{"id":626444,"date":"2023-09-12T02:03:32","date_gmt":"2023-09-12T02:03:32","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=626444"},"modified":"2023-09-12T02:03:40","modified_gmt":"2023-09-12T02:03:40","slug":"kc-sinha-exercise-12-1-mathematics-solution-class-12-chapter-12-dvitiy-koti-ka-avakalaj","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-12-1-mathematics-solution-class-12-chapter-12-dvitiy-koti-ka-avakalaj\/","title":{"rendered":"KC Sinha: Exercise 12.1- Mathematics Solution Class 12 Chapter 12 \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u0915\u094b\u091f\u093f \u0915\u093e \u0905\u0935\u0915\u0932\u091c"},"content":{"rendered":"\n

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

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928\u094b\u0902 \u0915\u0947 \u0926\u094d\u0935\u093f\u0924\u0940\u092f(\u0915\u094b\u091f\u093f) \u0915\u093e \u0905\u0935\u0915\u0932\u091c \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0902 \u0964
[Find the second order derivatives of the following functions]<\/strong><\/p>\n\n\n\n

(i)<\/strong> log x
Sol :
\u092e\u093e\u0928\u093e y=log x<\/p>\n\n\n\n

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

$\\frac{d y}{dx}=\\frac{1}{x}$<\/p>\n\n\n\n

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

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

(ii)<\/strong> x20<\/sup>
Sol :
\u092e\u093e\u0928\u093e y=x20<\/sup><\/p>\n\n\n\n

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

$\\frac{d y}{dx}=20 x^{19}$<\/p>\n\n\n\n

Again , differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=380 x^{18}$<\/p>\n\n\n\n

(iii)<\/strong> log(log x)
Sol :
\u092e\u093e\u0928\u093e y=log(log x)<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=\\frac{1}{\\log x} \\times \\frac{1}{x}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{1}{x \\log x}$<\/p>\n\n\n\n

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

$\\frac{d^{2}y}{d x^{2}}=\\frac{0 \\cdot x \\log x-1 \\cdot\\left[1 \\cdot \\log {x}+x \\times \\frac{1}{x}\\right]}{(x \\log x)^{2}}$<\/p>\n\n\n\n

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

(iv)<\/strong> x2<\/sup>+3x+2
Sol :
\u092e\u093e\u0928\u093e y=x2<\/sup>+3x+2<\/p>\n\n\n\n

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

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

Again, differentiating w.r.t x<\/p>\n\n\n\n

$\\frac{d^{2} y}{dx^{2}}=2$<\/p>\n\n\n\n

(v)<\/strong> xcosx
Sol :
\u092e\u093e\u0928\u093e y=xcosx<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=1 \\cdot \\cos x+x(-\\sin x)$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\cos x-x \\sin x$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=-\\sin x-[1 \\cdot \\operatorname{sin} x+x \\cos x]$<\/p>\n\n\n\n

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

=-xcosx-2sinx<\/p>\n\n\n\n

(vi)<\/strong> ex<\/sup>sin5x
Sol :
\u092e\u093e\u0928\u093e y=ex<\/sup>sin5x<\/p>\n\n\n\n

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

$\\frac{d y}{dx}=e^{x} \\cdot \\sin 5 x+e^{x} \\cdot \\cos 5x \\times 5$<\/p>\n\n\n\n

$\\frac{d y}{d x}=e^{x} \\sin 5 x+5e ^{2} \\cos x$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=e^{x} \\sin 5 x+e^{x} \\cos 5 x \\times 5+\\left[e^{x} \\cos 5 x+e^{2}(-\\sin 5x) \\times 5\\right]$<\/p>\n\n\n\n

=ex<\/sup>sin5x+5ex<\/sup>cos5x+5ex<\/sup>cos5x-25ex<\/sup>sin5x<\/p>\n\n\n\n

=10ex<\/sup>cos5x-24ex<\/sup>sin5x<\/p>\n\n\n\n

=2ex<\/sup>(5cos5x-12sin5x)<\/p>\n\n\n\n

(vii)<\/strong> sin(log x)
Sol :
\u092e\u093e\u0928\u093e y=sin(log x)<\/p>\n\n\n\n

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

$\\frac{d y}{dx}=\\cos (\\log x) \\times \\frac{1}{x}$<\/p>\n\n\n\n

$\\frac{d y}{dx}=\\frac{\\cos (\\log x)}{x}$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=\\frac{-\\sin (\\log x) \\times \\frac{1}{x} \\times x-\\cos (\\log x) \\times 1}{x^{2}}$<\/p>\n\n\n\n

$=-\\frac{\\sin\\left(\\log x\\right)+\\cos \\left(\\log x\\right)}{x^{2}}$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) y=sin(log x) , \u0938\u093f\u0926\u094d\u0927 \u0915\u0930\u0947 \u0915\u093f (prove that) $x^{2} \\frac{d^{2} y}{d x^{2}}+x \\frac{d y}{d x}+y=0$
<\/strong> Sol :
y=sin(log x)<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=\\cos (\\log x) \\times \\frac{1}{x}$<\/p>\n\n\n\n

$\\frac{dy}{dx}=\\frac{\\cos\\left(\\log x\\right)}{x}$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=\\frac{-\\sin \\left(\\log x\\right) \\times \\frac{1}{x}.x-\\cos \\left(\\log x\\right) \\cdot 1}{x^{2}}$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=-\\frac{[\\sin 4(\\log x)+\\cos (\\log x)]}{x^{2}}$<\/p>\n\n\n\n

$x^{2} \\frac{d^{2} y}{d x^{2}}=-\\sin (\\log x)-\\cos (\\log x)$<\/p>\n\n\n\n

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

$x^{2} \\frac{d^{2} y}{d x^{2}}+x \\frac{d y}{dx}+y=0$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) y=x3<\/sup>+tanx  , \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f(show that) $\\frac{d^{2} y}{d x^{2}}=6 x+2 \\sec ^{2} x \\tan x$
<\/strong> Sol :
y=x3<\/sup>+tanx<\/p>\n\n\n\n

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

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

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

$\\frac{d^{2} y}{d x^{2}}=6 x+2 \\sec x \\cdot \\sec x \\tan x$<\/p>\n\n\n\n

$=6 x+2 \\operatorname{sec}^{2} x \\tan x$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) y=sinx(sinx) , \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f(show that) $\\frac{d^{2} y}{d x^{2}}+\\tan x \\frac{d y}{d x}+y \\cos ^{2} x=0$
<\/strong> Sol :
y=sin(sinx)<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=\\cos (\\sin x)\\times \\cos x$<\/p>\n\n\n\n

$\\cos(\\sin x)=\\frac{1}{\\cos x} \\cdot \\frac{dy}{d x}$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=-\\sin (\\sin 3 x) \\cdot \\cos x \\cdot \\cos x+\\cos (\\sin x) \\times(-\\sin x)$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=-y \\cos ^{2} x \\cdot+\\frac{1}{\\cos x} \\cdot \\frac{d y}{d x}(-\\sin x)$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=-y \\cos ^{2} x-\\tan x \\frac{d y}{dx}$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}+\\tan x \\frac{d y}{d x}+y\\cos^{2} x=0$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) $y=[\\log (x+\\sqrt{x^{2}+a^{2}})]^{2}$ , \u0938\u093f\u0926\u094d\u0926 \u0915\u0930\u0947 \u0915\u093f(prove that)  $\\left(x^{2}+a^{2}\\right) y_{2}+x y_{1}=2$
<\/strong> Sol :
$y=\\left[\\log(x+\\sqrt{x^{2}+a^{2}})\\right]^{2}$<\/p>\n\n\n\n

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

$y_{1}=2\\left[\\log (x+\\sqrt{x^{2}+a^{2}}) \\cdot \\frac{1}{x+\\sqrt{x^{2}+a^{2}}} \\times\\left(1+\\frac{1}{2 \\sqrt{x^{2}+a^{2}}} \\times 2 x\\right.\\right)$<\/p>\n\n\n\n

$y_{1}=2 \\frac{\\left[\\log(x+\\sqrt{x^{2}+a^{2}}\\right]}{x+\\sqrt{x^{2}+a^{2}}}\\left[\\frac{\\sqrt{x^{2} + a^{2}}+x}{\\sqrt{x^{2}+a^{2}}}\\right]$<\/p>\n\n\n\n

$y_{1}=\\frac{2[\\log (x+\\sqrt{x^{2}+a^{2}})]}{\\sqrt{x^{2}+a^{2}}}$<\/p>\n\n\n\n

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

$y_{2}=\\frac{2 \\cdot \\frac{1}{x+\\sqrt{x^{2}+a^{2}}}\\left[1+\\frac{1}{2 \\sqrt{x^{2}+a^{2}}} \\times 2 x\\right] \\cdot \\sqrt{x^{2}+a^{2}}-2[\\log(x+\\sqrt{x^2+a^2})]\\times \\frac{1}{2\\sqrt{x^2+a^2}}\\times 2x }{(\\sqrt{x^{2}+a^{2}}]^{2}}$<\/p>\n\n\n\n

$y_{2}=\\frac{\\frac{2}{x+\\sqrt{x^{2}+a^{2}}}\\left[\\frac{\\sqrt{x^{2}+a^{2}}+x}{\\sqrt{x^{2}+a^{2}}}\\right] \\cdot \\sqrt{x^{2}+a^{2}}-\\frac{2 \\times \\log (x+\\sqrt{x^{2}+a^{2}}}{\\sqrt{x^{2}+a^{2}}}}{x^{2}+a^{2}}$<\/p>\n\n\n\n

$\\left(x^{2}+a^{2}\\right) y_{2}=2-xy_{1}$<\/p>\n\n\n\n

$\\left(x^{2}+a^{2}\\right) y_{2}+x y_{1}=2$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) $y=\\left(1-x^{2}\\right)^{3 \/ 2}$ ,  \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (show that) $\\left(1-x^{2}\\right) \\frac{d^{2} y}{d x^{2}}+x \\frac{d y}{d x}+3 y=0$
<\/strong> Sol :
$y=\\left(1-x^{2}\\right)^{\\frac{3}{2}}$<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=\\frac{3}{2}\\left(1-x^{2}\\right)^{\\frac{1}{2}} \\cdot[0-2 x]$<\/p>\n\n\n\n

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

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

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

$\\frac{d^{2} y}{d x^{2}}=-3 \\cdot\\left(1-x^{2}\\right)^{\\frac{1}{2}}+(-3 x) \\frac{1}{2}\\left(1-x^{2}\\right)^{-\\frac{1}{2}} \\times(-2 x)$<\/p>\n\n\n\n

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

Multiplying by $\\left(1-x^{2}\\right)$ in both sides<\/p>\n\n\n\n

$\\left(1-x^{2}\\right) \\cdot \\frac{d^{2} y}{d t^{2}}=-3\\left(1-x^{2}\\right)^{\\frac{3}{2}}+3 x^{2} \\cdot\\left(1-2^{2}\\right)^{\\frac{1}{2}}$<\/p>\n\n\n\n

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

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

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

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

\u092f\u0926\u093f(If) y=Asinx+Bcosx, \u0938\u093f\u0926\u094d\u0926 \u0915\u0930\u0947 \u0915\u093f (prove that) $\\frac{d^{2} y}{d x^{2}}+y=0$
<\/strong> Sol :
y=Asinx+Bcosx<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=A \\cos x+B(-\\sin x)$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=A(-\\sin x)-B\\cos x$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=-\\left[A \\sin x+B\\cos x\\right]$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}+y=0$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) y=5cosx-3sinx , \u0938\u093f\u0926\u094d\u0927 \u0915\u0930\u0947 \u0915\u093f (prove that) $\\frac{d^{2} y}{d x^{2}}+y=0$
<\/strong> Sol :
y=5cosx-3sinx<\/p>\n\n\n\n

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

$\\frac{d y}{dx}=-5\\sin x-3 \\cos x$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=-5 \\cos x-3(-\\sin x)$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=-[5\\cos x-3 \\sin x]$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=-y \\Rightarrow \\frac{d^{2} y}{d x^{2}}+y=0$<\/p>\n\n\n\n

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

A \u0914\u0930 B \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947 \u0924\u093e\u0915\u093f(Find A and B such that) y=Asin5x+Bcos5x \u0938\u092e\u0940\u0915\u0930\u0923 (satisfies the equation) $\\frac{d^{2} y}{d x^{2}}+\\frac{1}{5} \\cdot \\frac{d y}{d x}+15 y=101 \\sin 5 x$ \u0915\u094b \u0938\u0902\u0924\u0941\u0937\u094d\u091f \u0915\u0930\u0924\u093e \u0939\u0948 \u0964
<\/strong> Sol :
y=Asin5x+Bcos5x<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=5 A \\cos x-5B \\sin 5 x$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=-5 \\times 5 A \\sin 5 x-5 \\times 5 B \\cos 5 x$<\/p>\n\n\n\n

=-25Asin5x-25Bcosx<\/p>\n\n\n\n

\u2235$\\frac{d^{2} y}{d x^{2}}+\\frac{1}{5} \\cdot \\frac{d y}{d x}+15 y=101 \\mathrm{sin} 5 x$<\/p>\n\n\n\n

-25Asin5x-25Bcos5x+$\\frac{1}{5}$(5Acos5x-5Bsin5x)+15(Asin5x+Bcos5x)=101sin5x<\/p>\n\n\n\n

-25Asin5x-25Bcosx+Acos5x-Bsin5x+15Asin5x+15Bcos5x=101sin5x<\/p>\n\n\n\n

-10Asin5x-Bsin5x-10Bcos5x+Acos5x=101sin5x<\/p>\n\n\n\n

(-10A-B)sin5x+(-10B+A)cos5x=101sin5x+0.cos5x<\/p>\n\n\n\n

By Equating<\/p>\n\n\n\n

-10A-B=101..(i)\u00d71<\/p>\n\n\n\n

A-10B=0..(ii)\u00d710<\/p>\n\n\n\n

$\\begin{aligned}-10A-B=101\\\\10A-100B=0 \\\\\\hline -101B=101 \\end{aligned}$<\/p>\n\n\n\n

B=-1<\/p>\n\n\n\n

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

A-10(-1)=0<\/p>\n\n\n\n

A=-10<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) $y=3 e^{2 x}+2 e^{3 x}$ , \u0938\u093f\u0926\u094d\u0926 \u0915\u0930\u0947 \u0915\u093f (prove that) $\\frac{d^{2} y}{d x^{2}}-5 \\frac{d y}{d x}+6 y=0$
<\/strong> Sol:
$y=3 e^{2 x}+2 e^{3 x}$<\/p>\n\n\n\n

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

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

$\\frac{d y}{d x}=6 e^{2 x}+6 \\cdot e^{3 x}$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=6 \\cdot e^{2 x} \\cdot 2+6 \\cdot e^{3 x} \\cdot 3$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=12 e^{2 x}+18 e^{3 x}$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=30 e^{2 x}+30 e^{3 x}-18 e^{2 x}-12 e^{3 x}$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=5\\left(6 e^{2 x}+6 e^{3 x}\\right)-6\\left(3 e^{2 x}+2 e^{3 x}\\right)$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=5 \\frac{d y}{d x}-6 y$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}-\\frac{5 dy}{dx}+6 y=0$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) $y=\\mathrm{Ae}^{mx}+\\mathrm{Be}^{2x}$  , \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (show that) $\\frac{d^{2} y}{d x^{2}}-(m+n) \\frac{d y}{d x}+m n y=0$
<\/strong> Sol :
$y=\\mathrm{Ae}^{mx}+\\mathrm{Be}^{2x}$<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=A e^{m x} \\times m+B \\cdot e^{n x} \\cdot n$<\/p>\n\n\n\n

$\\frac{d y}{d x}=m A e^{m x}+n B e^{n x}$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=m A \\cdot e^{m x} \\cdot m+n B \\cdot e^{n x} \\cdot n$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=m^{2} A e^{m x}+n^{2} B e^{n x}$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=m^{2} A e^{m x}+m n B e^{n x}+m n A e^{m x}+n^2 Be^{nx}-mnAe^{mx}-mnBe^{nx}$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=m\\left(m A e^{m x}+n B e^{n x}\\right)+n\\left(m A e^{m}+n B e^{nx}\\right)-mn(Ae^{mx}+Be^{nx})$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=(m+n) \\cdot\\left(m A e^{m x}+n B e^{n x}\\right)-m n\\left(A e^{m x}+B e^{nx}\\right)$<\/p>\n\n\n\n

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

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

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

\u092f\u0926\u093f(If) $y=500 e^{7 x}+600 e^{-7 x}$ , \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (show that ) $\\frac{d^{2} y}{d x^{2}}=49 y$
<\/strong> Sol :
$y=500 e^{7 x}+600 e^{-7 x}$<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=500 e^{7 x} \\cdot 7+600 e^{-7 x} \\cdot(-7)$<\/p>\n\n\n\n

$\\frac{d y}{d x}=7 \\times 500 e^{7 x}-7 \\times 600 e^{-7 x}$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d x^{2}}=7 \\times 500 e^{7 x} \\times 7-7 \\times 600 e^{-7x} \\times(-7)$<\/p>\n\n\n\n

$\\frac{d^{2}y}{d^{2}x}=49\\left[500 e^{7 x}+600 e^{-7x}\\right]$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=49 y$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) $y=\\sin ^{-1} x$ , \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (show that) $\\left(1-x^{2}\\right) \\frac{d^{2} y}{d x^{2}}-x \\frac{d y}{d x}=0$
<\/strong> Sol :
$y=\\sin ^{-1} x$<\/p>\n\n\n\n

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

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

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

$\\frac{d^{2} y}{d x^{2}}=\\frac{0 \\cdot \\sqrt{1-x^{2}}-1 \\cdot \\frac{1}{2 \\sqrt{1-x^{2}}} \\times(-2x)}{(\\sqrt{1-x^{2}})^{2}}$<\/p>\n\n\n\n

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

$\\left(1-x^{2}\\right) \\frac{d^{2} y}{d x^{2}}=\\frac{x}{\\sqrt{1-x^{2}}}$<\/p>\n\n\n\n

$\\left(1-x^{2}\\right) \\frac{d^{2} y}{d x}=x \\cdot \\frac{d y}{d x}$<\/p>\n\n\n\n

$\\left(1-x^{2}\\right) \\frac{d^{2} y}{dx^{2}}-x \\frac{d y}{d x}=0$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) $y=\\left(\\tan ^{-1} x\\right)^{2}$ , \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (show that) $\\left(x^{2}+1\\right)^{2} y_{2}+2 x\\left(x^{2}+1\\right) y_{1}-2=0$
<\/strong> Sol :
$y=\\left(\\tan ^{-1} x\\right)^{2}$<\/p>\n\n\n\n

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

$y_{1}=2\\left(\\tan ^{-1} x\\right) \\times \\frac{1}{1+x^{2}}$<\/p>\n\n\n\n

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

$2 \\tan ^{-1} x=\\left(1+x^{2}\\right) y_{1}$<\/p>\n\n\n\n

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

$y_{2}=\\frac{2 \\times \\frac{1}{1+x^{2}} \\times\\left(1-x^{2}\\right)-2 \\tan ^{-1} x \\times 2 x}{\\left(1+x^{2}\\right)^{2}}$<\/p>\n\n\n\n

$y_{2}=\\frac{2-4 x \\tan ^{-1} x}{\\left(1+x^{2}\\right)^{2}}$<\/p>\n\n\n\n

$\\left(1+x^{2}\\right)^{2} \\cdot y_{2}=2-2 x \\cdot 2 \\tan ^{-1} x$<\/p>\n\n\n\n

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

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

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

\u092f\u0926\u093f (If) y=3cos(log x)+4sin(log x) \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (show that) $x^{2} y_{2}+x y_{1}+y=0$
<\/strong> Sol :
y=3cos(log x)+4sin(log x)<\/p>\n\n\n\n

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

$y_{1}=-3 \\cdot \\sin (\\log x) \\times \\frac{1}{x}+4 \\cos (\\log x) \\times \\frac{1}{x}$<\/p>\n\n\n\n

$y_{1}=\\frac{1}{x}[-3 \\sin (\\log x)+4 \\cos (\\log x)]$<\/p>\n\n\n\n

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

$y_{2}=\\frac{-1}{x^{2}}\\left[-3 \\sin\\left(\\log x\\right)+4 \\cos \\left(\\log x\\right)\\right]+\\frac{1}{x}\\left[\\frac{-3cos(log x)}{x}+\\frac{4\\{-sin(\\log x)\\}}{x}\\right]$<\/p>\n\n\n\n

$y_{2}=-\\frac{1}{x^{2}}\\left[-3 \\sin \\left(\\log {x}\\right)+4 \\cos (\\log x)\\right]+\\frac{1}{x^{2}}\\left[-3 \\cos \\left(\\log x\\right)-4sin(\\log x)\\right]$<\/p>\n\n\n\n

$x^{2} y_{2}=-\\left[-3 \\sin \\left(\\log x\\right)+4 \\cos (\\log x)\\right]-[3cos(\\log x)+4\\sin (\\log x)]$<\/p>\n\n\n\n

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

$x^{2} y_{2}+x y_{1}+y_{2}=0$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) $y=e^{-x} \\cos x$ , \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (show that) $\\frac{d^{4} y}{d x^{4}}+4 y=0$
<\/strong> Sol :
$y=e^{-x} \\cos x$<\/p>\n\n\n\n

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

$\\frac{d{y}}{d{x}}=e^{-x} \\times(-1) \\cdot \\cos x+e^{-x} \\times(-\\sin x)$<\/p>\n\n\n\n

$\\frac{d y}{d x}=-e^{-x} \\cos x-e^{-x} \\sin x$<\/p>\n\n\n\n

$\\frac{d y}{d x}=-e^{-x}[\\cos x+\\sin x]$<\/p>\n\n\n\n

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

$\\frac{d^2 y}{d x^{2}}=-e^{-x}(-1)(\\cos x+\\sin x)+\\left(-e^{-x}\\right)(-\\sin x+\\cos x)$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=e^{-x}[\\cos x+\\sin x+\\sin x-\\cos x]$<\/p>\n\n\n\n

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

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

$\\frac{d^{3} y}{d x^{3}}=2\\left[e^{-x} \\cdot(-1) \\sin x+e^{-x} \\cos x\\right]$<\/p>\n\n\n\n

$\\frac{d^{3} y}{d x^{3}}=2 e^{-x}(-\\sin x+\\cos x)$<\/p>\n\n\n\n

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

$\\frac{d^{4} y}{d x^{4}}=2\\left[e^{-x} \\times(-1)(-\\sin x+\\cos x)+e^{-x}(-\\cos x-\\sin x)\\right] $<\/p>\n\n\n\n

$\\frac{d^{4} y}{d x^{4}}=2 e^{-x}[\\sin x-\\cos x-\\cos x-\\sin x]$<\/p>\n\n\n\n

$\\frac{d^{4} y}{d x^{4}}=2 \\cdot e^{-x} \\times(-2 \\cos x)$<\/p>\n\n\n\n

$\\frac{d^{4} y}{d x^{4}}=-4 e^{-x} \\cos x$<\/p>\n\n\n\n

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

$\\frac{d^{4} y}{d x^{4}}+4 y=0$<\/p>\n\n\n\n

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

\u092f\u0926\u093f (If) $e^{y}(x+1)=1$ , \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (show that) $\\frac{d^{2} y}{d x^{2}}=\\left(\\frac{d y}{d x}\\right)^{2}$
<\/strong> Sol :
$e^{y}(x+1)=1$<\/p>\n\n\n\n

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

$e^{y} \\cdot \\frac{d y}{dx}(x+1)+e^{y} \\cdot 1=0$<\/p>\n\n\n\n

$e^{y}\\left[(x+1) \\cdot \\frac{d y}{d x}+1\\right]=0$<\/p>\n\n\n\n

$(x+1) \\frac{d y}{dx}+1=0$<\/p>\n\n\n\n

$\\left[ \\frac{d y}{d x}=\\frac{-1}{x+1} \\right]$<\/p>\n\n\n\n

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

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

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

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

$\\frac{d^{2} y}{dx^{2}}=\\frac{d y}{dx} \\cdot \\frac{dy}{dx}$<\/p>\n\n\n\n

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

Question 18<\/h4>\n\n\n\n

\u092f\u0926\u093f(If) $y=e^{a \\cos ^{-1} x}$ , -1\u2264x\u22641 \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f (show that) $\\left(1-x^{2}\\right) \\frac{d^{2} y}{d x^{2}}-x \\frac{d y}{d x}-a^{2} y=0$
<\/strong> Sol :
$y=e^{a \\cos ^{-1} x}$<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=e^{a \\cos ^{-1} x} \\times a \\cdot \\frac{-1}{\\sqrt{1-x^{2}}}$<\/p>\n\n\n\n

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

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

$\\frac{d^{2} y}{d x^{2}}=\\frac{-a \\cdot e^{a \\cos ^{-1} x}.a \\cdot \\frac{1}{\\sqrt{1-x^{2}}}\\times \\sqrt{1-x^{2}}-\\left.(-a e^{\\left.a \\cos ^{-1} x\\right.}) \\times \\frac{1}{2 \\sqrt{1-x^{2}}}\\right.\\times -2x}{(\\sqrt{1-x^{2}})^{2}}$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d x^{2}}=\\frac{a^{2} e^{a \\cos^{-1}}x-\\frac{a x \\cdot e^{a \\cos^{-1}}x}{\\sqrt{1-x^{2}}}}{1-x^{2}}$<\/p>\n\n\n\n

$\\left(1-x^{2}\\right) \\cdot \\frac{d^{2} y}{d x^{2}}=a^{2} \\cdot y+x \\cdot \\frac{d y}{d x}$<\/p>\n\n\n\n

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

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

\u092f\u0926\u093f(If) $y=e^{\\tan ^{-1} x}$ \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f(show that) $\\left(x^{2}+1\\right)^{2} \\frac{d^{2} y}{d x^{2}}+(2 x-1) \\frac{d y}{d x}=0$
<\/strong> Sol :
$y=e^{\\tan ^{-1} x}$<\/p>\n\n\n\n

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

$\\frac{d y}{d x}=e^{\\tan ^{-1} x}\\times\\frac{1}{1+x^{2}}$<\/p>\n\n\n\n

$\\frac{d y}{d x}=\\frac{e^{\\tan ^{-1} }x }{1+x^{2}}$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{dx^{2}}=\\frac{e^{\\tan ^{-1} x} \\cdot \\frac{x}{1+x^{2}} \\times\\left(1+x^{2}\\right)-e^{\\tan x} \\cdot 2 x}{\\left(1+x^{2}\\right)^{2}}$<\/p>\n\n\n\n

$\\left(1+x^{2}\\right)^{2} \\frac{d^{2} y}{d x^{2}}=-2 x e^{\\tan ^{-1} x}+e^{\\tan ^{-1} x}$<\/p>\n\n\n\n

$\\left(1+x^{2}\\right)^{2} \\frac{d^{2} y}{d x^{2}}=-e^{\\tan ^{-1} x}(2 x-1)$<\/p>\n\n\n\n

$\\left(1+x^{2}\\right) \\frac{d^{2} y}{dx^{2}}=-\\frac{e^{\\tan ^{-1} x}}{1+x^{2}}(2 x-1)$<\/p>\n\n\n\n

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

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

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

\u092f\u0926\u093f(If) $y=\\mathrm{A} e^{k t} \\cos (p t+\\alpha)$ , \u0926\u093f\u0916\u093e\u090f\u0901 \u0915\u093f(show that)
$\\frac{d^{2} y}{d t^{2}}-2 k \\frac{d y}{d t}+\\left(p^{2}+k^{2}\\right) y=0$
<\/strong> Sol :
$y=\\mathrm{A} e^{k t} \\cos (p t+\\alpha)$<\/p>\n\n\n\n

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

$\\frac{d y}{d t}=A e^{k t} \\times k \\cdot \\cos (p t+\\alpha)+A e^{k t} \\cdot\\left[-\\sin(p t+\\alpha)\\right] p$<\/p>\n\n\n\n

$\\frac{d y}{d t}=A k e^{k t} \\cos (p t+\\alpha)-A p e^{k t} \\operatorname{sin}(p t+\\alpha)$<\/p>\n\n\n\n

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

$\\frac{d^{2} y}{d t^{2}}=A k e^{k t} \\times k \\cos (p t+\\alpha)+A k e^{k t} \\cdot\\{-\\sin (p t+x)\\}p-Ape^{kt}\\times k\\sin(pt+\\alpha)-Ape^{kt}.\\cos(pt+\\alpha).p$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d t^{2}}=A k^{2} e^{k t} \\cos (p t+\\alpha)-A k p e^{k t} \\sin (p t+\\alpha)-A k p e^{k t} \\sin (p t+\\alpha)-A p^{2} e^{kt} \\cos (p t+\\alpha)$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d t^{2}}=2 A k^{2} e^{k t} \\cos (p t+t)-2 A k p e^{k t} \\sin (p t+\\alpha)-A k^{2} e^{k t} \\cos (p t+\\alpha)-A p^{2} e^{k t}\\cos(pt+\\alpha)$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d t^{2}}=2 k\\left[A k e^{k t} \\cos(p t+\\alpha)-A p e^{k t} \\sin (p t+\\alpha)\\right]-\\left(k^{2}+p^{2}\\right) A e^{k+1} \\cdot \\cos(p+\\alpha)$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d t^{2}}=2 k \\frac{d y}{d t}-\\left(k^{2}+p^{2}\\right) y$<\/p>\n\n\n\n

$\\frac{d^{2} y}{d t^{2}}-2 k \\frac{d y}{d t}+\\left(p^{2}+k^{2}\\right) y=0$<\/p>\n\n\n\n

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

\u092f\u0926\u093f(If) $y^{\\frac{1}{m}}+y^{-\\frac{1}{m}}=2 x$ , \u0938\u093f\u0926\u094d\u0926 \u0915\u0930\u0947 \u0915\u093f(prove that) $\\left(x^{2}-1\\right) y_{2}+x y_{1}-m^{2} y=0$
<\/strong> Sol :
$y^{\\frac{1}{m}}+y^{-\\frac{1}{m}}=2 x$<\/p>\n\n\n\n

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

$\\frac{1}{m} \\cdot y^{\\frac{1}{k}-1} \\cdot \\frac{d y}{dx}+\\left(\\frac{-1}{m}\\right) \\cdot y^{-\\frac{1}{m}-1} \\frac{dy}{d x}=2$<\/p>\n\n\n\n

$\\frac{1}{m} y^{-1}\\left[y^{\\frac{1}{m}}-y^{-\\frac{1}{m}}\\right] \\frac{d y}{dx}=2$<\/p>\n\n\n\n

$\\frac{1}{m{y}}\\left[y^{\\frac{1}{m}}-y^{-\\frac{1}{m}}\\right] \\frac{d y}{dx}=2$<\/p>\n\n\n\n

$\\left[y^{\\frac{1}{m}}-y^{-\\frac{1}{m}}\\right] \\frac{d y}{dx}=2 m y$<\/p>\n\n\n\n

\u0926\u094b\u0928\u094b \u0924\u0930\u092b \u0935\u0930\u094d\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

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

$\\left[y^{\\frac{2}{m}}+y^{-\\frac{2}{m}}-2 \\cdot y^{\\frac{1}{m}} \\cdot y^{-\\frac{1}{m}}\\right]\\left(\\frac{dy}{d x}\\right)^{2}=4 m^{2} y^{2}$<\/p>\n\n\n\n

$\\left[y^{\\frac{2}{m}}+y^{\\frac{-2}{m}}-2\\right]\\left(\\frac{dy}{dx}\\right)^{2}=4 m^{2} y^{2}$..(i)<\/p>\n\n\n\n

\u2235 $y^{\\frac{1}{m}}+y^{-\\frac{1}{m }}=2 x$<\/p>\n\n\n\n

\u0926\u094b\u0928\u094b \u0924\u0930\u092b \u0935\u0930\u094d\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930 ,<\/p>\n\n\n\n

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

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

\u2234$[4x^2-2-2]\\left(\\frac{dy}{dx}\\right)^2=4m^2y^2$ \u0938\u092e\u0940\u0915\u0930\u0923 (i) \u0938\u0947<\/p>\n\n\n\n

$\\left(x^{2}-1\\right) \\cdot\\left(\\frac{d y}{dx}\\right)^{2}=m^{2} y^{2}$<\/p>\n\n\n\n

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

$2 x \\cdot\\left(\\frac{d y}{d x}\\right)^{2}+\\left(x^{2}-1\\right) \\cdot 2\\left(\\frac{d y}{d x}\\right) \\frac{d^{2} y}{d x^{2}}=m^{2} \\cdot 2 y \\frac{d y}{dx}$<\/p>\n\n\n\n

$2 \\frac{d y}{d x}\\left[x\\left(\\frac{d y}{d x}\\right)+\\left(x^{2}-1\\right) \\cdot \\frac{d^{2} y}{dx^{2}}\\right]=m^{2} \\cdot 2 y \\frac{d y}{dx}$<\/p>\n\n\n\n

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

$\\left(x^{2}-1\\right) \\cdot y_{2}+x y_{1}-m^{2} y=0$<\/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":"

\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928\u094b\u0902 \u0915\u0947 \u0926\u094d\u0935\u093f\u0924\u0940\u092f(\u0915\u094b\u091f\u093f) \u0915\u093e \u0905\u0935\u0915\u0932\u091c \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0902 \u0964[Find the second order derivatives of the following functions] (i) log xSol :\u092e\u093e\u0928\u093e y=log x Differentiating w.r.t x $\\frac{d y}{dx}=\\frac{1}{x}$ Again, Differentiating w.r.t x $\\frac{d^{2} y}{d x^{2}}=-\\frac{1}{x^{2}}$ (ii) x20Sol :\u092e\u093e\u0928\u093e y=x20 Differentiating w.r.t x $\\frac{d y}{dx}=20 x^{19}$ Again , differentiating w.r.t x $\\frac{d^{2} y}{d x^{2}}=380 x^{18}$ (iii) […]<\/p>\n","protected":false},"author":302,"featured_media":626448,"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-626444","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-12","entry"],"yoast_head":"\nKC Sinha: Exercise 12.1- Mathematics Solution Class 12 Chapter 12 \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u0915\u094b\u091f\u093f \u0915\u093e \u0905\u0935\u0915\u0932\u091c - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928\u094b\u0902 \u0915\u0947 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Again,","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-12-1-mathematics-solution-class-12-chapter-12-dvitiy-koti-ka-avakalaj\/","og_locale":"en_US","og_type":"article","og_title":"KC Sinha: Exercise 12.1- Mathematics Solution Class 12 Chapter 12 \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u0915\u094b\u091f\u093f \u0915\u093e \u0905\u0935\u0915\u0932\u091c","og_description":"\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928\u094b\u0902 \u0915\u0947 \u0926\u094d\u0935\u093f\u0924\u0940\u092f(\u0915\u094b\u091f\u093f) \u0915\u093e \u0905\u0935\u0915\u0932\u091c \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0947\u0902 \u0964 (i) log xSol :\u092e\u093e\u0928\u093e y=log x Differentiating w.r.t x $frac{d y}{dx}=frac{1}{x}$ Again,","og_url":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-12-1-mathematics-solution-class-12-chapter-12-dvitiy-koti-ka-avakalaj\/","og_site_name":"IndCareer Schools","article_publisher":"https:\/\/www.facebook.com\/indcareer","article_published_time":"2023-09-12T02:03:32+00:00","article_modified_time":"2023-09-12T02:03:40+00:00","og_image":[{"width":1600,"height":901,"url":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-39-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":"9 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Sinha: Exercise 12.1- Mathematics Solution Class 12 Chapter 12 \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u0915\u094b\u091f\u093f \u0915\u093e \u0905\u0935\u0915\u0932\u091c - IndCareer Schools","isPartOf":{"@id":"https:\/\/www.indcareer.com\/schools\/#website"},"primaryImageOfPage":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-12-1-mathematics-solution-class-12-chapter-12-dvitiy-koti-ka-avakalaj\/#primaryimage"},"image":{"@id":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-12-1-mathematics-solution-class-12-chapter-12-dvitiy-koti-ka-avakalaj\/#primaryimage"},"thumbnailUrl":"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2023\/09\/indcareer-schools-2-39-scaled.jpg","datePublished":"2023-09-12T02:03:32+00:00","dateModified":"2023-09-12T02:03:40+00:00","description":"\u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092b\u0932\u0928\u094b\u0902 \u0915\u0947 \u0926\u094d\u0935\u093f\u0924\u0940\u092f(\u0915\u094b\u091f\u093f) 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