Calculus PYQ — Page 55
JEE Main Mathematics — Calculus previous year questions with solutions.
All Calculus Questions (1411)
Consider a rectangle whose length is increasing at the uniform rate of $2 \mathrm{~m} / \mathrm{sec}$, breadth is decreasing at the uniform rate of $3 \mathrm{~m} / \mathrm{sec}$ and the area is decreasing at the uniform rate of $5 \mathrm{~m}^2 / \mathrm{sec}$. If after some time the breadth of the rectangle is $2 \mathrm{~m}$ then the length of the rectangle is
Statement 1: The degrees of the differential equations $\frac{d y}{d x}+y^2=x$ and $\frac{d^2 y}{d x^2}+y=\sin x$ are equal. Statement 2: The degree of a differential equation, when it is a polynomial equation in derivatives, is the highest positive integral power of the highest order derivative involved in the differential equation, otherwise degree is not defined.
Let $f(x)$ be an indefinite integral of $\cos ^3 x$. Statement 1: $f(x)$ is a periodic function of period $\pi$. Statement 2: $\cos ^3 x$ is a periodic function.
If $f^{\prime}(x)=\sin (\log x)$ and $y=f\left(\frac{2 x+3}{3-2 x}\right)$, then $\frac{d y}{d x}$ equals
The area of the region enclosed by the curves $y=x, x=e, y=\frac{1}{x}$ and the positive $x$-axis is
The value of $\int_0^1 \frac{8 \log (1+x)}{1+x^2} d x$ is
The shortest distance between line $y-x=1$ and curve $x=y^2$ is
Let I be the purchase value of an equipment and $\mathrm{V}(\mathrm{t})$ be the value after it has been used for t years. The value $\mathrm{V}(\mathrm{t})$ depreciates at a rate given by differential equation $\frac{\mathrm{dV}(\mathrm{t})}{\mathrm{dt}}=-\mathrm{k}(\mathrm{T}-\mathrm{t})$, where $\mathrm{k}>0$ is a constant and $\mathrm{T}$ is the total life in years of the equipment. Then the scrap value $\mathrm{V}(\mathrm{T})$ of the equipment is
$$ \lim _{x \rightarrow 2}\left(\frac{\sqrt{1-\cos \{2(x-2)\}}}{x-2}\right)
The value of $p$ and $q$ for which the function $f(x)=\left\{\begin{array}{cl}\frac{\sin (p+1) x+\sin x}{x} & x < 0 \\ q & , x=0 \\ \frac{\sqrt{x+x^2}-\sqrt{x}}{x^{3 / 2}} & , x>0\end{array}\right.$ is continuous for all $\mathrm{x}$ in $\mathrm{R}$, is
$\frac{d^2 x}{d y^2}$ equals
For $x \in\left(0, \frac{5 \pi}{2}\right)$, define $f(x)=\int_0^x \sqrt{t} \sin t d t$. Then $f$ has
If $\frac{d y}{d x}=y+3>0$ and $y(0)=2$, then $y(\ln 2)$ is equal to
Let $f: R \rightarrow R$ be a continuous function defined by $f(x)=\frac{1}{e^x+2 e^{-x}}$. Statement-1: $f(c)=\frac{1}{3}$, for some $c \in R$. Statement-2: $0 < f(x) \leq \frac{1}{2 \sqrt{2}}$, for all $x \in R$
Let $f:(-1,1) \rightarrow R$ be a differentiable function with $f(0)=-1$ and $f^{\prime}(0)=1$. Let $g(x)=[f(2 f(x)+2)]^2$. Then $g^{\prime}(0)=$
Let $p(x)$ be a function defined on $R$ such that $p^{\prime}(x)=p^{\prime}(1-x)$, for all $x \in[0,1], p(0)=1$ and $p(1)=41$. Then $\int_0^1 p(x) d x$ equals
Let $f: R \rightarrow R$ be a positive increasing function with $\lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}=1$. Then $\lim _{x \rightarrow \infty} \frac{f(2 x)}{f(x)}=$
Solution of the differential equation $\cos x d y=y(\sin x-y) d x, 0 < x < \frac{\pi}{2}$ is
Let $f: R \rightarrow R$ be defined by $f(x)=\left\{\begin{array}{ll}k-2 x, & \text { if } x \leq-1 \\ 2 x+3, & \text { if } x>-1\end{array}\right.$. If $f$ has a local minimum at $x=-1$, then a possible value of $\mathrm{k}$ is
The area bounded by the curves $y=\cos x$ and $y=\sin x$ between the ordinates $x=0$ and $x=\frac{3 \pi}{2}$ is
$\int_0^\pi[\cot x] d x,[\bullet]$ denotes the greatest integer function, is equal to
Let $f(x)=x|x|$ and $g(x)=\sin x$. Statement-1 : gof is differentiable at $x=0$ and its derivative is continuous at that point. Statement-2 : gof is twice differentiable at $x=0$.
Let $y$ be an implicit function of $x$ defined by $x^{2 x}-2 x^x \cot y-1=0$. Then $y^{\prime}(1)$ equals
The shortest distance between the line $\mathrm{y}-\mathrm{x}=1$ and the curve $\mathrm{x}=\mathrm{y}^2$ is