Calculus PYQ — Page 32

If $y = \frac{\log_e x}{x}$, then $\frac{d^2y}{dx^2} =$

If $\frac{d}{dx}(f(x)) = 5x^4 - \frac{4}{x^5}$ such that $f(1) = 0$. Then $f(2) - 2f\left(\frac{1}{2}\right)$ is equal to :

$\int \sqrt{1 - 49x^2} \, dx$ is equal to

The area of the region bounded by the curve $x = y^2$ and the line $x = 4$ is equal to :

$\int_{\frac{1}{3}}^{1} \frac{(x - x^3)^{\frac{1}{3}}}{x^4} dx =$

Match List I with List II | List - I | List - II | |---|---| | A. $\int \frac{dx}{x^2 - a^2} =$ | I. $\log_e\left\lvert x + \sqrt{x^2 + a^2}\right\rvert + C$ | | B. $\int \frac{dx}{a^2 - x^2} =$ | II. $\frac{1}{2a}\log_e\left\lvert\frac{x-a}{x+a}\right\rvert + C$ | | C. $\int \frac{dx}{\sqrt{x^2 - a^2}} =$ | III. $\frac{1}{2a}\log_e\left\lvert\frac{a+x}{a-x}\right\rvert + C$ | | D. $\int \frac{dx}{\sqrt{x^2 + a^2}} =$ | IV. $\log_e\left\lvert x + \sqrt{x^2 - a^2}\right\rvert + C$ | Choose the correct answer from the options given below:

$\int \sqrt{x^2 - 4x + 5} \, dx =$

The line $ax + by = 7$ is a tangent to the curve $y = \frac{x-7}{(x-2)(x-3)}$ at the point where it cuts the x-axis A. The y-intercept of the line is $-0.7$ B. $b = -7$ C. $a = 1$ D. the line passes through the point $(-13, -1)$ E. $b = -20$ Choose the correct answer from the options given below:

If $f(x) = \begin{cases} 1 - 2x, & x \leq 0 \\ 1 + 2x, & x > 0 \end{cases}$, then $\int_{-1}^{1} f(x) \, dx =$

The tangent to the curve $y^2 + 2x - 5 = 0$ at the point (h, k) is parallel to the line $x + 2y = 4$, then the value of 'k' is:

The interval in which $f(x) = \frac{x}{2} + \frac{2}{x}$ is a decreasing function of $x$ is :

The smaller of the areas enclosed by the circle $x^2 + y^2 = 4$ and the line $x + y = 2$ is

If x is real, then minimum value of $x^2 - 8x + 17$ is :

Match List I with List II | List - I | List - II | |---|---| | A. $\int_{-\pi/2}^{\pi/2} \sin^7 x \, dx$ | I. $\frac{\pi}{2}$ | | B. $\int_{-\pi/2}^{\pi/2} \sin^2 x \, dx$ | II. $\frac{\pi}{4}$ | | C. $\int_0^{\pi/2} \frac{1}{1 + \tan x} \, dx$ | III. 0 | | D. $\int_0^{\pi} \lvert \cos x \rvert \, dx$ | IV. 2 |

General solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} = 0$ is A. $\tan^{-1} x + \tan^{-1} y = C$ B. $\sin^{-1} x - \cos^{-1} y = C$ C. $x\sqrt{1 - y^2} - y\sqrt{1 - x^2} = C$ D. $\sin^{-1} x + \sin^{-1} y = C$ E. $\cos^{-1} x + \cos^{-1} y = C$ (where C is arbitrary constant) Choose the correct answer from the options given below

If $x^y = e^{x-y}$, then $\frac{dy}{dx} =$

Based on above information answer the following question : The area of the flower bed $(A(x))$ is given by :

Derivative of $x^3 + 1$ with respect to $x^2 + 1$ is

The value of $\int_0^1 e^x (x + 1) \, dx$ is equal to

The interval in which the function $f(x) = 2x^3 + 3x^2 - 12x + 1$ is strictly increasing is -

$\int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx =$

The interval in which the function, $f(x) = 7 - 4x - x^2$ is strictly increasing is

The differential equation representing family of curves $y = e^{-2x}(a \cos x + b \sin x)$, where a and b are arbitrary constant, is :

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