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Let $f:[0,1] \rightarrow \mathbb{R}$ (the set of all real numbers) be a function. Suppose the function $f$ is twice differentiable, $f(0)=f(1)=0$ and satisfies $f^{\prime \prime}(\mathrm{x})-2 f^{\prime}(\mathrm{x})+f(\mathrm{x}) \geq \mathrm{e}^{\mathrm{x}}, \mathrm{x} \in[0,1]$b
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Q. Match the statements/expressions in Column-I with the open intervals in Column-II.
Ans. ( (A) p,q,s (B) p,t (C) p,q,r,t (D) S)
(i) $(\mathrm{A}) \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{(\mathrm{x}-3)^{2}}$ $\Rightarrow \ln y=\frac{1}{x-3}+c$ $\Rightarrow \mathrm{y}=\frac{1}{\mathrm{e}^{x-3}}, \mathrm{x} \neq 3$
(B) $\mathrm{I}=\int_{1}^{5}(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x} 4)(\mathrm{x}-5) \mathrm{d} \mathrm{x}$
Applying king $\mathrm{x} \rightarrow 6-\mathrm{x}$ $\mathrm{I}=\int_{1}^{5}(5-\mathrm{x})(4-\mathrm{x})(3-\mathrm{x})(2-\mathrm{x})(1-\mathrm{x}) \mathrm{d} \mathrm{x}=-\mathrm{I}$ $\Rightarrow \mathrm{I}=0$
(C) $f(x)=\cos ^{2} x+\sin x$ $f^{\prime}(x)=-2 \cos x \sin x+\cos x$ $\Rightarrow \cos x(-2 \sin x+1)=0$ $\cos x=0$ or $\sin x=\frac{1}{2}$
sign of $f^{\prime}(\mathrm{x})$ changes from -ve to +ve while $f(\mathrm{x})$ passes through $\mathrm{x}=\frac{\pi}{6}, \frac{5 \pi}{6}$
(D) $f(\mathrm{x})=\tan ^{-1}(\sin \mathrm{x}+\cos \mathrm{x})$ $f(x)=\frac{\cos x-\sin x}{1+(\sin x+\cos )^{2}}>0$
$\mathrm{x} \in(-3 \pi / 4, \pi / 4)$
(ii) $(\mathrm{A}) f(\mathrm{x})=\mathrm{xe}^{\sin \mathrm{x}}-\cos \mathrm{x}$
$f(0)=-1$
$f(\pi / 2)=\frac{\pi}{2} \mathrm{e}$
$f^{\prime}(x)=x e^{\sin x} \cos x+e^{\sin x}>0$
$\Rightarrow k(k-4)-4 c+8-2 k=0$
$\Rightarrow \mathrm{k}^{2}-4 \mathrm{k}+8-2 \mathrm{k}=0$
$\Rightarrow \mathrm{k}^{2}-6 \mathrm{k}+8=0$
$\Rightarrow \mathrm{k}=2,4$
(C) $|x-1|+|x-2|+|x+1|+|x+2|=4 k$
$4 k=8,12,16,20$$\quad\left\{\begin{array}{l}{\text { modulus denotes the }} \\ {\text { distance of from }} \\ {-2,-1,2}\end{array}\right.$
$\therefore \mathrm{k}=2,3,4,5$
(D) $\frac{d y}{y+1}=d x$
$\ln (\mathrm{y}+1)=\mathrm{ke}^{\mathrm{x}}$
$y+1=k e^{x}$
$y+1=2=k$
$y+1=2 e^{x}$
$y=\left(2 e^{x}-1\right)$
$y(\ln 2)=3$
$\mathrm{x} \in(-3 \pi / 4, \pi / 4)$
(ii) $(\mathrm{A}) f(\mathrm{x})=\mathrm{xe}^{\sin \mathrm{x}}-\cos \mathrm{x}$
$f(0)=-1$
$f(\pi / 2)=\frac{\pi}{2} \mathrm{e}$
$f^{\prime}(x)=x e^{\sin x} \cos x+e^{\sin x}>0$
$\Rightarrow k(k-4)-4 c+8-2 k=0$
$\Rightarrow \mathrm{k}^{2}-4 \mathrm{k}+8-2 \mathrm{k}=0$
$\Rightarrow \mathrm{k}^{2}-6 \mathrm{k}+8=0$
$\Rightarrow \mathrm{k}=2,4$
(C) $|x-1|+|x-2|+|x+1|+|x+2|=4 k$
$4 k=8,12,16,20$$\quad\left\{\begin{array}{l}{\text { modulus denotes the }} \\ {\text { distance of from }} \\ {-2,-1,2}\end{array}\right.$
$\therefore \mathrm{k}=2,3,4,5$
(D) $\frac{d y}{y+1}=d x$
$\ln (\mathrm{y}+1)=\mathrm{ke}^{\mathrm{x}}$
$y+1=k e^{x}$
$y+1=2=k$
$y+1=2 e^{x}$
$y=\left(2 e^{x}-1\right)$
$y(\ln 2)=3$
Q. Match the statements/expressions in Column-I with the values given in Column-II.
[JEE 2009, (2+2+2+2) × 2]
[JEE 2009, (2+2+2+2) × 2]
Ans. ( (A) P (B) q,s (C) q,r,s,t (D) r )
Given $\mathrm{y}=f(\mathrm{x})$
Tangent at point $\mathrm{P}(\mathrm{x}, \mathrm{y})$
$\mathrm{Y}-\mathrm{y}=\left(\frac{\mathrm{d} y}{\mathrm{dx}}\right)_{(\mathrm{x}, \mathrm{y})}(\mathrm{X}-\mathrm{x})$
Now y-intercept $\Rightarrow Y=y-x \frac{d y}{d x}$
Given that, $\quad \mathrm{y}-\mathrm{x} \frac{\mathrm{d} y}{\mathrm{dx}}=\mathrm{x}^{3}$
$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=-x^{2}$ is a linear differential equation
Q. Let f be a real valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y-intercept of the tangent at any point P(x,y) on the curve y =f(x) is equal to the cube of the abscissa of P, then the value of f(–3) is equal to
[JEE 2010, 3]
Ans. 9
Q. (A) Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function such that $f(1)=2 .$ If $6 \int_{1}^{x} f(t) d t=3 x f(x)-x^{3}$ for all $x \geq 1,$ then the value of $f(2)$ is
(B) Let $y^{\prime}(x)+y(x) g^{\prime}(x)=g(x) g^{\prime}(x), y(0)=0, x \in R,$ where $f^{\prime}(x)$ denotes $\frac{d f(x)}{d x}$ and $g(x)$ is a given non-constant differentiable function on $\mathrm{R}$ with $\mathrm{g}(0)=\mathrm{g}(2)=0 .$ Then the value of $\mathrm{y}(2)$ is
[JEE 2011, 4]
Ans. ( (A) Bonus (B) 0 )
(a) (Bonus)
(Comment : The given relation does not hold for x =1, therefore it is not an identity. Hence there is an error in given question. The correct identity must be-)
$6 \int_{1}^{x} f(t) d t=3 x f(x)-x^{3}-5, \forall x \geq 1$
Now applying Newton Leibnitz theorem
$6 f(x)=3 x f^{\prime}(x)-3 x^{2}+3 f(x)$
$\Rightarrow 3 f(x)=3 x f^{\prime}(x)-3 x^{2}$
Let $y=f(x)$
$\Rightarrow x \frac{d y}{d x}-y=x^{2}$
$\Rightarrow \frac{x d y-y d x}{x^{2}}=d x$
$\Rightarrow \quad \int d\left(\frac{y}{x}\right)=\int d x$
$\Rightarrow \quad \frac{y}{x}=x+C \quad$ (where $C$ is constant)
$\Rightarrow \quad y=x^{2}+C x$
$\therefore f(\mathrm{x})=\mathrm{x}^{2}+\mathrm{Cx}$
Given $f(1)=2$
$\Rightarrow \mathrm{C}=1$
$\therefore f(2)=2^{2}+2=6$
(B) Given $\mathrm{y}(0)=0, \mathrm{g}(0)=\mathrm{g}(2)=0$
\[
\begin{array}{l}
{\text { Let } y^{\prime}(x)+y(x) \cdot g^{\prime}(x)=g(x) g^{\prime}(x)} \\
{\Rightarrow y^{\prime}(x)+(y(x)-g(x)) g^{\prime}(x)=0} \\
{\Rightarrow \quad \frac{y^{\prime}(x)}{g^{\prime}(x)}+y(x)=g(x)} \end{array}
$\Rightarrow \quad \frac{d y(x)}{d g(x)}+y(x)=g(x)$\]
$\Rightarrow \quad$ I.F. $=e^{\int d(g(x))}=e^{g(x)}$
$\Rightarrow \quad y(x) \cdot e^{g(x)}=\int e^{g(x)} g(x) \cdot d g(x)$
$y(x) \cdot e^{g(x)}=g(x) \cdot e^{g(x)}-e^{g(x)}+c$
put $\mathrm{x}=0$
$\Rightarrow 0=0-1+c \Rightarrow c=1$
$\Rightarrow \mathrm{y}(2) \cdot \mathrm{e}^{\mathrm{g}(2)}=\mathrm{g}(2) \mathrm{e}^{\mathrm{g}(2)}-\mathrm{e}^{\mathrm{g}(2)}+1$
$\Rightarrow \mathrm{y}(2)=0-\mathrm{e}^{0}+1$
$\Rightarrow \quad \mathrm{y}(2)=0$
Q. If y(x) satisfies the differential equation y' – ytanx = 2x sec x and y(0) = 0, then
(A) $\mathrm{y}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}$
(B) $y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{18}$
(C) $\mathrm{y}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{9}$
(D) $\mathrm{y}^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}}$
[JEE 2012, 4M]
Ans. (A,D)
$\frac{\mathrm{dy}}{\mathrm{dx}}-y \tan \mathrm{x}=2 \mathrm{x} \sec \mathrm{x}$
$\mathrm{I.F.}=\mathrm{e}^{\int-\tan \mathrm{xdx}}=\cos \mathrm{x}$
$\therefore$ Equation reduces to
y. $\cos x=\int 2 x \cdot \sec x \cdot \cos x d x$
$\Rightarrow \mathrm{y} \cos \mathrm{x}=\mathrm{x}^{2}+\mathrm{C}$
$\therefore y \cos x=x^{2}$
$\Rightarrow \mathrm{y}(\mathrm{x})=\mathrm{x}^{2} \mathrm{secx}$
$\therefore y\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{16} \sqrt{2}=\frac{\pi^{2}}{8 \sqrt{2}}(\therefore(\mathrm{A}) \text { is correct })$
$y\left(\frac{\pi}{3}\right)=\frac{\pi^{2}}{9} \cdot 2=\frac{2 \pi^{2}}{9}(\therefore(\mathrm{C}) \text { is wrong })$
Also $\mathrm{y}^{\prime}(\mathrm{x})=2 \mathrm{x} \sec \mathrm{x}+\mathrm{x}^{2} \sec \mathrm{x} \tan \mathrm{x}$
$\Rightarrow y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi}{2} \cdot \sqrt{2}+\frac{\pi^{2} \sqrt{2}}{16}(\therefore(\mathrm{B}) \text { is wrong })$
and $y^{\prime}\left(\frac{\pi}{3}\right)=2 \cdot \frac{\pi}{3} \cdot 2+\frac{\pi^{2}}{9} \cdot 2 \cdot \sqrt{3}$
$=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}} \quad(\therefore(\mathrm{D}) \text { is correct })$
Q. Let $f:\left[\frac{1}{2}, 1\right] \rightarrow \mathrm{R}$ (the set of all real numbers) be a positive, non-constant and differentiable function such that $f^{\prime}(\mathrm{x})<2 f(\mathrm{x})$ and $f\left(\frac{1}{2}\right)=1 .$ Then the value of $\int_{1 / 2}^{1} f(\mathrm{x}) \mathrm{d} \mathrm{x}$ lies in the interval
(A) (2e – 1, 2e)
(B) (e – 1, 2e – 1)
(C) $\left(\frac{\mathrm{e}-1}{2}, \mathrm{e}-1\right)$
$(D)\left(0, \frac{e-1}{2}\right)$
[JEE(Advanced) 2013, 2M]
Ans. (D)
$f^{\prime}(x)-2 f(x)<0$
Multiply both side by e $^{-2 x}$
$e^{-2 x} f^{\prime}(x)-2 e^{-2 x} f(x)<0$
$\frac{d}{d x}\left(e^{-2 x} f(x)\right)<0$
$\mathrm{Now}, \mathrm{g}(\mathrm{x})=\mathrm{e}^{-2 \mathrm{x}} f(\mathrm{x})$
$g(x)$ is a decreasing function.
$x>\frac{1}{2}$
$g(x)
$\Rightarrow e^{-2 x} f(x)<\frac{1}{e}$
$\Rightarrow f(\mathrm{x})<\mathrm{e}^{2 \mathrm{x}-1}$
$\Rightarrow \int_{1 / 2}^{1} f(\mathrm{x}) \mathrm{d} \mathrm{x}<\frac{1}{\mathrm{e}} \int_{1 / 2}^{1} \mathrm{e}^{2 \mathrm{x}} \mathrm{d} \mathrm{x}$
$=\left[\frac{1}{2 e} e^{2 x}\right]_{1 / 2}^{1}=\frac{1}{2 e}\left(e^{2}-e\right)=\frac{1}{2}(e-1)$
Q. curve passes through the point $\left(1, \frac{\pi}{6}\right) .$ Let the slope of the curve at each point $(\mathrm{x}, \mathrm{y})$ be $\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0 .$ Then the equation of the curve is
(A) $\sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2}$
(B) $\csc \left(\frac{y}{x}\right)=\log x+2$
(C) $\sec \left(\frac{2 \mathrm{y}}{\mathrm{x}}\right)=\log \mathrm{x}+2$
(D) $\cos \left(\frac{2 y}{x}\right)=\log x+\frac{1}{2}$
[JEE(Advanced) 2013, 2M]
Ans. (A)
$\frac{d y}{d x}=\frac{y}{x}+\sec \frac{y}{x}$
Let $y=v x$
$\frac{d y}{d x}=v+x \frac{d v}{d x}$
$\mathrm{v}+\frac{\mathrm{xdv}}{\mathrm{dx}}=\mathrm{v}+\sec \mathrm{v}$
$\cos v d v=\frac{d x}{x}$
$\sin \mathrm{v}=\ell \mathrm{nx}+\mathrm{c}$
$\sin \left(\frac{y}{x}\right)=\ell n x+c$
$\because$ passing through $\left(1, \frac{\pi}{6}\right)$
$\Rightarrow \sin \frac{\pi}{6}=c \Rightarrow c=\frac{1}{2}$
$\therefore \sin \frac{y}{x}=\ell n x+\frac{1}{2}$
Q. If the function $\mathrm{e}^{-\mathrm{x}} f(\mathrm{x})$ assumes its minimum in the interval $[0,1]$ at $\mathrm{x}=\frac{1}{4},$ which of the following is true?
(A) $f^{\prime}(x)
(B) $f^{\prime}(x)>f(x), 0
(C) $f^{\prime}(x)
(D) $f^{\prime}(x)
[JEE(Advanced) 2013, 3, (–1)]
Ans. (C)
$\mathrm{e}^{-\mathrm{x}}\left(f^{\prime \prime}(\mathrm{x})-2 f^{\prime}(\mathrm{x})+f(\mathrm{x})\right) \geq 1$
$\mathrm{D}\left(\left(f^{\prime}(\mathrm{x})-f(\mathrm{x})\right) \mathrm{e}^{-\mathrm{x}}\right) \geq 1$
$\Rightarrow \mathrm{D}\left(\left(f^{\prime}(\mathrm{x})-f(\mathrm{x}) \mathrm{e}^{-\mathrm{x}}\right) \geq 0\right.$
$\Rightarrow\left(f^{\prime}(\mathrm{x})-f(\mathrm{x})\right) \mathrm{e}^{-\mathrm{x}}$ is an increasing function.
As we know that $\mathrm{e}^{-\mathrm{x}} f(\mathrm{x})$ has local minima at $\mathrm{x}=\frac{1}{4}$
$\mathrm{e}^{-\mathrm{x}}\left(f^{\prime}(\mathrm{x})-f(\mathrm{x})\right)=0$ at $\mathrm{x}=\frac{1}{4}$
Let $\mathrm{F}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}}\left(f^{\prime}(\mathrm{x})-f(\mathrm{x})\right)$
$\mathrm{F}(\mathrm{x})<0$ in $\left(0, \frac{1}{4}\right)$
$\mathrm{e}^{-\mathrm{x}}\left(f^{\prime}(\mathrm{x})-f(\mathrm{x})<0 \text { in }\left(0, \frac{1}{4}\right)\right.$
$f^{\prime}(x)
option $\mathrm{C}$
Q. Which of the following is true for 0 < x < 1 ?
(A) $0
(B) $-\frac{1}{2}
(C) $-\frac{1}{4}
(D) $-\infty
[JEE(Advanced) 2013, 3, (–1)]
Ans. (D)
$\mathrm{D}\left(\mathrm{e}^{-\mathrm{x}}\left(f^{\prime}(\mathrm{x})-f(\mathrm{x})\right) \geq 0 \quad \forall \mathrm{x} \in(0,1)\right.$
$\mathrm{D}\left(\mathrm{D}\left(\mathrm{e}^{-\mathrm{x}} f(\mathrm{x})\right) \geq 0 \forall \mathrm{x} \in(0,1)\right.$
$\mathrm{D}^{2}\left(\mathrm{e}^{-\mathrm{x}} f(\mathrm{x})\right) \geq 0$
Let $\mathrm{F}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} f(\mathrm{x})$
$\mathrm{F}^{\prime \prime}(\mathrm{x})>0$ means it is concave upward.
$\mathrm{F}(0)=\mathrm{F}(1)=0$
$\mathrm{F}(\mathrm{x})<0 \forall \mathrm{x} \in(0,1)$
$\mathrm{e}^{-\mathrm{x}} f(\mathrm{x})<0 \forall \mathrm{x} \in(0,1)$
$f(x)<0$
Option D is possible
$\mathrm{F}(0)=\mathrm{F}(1)=0$
$\mathrm{F}(\mathrm{x})<0 \forall \mathrm{x} \in(0,1)$
$\mathrm{e}^{-\mathrm{x}} f(\mathrm{x})<0 \forall \mathrm{x} \in(0,1)$
$f(x)<0$
Option D is possible
Q. The function y = ƒ(x) is the solution of the differential equation \frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}} in (–1, 1) satisfying ƒ(0) = 0. Then \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(\mathrm{x}) \mathrm{d} \mathrm{x} \text { is }
(A) $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$
(B) $\frac{\pi}{3}-\frac{\sqrt{3}}{4}$
(C) $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$
(D) $\frac{\pi}{6}-\frac{\sqrt{3}}{2}$
[JEE(Advanced)-2014, 3(–1)]
Ans. (B)
$\left(\frac{d y}{d x}-\frac{x y}{\left(1-x^{2}\right)}\right) \sqrt{1-x^{2}}=x^{4}+2 x$
$\Rightarrow \sqrt{1-x^{2}} d y+d(\sqrt{1-x^{2}}) y=\left(x^{4}+2 x\right) d x$
$\Rightarrow y \sqrt{1-x^{2}}=\frac{x^{5}}{5}+x^{2}+c$
by $(0,0) \mathrm{c}=0$
$y=\frac{x^{5}}{5 \sqrt{1-x^{2}}}+\frac{x^{2}}{\sqrt{1-x^{2}}}$
$=\int_{-\sqrt{3} / 2}^{\sqrt{3} / 2}\left(\frac{x^{5}}{5 \sqrt{1-x^{2}}}+\frac{x^{2}}{\sqrt{1-x^{2}}}\right) d x$
$=2 \int_{0}^{\sqrt{3} / 2} \frac{x^{2}}{\sqrt{1-x^{2}}} d x$ put $x=\sin \theta$
$=2 \int_{0}^{\pi / 3} \sin ^{2} \theta \mathrm{d} \theta=\int_{0}^{\pi / 3}(1-\cos 2 \theta) \mathrm{d} \theta$
$=\left(\theta-\frac{\sin 2 \theta}{2}\right)_{0}^{\pi / 3}=\frac{\pi}{3}-\frac{\sqrt{3}}{4}$
Q. Let $\mathrm{y}(\mathrm{x})$ be a solution of the differential equation $\left(1+\mathrm{e}^{\mathrm{x}}\right) \mathrm{y}^{\prime}+\mathrm{ye}^{\mathrm{x}}=1 .$ If $\mathrm{y}(0)=2,$ then which of the following statements is(are) true ?
(A) y(–4) = 0
(B) y(–2) = 0
(C) y(x) has a critical point in the interval (–1,0)
(D) y(x) has no critical point in the interval (–1,0)
[JEE 2015, 4M, –2M]
Ans. (A,C)
$y^{\prime}+e^{x} y^{\prime}+y e^{x}=1$
$\Rightarrow \mathrm{dy}+\mathrm{d}\left(\mathrm{e}^{\mathrm{x}} \mathrm{y}\right)=\mathrm{d} \mathrm{x}$
$\Rightarrow \mathrm{y}+\mathrm{e}^{\mathrm{x}} \mathrm{y}=\mathrm{x}+\mathrm{c}$
$\because \mathrm{y}(0)=2$
$\Rightarrow \mathrm{c}=4$
$\Rightarrow y=\frac{x+4}{1+e^{x}}$
$\therefore \mathrm{y}(-4)=0$
for critical point given
$\frac{d y}{d x}=\frac{1-y e^{x}}{1+e^{x}}=\frac{1-\left(\frac{x+4}{1+e^{x}}\right) e^{x}}{1+e^{x}}=\frac{1-(x+3) e^{x}}{\left(1+e^{x}\right)^{2}}$
$\Rightarrow \quad x+3=e^{-x}$
$\mathrm{y}(\mathrm{x})$ has a critical point in the interval $(-1,0)$
$\Rightarrow \quad x+3=e^{-x}$
$\mathrm{y}(\mathrm{x})$ has a critical point in the interval $(-1,0)$
Q. Consider the family of all circles whose centers lie on the straight line y = x. If this family of circles is represented by the differential equation Py" + Qy' + 1 =
0, where P,Q are functions of x,y and y' (here $y^{\prime}=\frac{d y}{d x}, y^{\prime \prime}=\frac{d^{2} y}{d x^{2}},$ then which of the following statements is (are) true?
(A) P = y + x
(B) P = y – x
(C) $\mathrm{P}+\mathrm{Q}=1-\mathrm{x}+\mathrm{y}+\mathrm{y}^{\prime}+\left(\mathrm{y}^{\prime}\right)^{2}$
(D) P – Q = x + y – y' – $\left(y^{\prime}\right)^{2}$
[JEE 2015, 4M, –2M]
Ans. (B,C)
Let Circle $\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{ax}-2 \mathrm{ay}+\mathrm{c}=0$
On differentiation
$2 \mathrm{x}+2 \mathrm{yy}^{\prime}-2 \mathrm{a}-2 \mathrm{ay}^{\prime}=0$
$\Rightarrow \mathrm{x}+\mathrm{yy}^{\prime}-\mathrm{a}\left(1+\mathrm{y}^{\prime}\right)=0$
$\Rightarrow a=\frac{x+y y^{\prime}}{1+y^{\prime}}$
again differentiation
$\frac{\left(1+(y)^{2}+y y^{\prime \prime}(1+y)-(x+y y)\left(y^{\prime \prime}\right)\right.}{(1+y)^{2}}=0$
$\Rightarrow 1+\mathrm{y}^{\prime}\left(\left(\mathrm{y}^{\prime}\right)^{2}+\mathrm{y}^{\prime}+1\right)+\mathrm{y}^{\prime \prime}(\mathrm{y}-\mathrm{x})=0$
$\therefore P=y-x$
$\mathrm{Q}=1+\mathrm{y}^{\prime}+\left(\mathrm{y}^{\prime}\right)^{2}$
Q. A solution curve of the differential equation $\left(\mathrm{x}^{2}+\mathrm{xy}+4 \mathrm{x}+2 \mathrm{y}+4\right) \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}^{2}=0, \mathrm{x}>0$ passes through the point $(1,3) .$ The the solution curve-
(A) intersects y = x + 2 exactly at one point
(B) intersects y = x + 2 exactly at two points
(C) intersects $y=(x+2)^{2}$
(D) does NOT intersect $y=(x+3)^{2}$
[JEE(Advanced)-2016]
Ans. (A,D)
$\left(\mathrm{x}^{2}+\mathrm{xy}+4 \mathrm{x}+2 \mathrm{y}+4\right) \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}^{2}=0$
$\left((x+2)^{2}+y(x+2)\right) \frac{d y}{d x}=y^{2}$
Let $\mathrm{x}+2=\mathrm{X}, \mathrm{y}=\mathrm{Y}$
$(\mathrm{X})(\mathrm{X}+\mathrm{Y}) \frac{\mathrm{d} \mathrm{Y}}{\mathrm{dX}}=\mathrm{Y}^{2}$
$-\mathrm{X}^{2} \mathrm{d} \mathrm{Y}=\mathrm{X} \mathrm{YdY}-\mathrm{Y}^{2} \mathrm{d} \mathrm{X}$
$-\mathrm{X}^{2} \mathrm{d} \mathrm{Y}=\mathrm{Y}(\mathrm{XdY}-\mathrm{YdX})$
$-\frac{\mathrm{d} \mathrm{Y}}{\mathrm{Y}}=\frac{\mathrm{XdY}-\mathrm{YdX}}{\mathrm{X}^{2}}$
Q. Let $f:(0, \infty) \rightarrow \mathrm{R}$ be a differentiable function such that $f^{\prime}(\mathrm{x})=2-\frac{f(\mathrm{x})}{\mathrm{x}}$ for all $\mathrm{x} \in$ $(0, \infty)$ and $f(1) \neq 1 .$ Then
(A) $\lim _{x \rightarrow 0^{+}} f^{\prime}\left(\frac{1}{x}\right)=1$
(B) $\lim _{x \rightarrow 0^{+}} x f\left(\frac{1}{x}\right)=2$
(C) $\lim _{x \rightarrow 0^{+}} x^{2} f^{\prime}(x)=0$
(D) $|f(x)| \leq 2$ for all $x \in(0,2)$
[JEE(Advanced)-2016]
Ans. (A)
Let $\mathrm{y}=f(\mathrm{x})$
$\frac{d y}{d x}+\frac{y}{x}=2$ (linear differential equation)
$\therefore \mathrm{y.e}^{\int \frac{\mathrm{dx}}{\mathrm{x}}}=2 \int \mathrm{e}^{\int \frac{\mathrm{dx}}{\mathrm{x}}=2 \int \mathrm{e}^{\int \frac{\mathrm{dx}}{\mathrm{x}}} \mathrm{dx}+\mathrm{c}}$
$\Rightarrow \mathrm{yx}=2 \int \mathrm{xdx}+\mathrm{c}$
$\therefore \mathrm{yx}=\mathrm{x}^{2}+\mathrm{c}$
$\Rightarrow f(x)=x+\frac{c}{x} ;$ As $f(1) \neq 1 \Rightarrow c \neq 0$
$\Rightarrow f^{\prime}(x)=1-\frac{c}{x^{2}}, c \neq 0$
(A) $\lim _{x \rightarrow 0^{+}} f^{\prime}\left(\frac{1}{x}\right)=\lim _{x \rightarrow 0^{+}}\left(1-c x^{2}\right)=1$
(B) $\lim _{x \rightarrow 0^{+}} x f\left(\frac{1}{x}\right)=\lim _{x \rightarrow 0^{+}} x\left(\frac{1}{x}+c x\right)=\lim _{x \rightarrow 0^{+}}\left(1+c x^{2}\right)=1$
(C) $\lim _{x \rightarrow 0^{+}} x^{2} f^{\prime}(x)=\lim _{x \rightarrow 0^{+}} x^{2}\left(1-\frac{c}{x^{2}}\right)=\lim _{x \rightarrow 0^{+}}\left(x^{2}-c\right)=-c$
(D) $f(x)=x+\frac{c}{x}, c \neq 0$
for $\mathrm{c}>0$
$\therefore \lim _{x \rightarrow 0^{+}} f(x)=\infty \Rightarrow$ function is not bounded in $(0,2)$
Q. If y = y(x) satisfies the differential equation
$8 \sqrt{x}(\sqrt{9+\sqrt{x}}) d y=(\sqrt{4+\sqrt{9+\sqrt{x}}})^{-1} d x, \quad x>0$ and $y(0)=\sqrt{7},$ then $y(256)=$
(A) 80 (B) 3 (C) 16 (D) 9
[JEE(Advanced)-2016]
Ans. (B)
$y=\frac{1}{8} \int \frac{d x}{\sqrt{4+\sqrt{9+x} \cdot \sqrt{x} \cdot \sqrt{9+\sqrt{x}}}}$
put $\sqrt{9+\sqrt{\mathrm{x}}}=\mathrm{t}$
$\Rightarrow \frac{d x}{\sqrt{x} \cdot \sqrt{9+\sqrt{x}}}=4 d t$
$\therefore \mathrm{y}=\frac{4}{8} \int \frac{\mathrm{dt}}{\sqrt{4+\mathrm{t}}}$
$\Rightarrow \quad y=\sqrt{4+t}+C$
$\Rightarrow \mathrm{y}(\mathrm{x})=\sqrt{4+\sqrt{9+\sqrt{\mathrm{x}}}}+\mathrm{C}$
at $x=0: y(0)=\sqrt{7}$
$\Rightarrow \mathrm{C}=0$
$\therefore \mathrm{y}(\mathrm{x})=\sqrt{4+\sqrt{9+\sqrt{\mathrm{x}}}}$
$\Rightarrow \mathrm{y}(256)=3$
Q. If $f: \square \rightarrow \square$ is a differentiable function such that $f^{\prime}(\mathrm{x})>2 f(\mathrm{x})$ for all $\mathrm{x} \in \mathbb{U},$ and $f(0)$ $=1,$ then
(A) $f(\mathrm{x})>\mathrm{e}^{2 \mathrm{x}}$ in $(0, \infty)$
(B) $f(x)$ is decreasing in $(0, \infty)$
(C) $f(x)$ is increasing in $(0, \infty)$
(D) $f^{\prime}(x)
[JEE(Advanced)-2017]
Ans. (A,C)
Given that,
$\mathrm{f}^{\prime}(\mathrm{x})>2 \mathrm{f}(\mathrm{x}) \forall \mathrm{x} \in \mathrm{R}$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})-2 \mathrm{f}(\mathrm{x})>0$
$\forall \mathrm{x} \in \mathrm{R}$
$\therefore \mathrm{e}^{-2 \mathrm{x}}\left(\mathrm{f}^{\prime}(\mathrm{x})-2 \mathrm{f}(\mathrm{x})\right)>0 \forall \mathrm{x} \in \mathrm{R}$
$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{-2 \mathrm{x}} \mathrm{f}(\mathrm{x})\right)>0 \forall \mathrm{x} \in \mathrm{R}$
Let $\mathrm{g}(\mathrm{x})=\mathrm{e}^{-2 \mathrm{x}} \mathrm{f}(\mathrm{x})$
Now, $g^{\prime}(x)>0 \forall x \in R$
$\Rightarrow \mathrm{g}(\mathrm{x})$ is strictly increasing $\forall \mathrm{x} \in \mathrm{R}$
Also, $\mathrm{g}(0)=1$
$\therefore \forall \mathrm{x}>0$
$\Rightarrow \mathrm{g}(\mathrm{x})>\mathrm{g}(0)=1$
$\therefore \mathrm{e}^{-2 \mathrm{x}} \cdot \mathrm{f}(\mathrm{x})>1 \forall \mathrm{x} \in(0, \infty)$
$\Rightarrow f(\mathrm{x})>\mathrm{e}^{2 \mathrm{x}} \forall \mathrm{x} \in(0, \infty)$
$\therefore$ option $(\mathrm{A})$ is correct
As, $\mathrm{f}^{\prime}(\mathrm{x})>2 \mathrm{f}(\mathrm{x})>2 \mathrm{e}^{2 \mathrm{x}}>2 \forall \mathrm{x} \in(0, \infty)$
$\Rightarrow \mathrm{f}(\mathrm{x})$ is strictly increasing on $\mathrm{x} \in(0, \infty)$
$\Rightarrow$ option $(\mathrm{C})$ is correct
As, we have proved above that $\mathrm{f}^{\prime}(\mathrm{x})>2 \cdot \mathrm{e}^{2 \mathrm{x}} \forall \mathrm{x} \in(0, \infty)$
$\Rightarrow$ option $(\mathrm{D})$ is incorrect
options $(\mathrm{A})$ and $(\mathrm{C})$ are correct
Q. Let $f: \square \rightarrow \square$ and $g: \square \rightarrow \square$ be two non-constant differentiable functions. If $f^{\prime}(x)=\left(e^{(f(x)}\right)$ $-g(x)) g^{\prime}(x)$ for all $x \in \square,$ and $f(1)=g(2)=1,$ then which of the following statement(s) is (are) TRUE?
(A) $f(2)<1-\log _{e} 2$
(B) $f(2)>1-\log _{e} 2$
(C) $\mathrm{g}(1)>1-\log _{\mathrm{e}} 2$
(D) $\mathrm{g}(1)<1-\log _{\mathrm{e}} 2$
[JEE(Advanced)-2018,4(-2)]
Ans. (B,C)
$f^{\prime}(x)=e^{(f(x)-g(x))} g^{\prime}(x) \forall x \in \square$
$\Rightarrow \quad \mathrm{e}^{-f(\mathrm{x})} \cdot f^{\prime}(\mathrm{x})-\mathrm{e}^{-\mathrm{g}(\mathrm{x})} \mathrm{g}^{\prime}(\mathrm{x})=0$
$\Rightarrow \int\left(\mathrm{e}^{-f(x)} f^{\prime}(x)-e^{-g(x)} \cdot g^{\prime}(x)\right) d x=C$
$\Rightarrow \quad-\mathrm{e}^{-f(\mathrm{x})}+\mathrm{e}^{-\mathrm{g}(\mathrm{x})}=\mathrm{C}$
$\Rightarrow \quad-\mathrm{e}^{-f(1)}+\mathrm{e}^{-\mathrm{g}(1)}=-\mathrm{e}^{-f(2)}+\mathrm{e}^{-\mathrm{g}(2)}$
$\Rightarrow \quad-\frac{1}{\mathrm{e}}+\mathrm{e}^{-\mathrm{g}(1)}=-\mathrm{e}^{-f(2)}+\frac{1}{\mathrm{e}}$
$\Rightarrow \quad \mathrm{e}^{-f(2)}+\mathrm{e}^{-\mathrm{g}(1)}=\frac{2}{\mathrm{e}}$
$\therefore \quad \mathrm{e}^{-f(2)}<\frac{2}{\mathrm{e}}$ and $\mathrm{e}^{-\mathrm{g}(1)}<\frac{2}{\mathrm{e}}$
$\Rightarrow \quad-f(2)<\ln 2-1$ and $-\mathrm{g}(1)<\ln 2-1$
$\Rightarrow f(2)>1-\ln 2$ and $g(1)>1-\ln 2$
Q. Let $f:(0, \pi) \rightarrow \square$ be a twice differentiable function such that $\lim _{t \rightarrow \pi} \frac{f(x) \sin t-f(t) \sin x}{t-x}=\sin ^{2} x$ for all $\mathrm{x} \in(0, \pi) .$ If $f\left(\frac{\pi}{6}\right)=-\frac{\pi}{12},$ then which of the following statement(s) is (are) TRUE?
(A) $f\left(\frac{\pi}{4}\right)=\frac{\pi}{4 \sqrt{2}}$
(B) $f(\mathrm{x})<\frac{\mathrm{x}^{4}}{6}-\mathrm{x}^{2}$ for all $\mathrm{x} \in(0, \pi)$
(C) There exists $\alpha \in(0, \pi)$ such that $f^{\prime}(\alpha)=0$
(D) $f^{\prime \prime}\left(\frac{\pi}{2}\right)+f\left(\frac{\pi}{2}\right)=0$
[JEE(Advanced)-2018,4(-2)]
Ans. (B,C,D)
Q. Let $f: \square \rightarrow \square$ be a differentiable function with $f(0)=0 .$ If $\mathrm{y}=f(\mathrm{x})$ satisfies the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=(2+5 \mathrm{y})(5 \mathrm{y}-2),$ then the value of $\lim _{x \rightarrow-\infty} f(\mathrm{x})$ is
[JEE(Advanced)-2018,3(0)]
Ans. 0.4
