Calculus PYQ — Page 31

$\int x\sqrt{x + 2} \, dx$ is equal to :

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

$\int \frac{x}{(x^2+3)(x^2+4)} dx =$

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

Match List I with List II | List I | List II : Order and degree respectively | |---|---| | A. $\frac{dy}{dx} - (x^2+3) = 0$ | I. 2 and 1 | | B. $2x^2 \frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ | II. 2 and 3 | | C. $y''' + y^2 + e^{y'} = 0$ | III. 1 and 1 | | D. $\left(\frac{ds}{dt}\right)^4 + 3s\left(\frac{d^2s}{dt^2}\right)^3 = 0$ | IV. 3 and not defined | Choose the correct answer from the options given below:

The slope of normal to the curve $y = kx^2 - 3x + 2$ at $x = \frac{1}{2}$ is 5. The value of 'k' is

Solution of the differential equation $\frac{dy}{dx} = x + xy - (1 + y)$ is:

Match List I with List II | List I | List II | |---|---| | A. The number of arbitrary constants in the particular solution of differential equation of order 2 | I. 1 | | B. The number of arbitrary constants in the general solution of differential equation of order 2 | II. 0 | | C. The integrating factor of differential equation $\frac{dy}{dx} + \frac{1}{x}y = 3, x > 0$, is | III. 2 | | D. For differential equation, $x^2 \frac{dy}{dx} + x = xy, x > 0, \lim_{x \to 0^+} y(x)$ is equal to | IV. $x$ | Choose the correct answer from the options given below:

If $f$ is a function defined by $f(x) = \begin{cases} 5x^2 - x + 3, & x < 1 \\ 3x + 4, & x \geq 1 \end{cases}$, then, at $x = 1$, $f$ is

If the solution curve of the differentiable equation $\frac{dy}{dx} + 2y = e^{3x}$, passes through the point $\left(0, \frac{6}{5}\right)$, then the value of $y(\log_e 2)$ is:

Evaluate $\int_0^{\frac{1}{3}} y \, dx$

If $\int(\sqrt{x+1} + \sqrt{x-1})^2 dx = \alpha x^2 + \beta x\sqrt{x^2-1} + \gamma \log|x+\sqrt{x^2-1}| + C$, then value of $\alpha + \beta - 2\gamma$ is :

Based on above information answer the following question : The maximum area (in $m^2$) of the flower bed is :

Based on above information answer the following question : If area of the flower bed is maximum, then area (in $m^2$) of the garden, which is outside the flower bed is :

$f(x) = [x]$, where $[\,]$ represents greatest integer function. (A) For $2 \leq x < 3$, $[x] = 3$ (B) For $2 \leq x < 3$, $[x] = 2$ (C) Right hand derivative of f at $x = 2$ is not defined (D) Left hand derivative of f at $x = 2$ is zero (E) $f(x)$ is not differentiable at $x = 2$ Choose the correct answer from the options given below :

For $a, b \in R$ and $a < b$, $\int_{a}^{b} \frac{f(x)}{f(x) + f(a + b - x)} \, dx =$

If curve represented by differential equation $x \frac{dy}{dx} + y = e^x$ passes through (1, 1), then $y(-1)$ is :

The derivative of $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ w.r.t. $\tan^{-1}\left(\frac{2x}{1-x^2}\right)$ is

If the order and the degree of the differential equation $\left(\frac{dy}{dx}\right)^{\frac{1}{2}} = \left(\frac{d^2y}{dx^2}\right)^{\frac{1}{5}}$ are O and S respectively, then $S - O$ is equal to

Match List - I with List - II | List - I (Differential Equation) | List - II (Degree) | |---|---| | A. $\left[1 + (y')^2\right]^2 = y''$ | I. $2$ | | B. $\left[1 + (y'')^3\right]^{\frac{1}{2}} = (y')^3$ | II. $4$ | | C. $(y''')^2 + y'' + 3y' + 5y = e^x$ | III. $1$ | | D. $\left[1 + (y')^3\right]^{\frac{1}{2}} = (y'')^2$ | IV. $3$ | Choose the correct answer from the option given below:

If $x = e^{y + e^{y + e^{y + \ldots \infty}}}$, $x > 0$, then $\frac{dy}{dx}$ is equal to

The value of $x$ for which $\frac{dy}{dx} = 0$, is

$\int \frac{x^2 + 1}{x^4 + 1} dx =$

The area enclosed by the curve $y^2 = 4ax$ and its latus-rectum is