Calculus PYQ — Page 33

If $\int_0^{\pi/2} \sqrt{\tan x} \, dx = \frac{\lambda}{\sqrt{2}}$, then the value of $\lambda$ is

$\int_0^{4\pi} \frac{x}{1 + |\cos x|} dx =$

The value of $\int_{\pi/2}^{\pi} \frac{1}{1 + \cot x} dx$ is equal to:

If $U(x) = x + \sqrt{1 + x^2}$, then solution of the differential equation $\frac{dy}{dx} + \sqrt{\frac{1 + y^2}{1 + x^2}} = 0$, is :

If $y = x^{\sin x}$, then value of $x^{-\sin x} \frac{dy}{dx} - \cos x \log x$ is :

If $2x + y = 6$ then the maximum value of $x^2 y$ is :

If $y = \frac{\sqrt{x+1} + \sqrt{x-1}}{\sqrt{x+1} - \sqrt{x-1}}$, then $(x^2-1)^{3/2} \frac{d^2y}{dx^2} =$

Match List I with List - II | List - I | List - II | |---|---| | A. An even function | I. $x^2 + \cos x$ | | B. For an even function, $\int_{-a}^{a} f(x)dx =$ | II. 0 | | C. If $f(2a-x) = -f(x)$, then $\int_{0}^{2a} f(x)dx =$ | III. $2\int_{0}^{a} f(x)dx$ | | D. An odd function | IV. $x^3 + \sin x$ | Choose the correct answer from the options given below:

The line $y = x$, partition the area of the circle $(x-1)^2 + y^2 = 1$, into two segments. The area of the major segment is

If $f'(x) = 4x^5 - 6x$ and $f(0) = 3$, then $f(3)$ is equal to

$\int \frac{x}{(x-1)(x-2)} dx =$ (where c is an arbitrary constant)

$\int \frac{dx}{\sin^2 x \cos^2 x} =$

Consider the function $f(x) = x^{\frac{1}{x}}$. Its

If $x = at^2$ and $y = 2at$, then the value of $\frac{d^2y}{dx^2}$ is ( where t is a parameter )

The equation of the normal to the curve $y = x - \frac{1}{x}$ at $(1,0)$ is:

Let $\int \frac{dx}{\left(\sqrt{x} - \sqrt{x-1}\right)^2} = \alpha u(x) + \beta v(x) + C$, where $u(x) = x^2 - x + \left(x - \frac{1}{2}\right)\sqrt{x^2 - x}$ and $v(x) = \log\left|x - \frac{1}{2} + \sqrt{x^2 - x}\right|$. The value of $\alpha + \beta$ is :

The value of $\int_{-3}^{3} \log_e\left(\frac{4+x}{4-x}\right) dx$ is

$\int \frac{x^2 + 4}{(x^2 + 3)(x^2 + 5)} \, dx = u \tan^{-1}\left(\frac{x}{\sqrt{3}}\right) + v \tan^{-1}\left(\frac{x}{\sqrt{5}}\right) + c$, where c is arbitrary constant, then value of $\frac{1}{u^2} + \frac{1}{v^2}$ is equal to :

The area of the smaller region bounded by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$ is :

The area of the region bounded by $f(x) = x|x|$, x-axis and from $x = -1$ to $x = 1$ is :

The integrating factor of the differential equation $x\frac{dy}{dx} - y = 2x^2$ is :

$\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} dx =$

The value of $\frac{dy}{dx}$ when $x = \frac{1}{6}$, is

The value of $\int_{-4}^{4} \log_e \left( \frac{1-x}{1+x} \right) dx$ is equal to