasing function on (0,\u03c0). Also show that]
(i)<\/strong> (0,\u03c0) \u092e\u0947\u0902 f<\/em> \u0928\u093f\u0930\u0902\u0924\u0930 \u0939\u094d\u0930\u0938\u092e\u093e\u0928 \u0939\u0948 \u0964
[f<\/em> is strictly decreasing on (0,\u03c0)]
(ii)<\/strong> (\u03c0,2\u03c0) \u092e\u0947\u0902 f<\/em> \u0928\u093f\u0930\u0902\u0924\u0930 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0939\u0948 \u0964
[f<\/em> is increasing in (\u03c0,2\u03c0)]
Sol :
f<\/em>(x)=cosx<\/p>\n\n\n\n

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

f<\/em> ‘(x)=-sinx<\/p>\n\n\n\n

\u092f\u0926\u093f x\u2208(0,\u03c0)
\u21d2-sinx<0
\u21d2f<\/em> ‘(x)<0<\/p>\n\n\n\n

\u2234f<\/em>(x) is decreasing function on (0,\u03c0)<\/p>\n\n\n\n

(i) f<\/em>(x) is one-one function on (0,\u03c0)
\u2234f<\/em>(x) is strictly decreasing function on (0,\u03c0)<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 (\u03c0,2\u03c0) \u092e\u0947 , f <\/em>‘(x)>0
and f<\/em>(x) is one-one function on (\u03c0,2\u03c0)
\u2234f<\/em>(x) is strictly increasing on (\u03c0,2\u03c0)<\/p>\n\n\n\n

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

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f f<\/em>(x)=sinx \u0938\u0947 \u092a\u094d\u0930\u0926\u0924\u094d\u0924 \u092b\u0932\u0928
[Show that the function given by f<\/em>(x)=sinx]
(i)<\/strong> $\\left(0,\\frac{\\pi}{2}\\right)$ \u092e\u0947 \u0928\u093f\u0930\u0902\u0924\u0930 \u0935\u0930\u094d\u0918\u092e\u093e\u0928 \u0939\u0948 \u0964
[Strictly increasing in $\\left(0,\\frac{\\pi}{2}\\right)$]<\/p>\n\n\n\n

(ii)<\/strong> $\\left(\\frac{\\pi}{2},\\pi \\right)$ \u092e\u0947 \u0928\u093f\u0930\u0902\u0924\u0930 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u0939\u0948 \u0964
[Strictly decreasing in $\\left(\\frac{\\pi}{2},\\pi \\right)$]<\/p>\n\n\n\n

(iii)<\/strong> (0,\u03c0) \u092e\u0947 \u0928 \u0924\u094b \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0939\u0948 \u0914\u0930 \u0928 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u0939\u0948 \u0964
[Neither increasing nor decreasing (0,\u03c0)]
Sol :
f<\/em>(x)=sinx<\/p>\n\n\n\n

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

f <\/em>‘(x)=cosx<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 $\\left(0,\\frac{\\pi}{2}\\right)$ \u092e\u0947 , f<\/em> ‘(x)>0
\u2235 f<\/em>(x) is one-one function on $\\left(0,\\right)$
\u2234f<\/em>(x) is strictly increasing in $\\left(0,\\frac{\\pi}{2}\\right)$<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 $\\left(\\frac{\\pi}{2},\\pi\\right)$ \u092e\u0947\u0902 , f <\/em>‘(x)<0
\u2235f<\/em>(x) is one-one function on $\\left(\\frac{\\pi}{2},\\pi\\right)$
\u2234f<\/em>(x) is strictly decreasing in$\\left(\\frac{\\pi}{2},\\pi\\right)$<\/p>\n\n\n\n

(iii) \u2234 f<\/em>(x) is neither increasing nor decreasing in (0,\u03c0)<\/p>\n\n\n\n

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

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f \u092b\u0932\u0928 f<\/em> \u0915\u094b f<\/em>(x)=log sinx \u0938\u0947 \u092a\u094d\u0930\u0926\u0924\u094d\u0924 \u0939\u0948 $\\left(0,\\frac{\\pi}{2}\\right)$ \u092e\u0947 \u0928\u093f\u0930\u0902\u0924\u0930 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0939\u0948 \u0924\u0925\u093e $\\left(\\frac{\\pi}{2},\\pi \\right)$ \u092e\u0947\u0902 \u0928\u093f\u0930\u0902\u0924\u0930 \u0939\u094d\u0930\u093e\u0938\u092e\u0928 \u0939\u0948 \u0964
Sol :
f<\/em>(x)=log(sinx)<\/p>\n\n\n\n

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

$f'(x)=\\frac{1}{\\sin }\\times \\cos x$<\/p>\n\n\n\n

f <\/em>‘(x)=cot x<\/p>\n\n\n\n

(i) \u0905\u0924\u0902\u0930\u093e\u0932 $\\left(0,\\frac{\\pi}{2}\\right)$ \u092e\u0947\u0902  , f<\/em> ‘(x)>0
\u2235 f<\/em>(x) is one-one function in $\\left(0,\\frac{\\pi}{2}\\right)$
\u2234f<\/em>(x) is strictly increasing in $\\left(0,\\frac{\\pi}{2}\\right)$<\/p>\n\n\n\n

(ii) \u0905\u0924\u0902\u0930\u093e\u0932 $\\left(\\frac{\\pi}{2},\\pi\\right)$ \u092e\u0947\u0902 , f<\/em> ‘(x)<0
\u2235 f<\/em>(x) is one-one function in $\\left(\\frac{\\pi}{2},\\pi \\right)$
\u2234f<\/em>(x) is strictly decreasing in $\\left(\\frac{\\pi}{2}, \\pi\\right)$<\/p>\n\n\n\n

Question 13<\/h4>\n\n\n\n

\u0935\u0939 \u0905\u0928\u094d\u0924\u0930\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u092e\u0947\u0902 \u092b\u0932\u0928 f<\/em>  \u091c\u094b \u0928\u093f\u092e\u094d\u0928 \u092a\u094d\u0930\u0915\u093e\u0930 \u092a\u094d\u0930\u0926\u0924\u094d\u0924 \u0939\u0948 \u0964
[Find the intervals in which the function f<\/em> given by]
f<\/em>(x)=sinx+cosx , 0\u2264x\u2264\u03c0<\/p>\n\n\n\n

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

f<\/em> ‘(x)=cosx-sinx<\/p>\n\n\n\n

\u2234f<\/em> ‘(x)=0<\/p>\n\n\n\n

cosx-sinx=0<\/p>\n\n\n\n

-sinx=-cosx<\/p>\n\n\n\n

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

$\\tan x=\\frac{\\pi}{4}$<\/p>\n\n\n

[tan\u03b8=tan\u03b1
\n\u03b8=n\u03c0+\u03b1 , n\u2208z]<\/p>\n\n\n\n

\u2234$x=n\\pi+\\frac{\\pi}{4}$ , n\u2208z<\/p>\n\n\n\n

\u22350\u2264x\u22642\u03c0<\/p>\n\n\n\n

$x=\\frac{\\pi}{4}, \\frac{5\\pi}{4}$<\/p>\n\n\n\n

f<\/em> ‘(x) \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e<\/p>\n\n\n\n

Diagram<\/p>\n\n\n\n

(i) \u0905\u0902\u0924\u0930\u093e\u0932 $\\left(0,\\frac{\\pi}{4} \\right)$ \u0914\u0930 $\\left(\\frac{5\\pi}{4},2\\pi\\right)$ \u092e\u0947 , f<\/em> ‘(x)>0
\u2234f<\/em>(x) is strictly increasing in $\\left[0,\\frac{\\pi}{4} \\right)$\u22c3$\\left(\\frac{5\\pi}{4},2\\pi\\right]$<\/p>\n\n\n\n

(ii) \u0905\u0902\u0924\u0930\u093e\u0932 $\\left(\\frac{\\pi}{4},\\frac{5\\pi}{4}\\right)$ \u092e\u0947\u0902 , f <\/em>‘(x)<0
\u2234f<\/em>(x) is strictly decreasing in $\\left(\\frac{\\pi}{4},\\frac{5\\pi}{4}\\right)$<\/p>\n\n\n\n

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

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f \u092b\u0932\u0928 $y=\\frac{4\\sin \\theta}{2+\\cos \\theta}-\\theta$ , $\\left[0,\\frac{\\pi}{2}\\right]$ \u092e\u0947 \u03b8 \u0915\u093e \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964
Prove that $y=\\frac{4\\sin \\theta}{2+\\cos \\theta}-\\theta$ is an increasing function of \u03b8 in $\\left[0,\\frac{\\pi}{2}\\right]$
Sol :
$y=\\frac{4\\sin \\theta}{2+\\cos \\theta}-\\theta$<\/p>\n\n\n\n

Differentiating w.r.t \u03b8<\/p>\n\n\n\n

$\\frac{dy}{d \\theta}=\\frac{4\\cos \\theta (2+\\cos \\theta)-4\\sin (-\\sin \\theta)}{(2+\\cos \\theta)^2}-1$<\/p>\n\n\n\n

$=\\frac{8\\cos\\theta+4\\cos^2 \\theta+4\\sin^2 \\theta}{(2+\\cos \\theta)^2}-1$<\/p>\n\n\n\n

$=\\frac{8\\cos \\theta+4(\\cos^2 \\theta +\\sin^2 \\theta)-(2+\\cos \\theta)^2}{(2+\\cos \\theta)^2}$<\/p>\n\n\n\n

$=\\frac{8\\cos \\theta +4(1)-(2^2+2.2.\\cos \\theta +\\cos ^2 \\theta)}{(2+\\cos \\theta)^2}$<\/p>\n\n\n\n

$\\frac{dy}{d\\theta}=\\frac{8\\cos \\theta+4-4\\cos \\theta -\\cos^2 \\theta}{(2+\\cos \\theta)^2}$<\/p>\n\n\n\n

$\\frac{dy}{d\\theta}=\\frac{4\\cos \\theta-\\cos^2\\theta}{(2+\\cos\\theta)^2}$<\/p>\n\n\n\n

$\\frac{dy}{d\\theta}=\\frac{\\cos \\theta (4-\\cos \\theta)}{(2+\\cos \\theta)^2}>0 ,\\forall \\theta \\in \\left(0,\\frac{\\pi}{2}\\right) $<\/p>\n\n\n\n

[\u2235cos\u03b8>0 ,4-cos\u03b8>0
(2+cos\u03b8)2<\/sup>>0]<\/p>\n\n\n\n

\u2234\u092b\u0932\u0928 $y=\\frac{4\\sin \\theta}{2+\\cos \\theta}-\\theta$<\/p>\n\n\n\n

\u0905\u0902\u0924\u0930\u093e\u0932 $\\left[0,\\frac{\\pi}{2}\\right]$ \u092e\u0947 \u090f\u0915 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

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

\u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f \u092b\u0932\u0928 f<\/em>(x)=x+cosx-a \u0915\u0947 \u0938\u092d\u0940 \u092e\u093e\u0928\u094b \u0915\u0947 \u0932\u093f\u090f R \u092a\u0930 \u090f\u0915 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964
[Show that f<\/em>(x)=x+cosx-a is an increasing function on R for all values of a]
Sol :
f<\/em>(x)=x+cosx-a<\/p>\n\n\n\n

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

f<\/em> ‘(x)=1-sinx\u22650 ,\u2200x\u2208R<\/p>\n\n\n\n

\u2234a \u0915\u0947 \u0938\u092d\u0940 \u092e\u093e\u0928\u094b \u0915\u0947 \u0932\u093f\u090f R \u092a\u0930 , f<\/em>(x) \u090f\u0915 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

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

a \u0915\u093e \u0928\u094d\u092f\u0942\u0928\u094d\u0924\u092e \u092e\u093e\u0928 \u0928\u093f\u0915\u093e\u0932\u0947 \u0924\u093e\u0915\u093f \u092b\u0932\u0928 f<\/em>(x)=x2<\/sup>+ax-1 , [1,2] \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0939\u0948 \u0964
[Find the least value of a for which the function f<\/em>(x)=x2<\/sup>+ax-1 is increasing in{1,2}]
Sol :
f<\/em>(x)=x2<\/sup>+ax-1<\/p>\n\n\n\n

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

f<\/em> ‘(x)=2x+a<\/p>\n\n\n\n

\u2235f<\/em>(x) , \u0905\u0902\u0924\u0930\u093e\u0932 [1,2] \u092e\u0947 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092b\u0932\u0928 \u0939\u0948 \u0964<\/p>\n\n\n\n

f<\/em> ‘(x)\u22650 , \u2200\u2208(1,2)<\/p>\n\n\n\n

2x+a\u22650<\/p>\n\n\n\n

\u2235a \u0915\u0947 \u0928\u094d\u092f\u0942\u0928\u094d\u0924\u092e \u092e\u093e\u0928 \u0915\u0947 \u0932\u093f\u090f,<\/p>\n\n\n\n

\u22342x+a \u0915\u093e \u092d\u0940 \u0928\u094d\u092f\u0942\u0928\u0924\u092e \u092e\u093e\u0928 \u0932\u0947\u0928\u093e \u0939\u094b\u0917\u093e \u0964<\/p>\n\n\n\n

\u2234 2\u00d71+a \u22650
2+a\u22650
a\u2265-2<\/p>\n\n\n\n

\u0905\u092d\u093f\u0937\u094d\u091f \u092e\u093e\u0928 : a=-2 [\u2235-2\u2264a\u2264\u221e]<\/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\u0928\u094d\u0924\u0930\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u092e\u0947 \u0928\u093f\u092e\u094d\u0930\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092f\u093e \u0939\u093e\u0938\u092e\u093e\u0928 \u0939\u0947[Find the intervals in which the following functions are increasing or decreasing](i) $f(x)=10-6 x-2 x^{2}$(ii) $f(x)=2 x^{3}-12 x^{2}+18 x+15$(iii) f(x)=5+36x+3×2-2×3(iv) f(x)=8+36+3×2-2×3(v)  f(x)=5×3-15×2-120x+3(vi) f(x)=x3-6×2-36x+2(xi) f(x)=2×3-24x+107(xii) f(x)=(x-1)(x-2)2(xv) f(x)=[x(x-2)]2Sol :(i) $f(x)=10-6 x-2 x^{2}$Sol : Differentiating w.r.t x $f^{\\prime}(x)=-6-4 x$ =-2(3+2 x) 2 x=-3 $x=\\frac{-3}{2}$ $f^{\\prime}(x)$ \u0915\u0940 \u091a\u093f\u0928\u094d\u0939 \u092f\u094b\u091c\u0928\u093e: Diagram (i) \u0905\u0902\u0924\u0930\u093e\u0932 $\\left(-\\infty,-\\frac{3}{2}\\right)$ , \u092e\u0947 f ‘(x)>0 \u2234f(x) […]<\/p>\n","protected":false},"author":302,"featured_media":626459,"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-626453","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-12","entry"],"yoast_head":"\nKC Sinha: Exercise 15.1- Mathematics Solution Class 12 Chapter 15 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0914\u0930 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928 - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Question 1 \u0905\u0928\u094d\u0924\u0930\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 \u091c\u093f\u0938\u092e\u0947 \u0928\u093f\u092e\u094d\u0930\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u092f\u093e \u0939\u093e\u0938\u092e\u093e\u0928 \u0939\u0947(i) $f(x)=10-6 x-2 x^{2}$(ii) $f(x)=2 x^{3}-12 x^{2}+18\" \/>\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-15-1-mathematics-solution-class-12-chapter-15-vardhaman-aur-hasman\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 15.1- Mathematics Solution Class 12 Chapter 15 \u0935\u0930\u094d\u0927\u092e\u093e\u0928 \u0914\u0930 \u0939\u094d\u0930\u093e\u0938\u092e\u093e\u0928 \u092b\u0932\u0928\" \/>\n<meta property=\"og:description\" content=\"Question 1 \u0905\u0928\u094d\u0924\u0930\u093e\u0932 \u0928\u093f\u0915\u093e\u0932\u0947 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