Differential Equation – JEE Advanced Previous Year Questions with Solutions
JEE Advanced Previous Year Questions of Math with Solutions are available at eSaral. Practicing JEE Advanced Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.eSaral helps the students in clearing and understanding each topic in a better
Q. Match the statements/expressions in Column-I with the open intervals in Column-II.

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. ( (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)$$y(\ln 2)=3 Q. Match the statements/expressions in Column-I with the values given in Column-II. [JEE 2009, (2+2+2+2) × 2] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. ( (A) P (B) q,s (C) q,r,s,t (D) r )Given \mathrm{y}=f(\mathrm{x})Tangent at point \mathrm{P}(\mathrm{x}, \mathrm{y})$$\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}$$\Rightarrow \frac{d y}{d x}-\frac{y}{x}=-x^{2} is a linear differential equation Q. Let f be a real valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If the y-intercept of the tangent at any point P(x,y) on the curve y =f(x) is equal to the cube of the abscissa of P, then the value of f(–3) is equal to [JEE 2010, 3] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. 9 Q. (A) Let f:[1, \infty) \rightarrow[2, \infty) be a differentiable function such that f(1)=2 . If 6 \int_{1}^{x} f(t) d t=3 x f(x)-x^{3} for all x \geq 1, then the value of f(2) is(B) Let y^{\prime}(x)+y(x) g^{\prime}(x)=g(x) g^{\prime}(x), y(0)=0, x \in R, where f^{\prime}(x) denotes \frac{d f(x)}{d x} and g(x) is a given non-constant differentiable function on \mathrm{R} with \mathrm{g}(0)=\mathrm{g}(2)=0 . Then the value of \mathrm{y}(2) is [JEE 2011, 4] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. ( (A) Bonus (B) 0 )(a) (Bonus) (Comment : The given relation does not hold for x =1, therefore it is not an identity. Hence there is an error in given question. The correct identity must be-)6 \int_{1}^{x} f(t) d t=3 x f(x)-x^{3}-5, \forall x \geq 1Now applying Newton Leibnitz theorem6 f(x)=3 x f^{\prime}(x)-3 x^{2}+3 f(x)$$\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)$$y(x) \cdot e^{g(x)}=g(x) \cdot e^{g(x)}-e^{g(x)}+cput \mathrm{x}=0$$\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$$\Rightarrow \quad \mathrm{y}(2)=0 Q. If y(x) satisfies the differential equation y’ – ytanx = 2x sec x and y(0) = 0, then(A) \mathrm{y}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{8 \sqrt{2}}(B) y^{\prime}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{18}(C) \mathrm{y}\left(\frac{\pi}{4}\right)=\frac{\pi^{2}}{9}(D) \mathrm{y}^{\prime}\left(\frac{\pi}{3}\right)=\frac{4 \pi}{3}+\frac{2 \pi^{2}}{3 \sqrt{3}} [JEE 2012, 4M] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,D)\frac{\mathrm{dy}}{\mathrm{dx}}-y \tan \mathrm{x}=2 \mathrm{x} \sec \mathrm{x}$$\mathrm{I.F.}=\mathrm{e}^{\int-\tan \mathrm{xdx}}=\cos \mathrm{x}$$\therefore Equation reduces toy. \cos x=\int 2 x \cdot \sec x \cdot \cos x d x$$\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)$$(D)\left(0, \frac{e-1}{2}\right) [JEE(Advanced) 2013, 2M] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D)f^{\prime}(x)-2 f(x)<0Multiply both side by e ^{-2 x}$$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)<g\left(\frac{1}{2}\right)$$\Rightarrow e^{-2 x} f(x)<\frac{1}{e}$$\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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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}$

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
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) $f^{\prime}(x)<f(x), \frac{1}{4}<x<\frac{3}{4}$(B) $f^{\prime}(x)>f(x), 0<x<\frac{1}{4}$(C) $f^{\prime}(x)<f(x), 0<x<\frac{1}{4}$(D) $f^{\prime}(x)<f(x), \frac{3}{4}<x<1$ [JEE(Advanced) 2013, 3, (–1)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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.$$f^{\prime}(x)<f(x) in \left(0, \frac{1}{4}\right)option \mathrm{C} Q. Which of the following is true for 0 < x < 1 ?(A) 0<f(x)<\infty(B) -\frac{1}{2}<f(x)<\frac{1}{2}(C) -\frac{1}{4}<f(x)<1(D) -\infty<f(x)<0 [JEE(Advanced) 2013, 3, (–1)] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (D)\mathrm{D}\left(\mathrm{e}^{-\mathrm{x}}\left(f^{\prime}(\mathrm{x})-f(\mathrm{x})\right) \geq 0 \quad \forall \mathrm{x} \in(0,1)\right.$$\mathrm{D}\left(\mathrm{D}\left(\mathrm{e}^{-\mathrm{x}} f(\mathrm{x})\right) \geq 0 \forall \mathrm{x} \in(0,1)\right.$$\mathrm{D}^{2}\left(\mathrm{e}^{-\mathrm{x}} f(\mathrm{x})\right) \geq 0Let \mathrm{F}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} f(\mathrm{x})$$\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)$$f(x)<0Option D is possible Q. The function y = ƒ(x) is the solution of the differential equation \frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}} in (–1, 1) satisfying ƒ(0) = 0. Then \int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(\mathrm{x}) \mathrm{d} \mathrm{x} \text { is }(A) \frac{\pi}{3}-\frac{\sqrt{3}}{2}(B) \frac{\pi}{3}-\frac{\sqrt{3}}{4}(C) \frac{\pi}{6}-\frac{\sqrt{3}}{4}(D) \frac{\pi}{6}-\frac{\sqrt{3}}{2} [JEE(Advanced)-2014, 3(–1)] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (B)\left(\frac{d y}{d x}-\frac{x y}{\left(1-x^{2}\right)}\right) \sqrt{1-x^{2}}=x^{4}+2 x$$\Rightarrow \sqrt{1-x^{2}} d y+d(\sqrt{1-x^{2}}) y=\left(x^{4}+2 x\right) d x$$\Rightarrow y \sqrt{1-x^{2}}=\frac{x^{5}}{5}+x^{2}+cby (0,0) \mathrm{c}=0$$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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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 $(-1,0)$

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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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$$\mathrm{Q}=1+\mathrm{y}^{\prime}+\left(\mathrm{y}^{\prime}\right)^{2} Q. A solution curve of the differential equation \left(\mathrm{x}^{2}+\mathrm{xy}+4 \mathrm{x}+2 \mathrm{y}+4\right) \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}^{2}=0, \mathrm{x}>0 passes through the point (1,3) . The the solution curve-(A) intersects y = x + 2 exactly at one point(B) intersects y = x + 2 exactly at two points(C) intersects y=(x+2)^{2}(D) does NOT intersect y=(x+3)^{2} [JEE(Advanced)-2016] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,D)\left(\mathrm{x}^{2}+\mathrm{xy}+4 \mathrm{x}+2 \mathrm{y}+4\right) \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}^{2}=0$$\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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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$$\Rightarrow f^{\prime}(x)=1-\frac{c}{x^{2}}, c \neq 0(A) \lim _{x \rightarrow 0^{+}} f^{\prime}\left(\frac{1}{x}\right)=\lim _{x \rightarrow 0^{+}}\left(1-c x^{2}\right)=1(B) \lim _{x \rightarrow 0^{+}} x f\left(\frac{1}{x}\right)=\lim _{x \rightarrow 0^{+}} x\left(\frac{1}{x}+c x\right)=\lim _{x \rightarrow 0^{+}}\left(1+c x^{2}\right)=1(C) \lim _{x \rightarrow 0^{+}} x^{2} f^{\prime}(x)=\lim _{x \rightarrow 0^{+}} x^{2}\left(1-\frac{c}{x^{2}}\right)=\lim _{x \rightarrow 0^{+}}\left(x^{2}-c\right)=-c(D) f(x)=x+\frac{c}{x}, c \neq 0for \mathrm{c}>0$$\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]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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}}}}$$\Rightarrow \mathrm{y}(256)=3 Q. If f: \square \rightarrow \square is a differentiable function such that f^{\prime}(\mathrm{x})>2 f(\mathrm{x}) for all \mathrm{x} \in \mathbb{U}, and f(0) =1, then(A) f(\mathrm{x})>\mathrm{e}^{2 \mathrm{x}} in (0, \infty)(B) f(x) is decreasing in (0, \infty)(C) f(x) is increasing in (0, \infty)(D) f^{\prime}(x)<e^{2 x} in (0, \infty) [JEE(Advanced)-2017] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (A,C)Given that,\mathrm{f}^{\prime}(\mathrm{x})>2 \mathrm{f}(\mathrm{x}) \forall \mathrm{x} \in \mathrm{R}$$\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)$$\therefore option (\mathrm{A}) is correctAs, \mathrm{f}^{\prime}(\mathrm{x})>2 \mathrm{f}(\mathrm{x})>2 \mathrm{e}^{2 \mathrm{x}}>2 \forall \mathrm{x} \in(0, \infty)$$\Rightarrow \mathrm{f}(\mathrm{x})$ is strictly increasing on $\mathrm{x} \in(0, \infty)$$\Rightarrow option (\mathrm{C}) is correctAs, we have proved above that \mathrm{f}^{\prime}(\mathrm{x})>2 \cdot \mathrm{e}^{2 \mathrm{x}} \forall \mathrm{x} \in(0, \infty)$$\Rightarrow$ option $(\mathrm{D})$ is incorrectoptions $(\mathrm{A})$ 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)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. (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)]

Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...

Sol. 0.4

way. eSaral also provides complete chapter-wise notes of Class 11th and 12th both for all subjects.Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more.Download eSaral app for free study material and video tutorials.

• September 7, 2021 at 11:52 am

I get 91 percentile in jee mains can I get any nit gen ews

• September 25, 2021 at 10:45 pm

no

• August 7, 2021 at 12:01 pm

Can i clear jee in one month?

• September 24, 2020 at 4:20 pm

Thankuuu beta… ❤

• August 23, 2020 at 11:28 pm

Yeh Prateek sir ko kuch nahi ata . Sirf family hai isliye kam pe liya hai. nahi to isko kaun kam pe leta

• May 29, 2021 at 8:25 am

aree kehna kya chahte ho