Calculus PYQ — Page 3

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be such that $f(xy) = f(x)f(y)$, for all $x, y \in \mathbb{R}$ and $f(0) \neq 0$. Let $g: [1, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $x^2 g(x) = \int\limits_1^x (t^2 f(t) - tg(t))\,dt$. Then $g(2)$ is equal to :

Let $f(x) = \begin{cases} x^3 + 8 ; & x < 0 \\ x^2 - 4 ; & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} (x-8)^{1/3} ; & x < 0 \\ (x+4)^{1/2} ; & x \geq 0 \end{cases}$. Then the number of points, where the function $g \circ f$ is discontinuous, is __________.

Let $y = y(x)$ be the solution of the differential equation $x\sin\left(\dfrac{y}{x}\right)dy = \left(y\sin\left(\dfrac{y}{x}\right) - x\right)dx$, $y(1) = \dfrac{\pi}{2}$ and let $\alpha = \cos\left(\dfrac{y(e^{12})}{e^{12}}\right)$. Then the number of integral values of $p$, for which the equation $x^2 + y^2 - 2px + 2py + \alpha + 2 = 0$ represents a circle of radius $r \leq 6$, is __________.

Let $f(t)=\int\left(\frac{1-\sin \left(\log _{e} t\right)}{1-\cos \left(\log _{e} t\right)}\right) d t, t>1$. If $f\left(e^{\pi / 2}\right)=-e^{\pi / 2}$ and $f\left(e^{\pi / 4}\right)=\alpha e^{\pi / 4}$, then $\alpha$ equals

Let $f(x)=\int \frac{\left(2-x^{2}\right) \cdot \mathrm{e}^{x}}{(\sqrt{1+x})(1-x)^{3 / 2}} \mathrm{~d} x$. If $f(0)=0$, then $f\left(\frac{1}{2}\right)$ is equal to:

The value of $\lim _{x \rightarrow 0} \frac{\log _{e}\left(\sec (e x) \cdot \sec \left(e^{2} x\right) \cdot \ldots \cdot \sec \left(e^{10} x\right)\right)}{e^{2}-e^{2 \cos x}}$ is equal to

The value of the integral $\displaystyle\int_{\pi/6}^{\pi/3} \left(\dfrac{4 - \csc^2 x}{\cos^4 x}\right) dx$ is:

If $\alpha = \displaystyle\int_0^{2\sqrt{3}} \log_2(x^2 + 4)\,dx + \displaystyle\int_2^4 \sqrt{2^x - 4}\,dx$, then $\alpha^2$ is equal to _______.

If $\int\left(\frac{1-5 \cos ^{2} x}{\sin ^{5} x \cos ^{2} x}\right) d x=f(x)+\mathrm{C}$, where C is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to

If $f(x)$ satisfies the relation $f(x)=e^{x}+\int_{0}^{1}\left(y+x e^{x}\right) f(y) d y$, then $e+f(0)$ is equal to $\_\_\_\_$.

Let $\mathrm{I}(x)=\int \frac{3 d x}{(4 x+6)\left(\sqrt{4 x^{2}+8 x+3}\right)}$ and $\mathrm{I}(0)=\frac{\sqrt{3}}{4}+20$. If $\mathrm{I}\left(\frac{1}{2}\right)=\frac{a \sqrt{2}}{b}+\mathrm{c}$, where $a, b, \mathrm{c} \in \mathrm{N}, \operatorname{gcd}(a, b)=1$, then $a+b+c$ is equal to

The area of the region $\{(x, y) : x^2 - 8x \leq y \leq -x\}$ is :

The area of the region $\{(x, y): y \leq \pi - |x|, y \leq |x \sin x|, y \geq 0\}$ is:

Let $\mathrm{A}_{1}$ be the bounded area enclosed by the curves $y=x^{2}+2, x+y=8$ and $y$-axis that lies in the first quadrant. Let $\mathrm{A}_{2}$ be the bounded area enclosed by the curves $y=x^{2}+2, y^{2}=x, x=2$, and $y$-axis that lies in the first quadrant. Then $\mathrm{A}_{1}-\mathrm{A}_{2}$ is equal to

The area of the region $\mathrm{R}=\left\{(x, y): x y \leq 8,1 \leq y \leq x^{2}, x \geq 0\right\}$ is

The area of the region $\mathrm{A}=\left\{(x, y): 4 x^{2}+y^{2} \leqslant 8\right.$ and $\left.y^{2} \leqslant 4 x\right\}$ is:

Let $f$ be a twice differentiable non-negative function such that $(f(x))^{2}=25+\int_{0}^{x}\left((f(\mathrm{t}))^{2}+\left(f^{\prime}(\mathrm{t})\right)^{2}\right) \mathrm{dt}$. Then the mean of $f\left(\log _{\mathrm{e}}(1)\right), f\left(\log _{\mathrm{e}}(2)\right), \ldots.., f\left(\log _{\mathrm{e}}(625)\right)$ is equal to $\_\_\_\_$.

The value of $\int_0^{20\pi} (\sin^4 x + \cos^4 x) \, dx$ is equal to:

If $\displaystyle\int_{\pi/6}^{\pi/4}\left(\cot\left(x-\dfrac{\pi}{3}\right)\cot\left(x+\dfrac{\pi}{3}\right)+1\right)dx = \alpha\log_e(\sqrt{3}-1)$, then $9\alpha^2$ is equal to ________.

Let $P_{1}: y=4 x^{2}$ and $P_{2}: y=x^{2}+27$ be two parabolas. If the area of the bounded region enclosed between $P_{1}$ and $P_{2}$ is six times the area of the bounded region enclosed between the line $y=\alpha x, \alpha>0$ and $P_{1}$, then $\alpha$ is equal to :

The area of the region $R = \{(x, y): xy \leq 27, 1 \leq y \leq x^2\}$ is equal to:

The area of the region, inside the ellipse $x^{2}+4 y^{2}=4$ and outside the region bounded by the curves $y=|x|-1$ and $y=1-|x|$, is :

Let $y = y(x)$ be the solution of the differential equation $\dfrac{dy}{dx} = (1 + x + x^2)(1 - y + y^2)$, $y(0) = \dfrac{1}{2}$. Then $(2y(1) - 1)$ is equal to:

If $\lim _{x \rightarrow 0} \frac{\mathrm{e}^{(\mathrm{a}-1) x}+2 \cos \mathrm{~b} x+(\mathrm{c}-2) \mathrm{e}^{-x}}{x \cos x-\log _{\mathrm{e}}(1+x)}=2$, then $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}$ is equal to :