Answer: Option B. -> -  xcosx - 2sinx   
:
B
Using the Product Rule, we have f'(x)=x$\frac{d}{dx}$(cos x)+cos x$\frac{d}{dx}$(x)=-xsinx+cosx
To find f"(x) we differentiate f'(x):
f"(x)=$\frac{d}{dx}$(-x sin x+cos x)=-x$\frac{d}{dx}$(sin x)+sin x $\frac{d}{dx}$(-x)+$\frac{d}{dx}$(cos x)
=-x cos x-sin x-sin x = -x cos x-2 sin x

Answer: Option C. -> e2x2−1x+c
:
C
$y=e^{2x}+\frac{1}{x^2}$
Integration both side w.r.t. 'x',
$I=\int y.dx=\int ⟮e^{2x}+\frac{1}{x^2}⟯dx=\frac{e^{2x}}{2}+\frac{x^{-2+1}}{-2+1}+c[\because \int e^{ax}=\frac{e^{ax}}{a}+c]$
$I=\frac{e^{2x}}{2}-\frac{1}{x}+c$

Answer: Option A. -> 2xcos(x2−4)
:
A
We now know how to differentiate sin x and $x^2-4$, but how do we differentiate a composite like sin$(x^2-4)$? the answer is, with the Chain Rule, which says that the derivatives of the composite of two differentiable fuctions is the product of their derivatives evaluated at appropriate points. The Chain Rule is probably the most used differentiation rule in mathematics.
Let $u=x^2-4$
Then y=sin u
We know the differentiation of u w.r.t x and differentiation of y w.r.t u.
$\frac{du}{dx}=2x\&\frac{dy}{du}=cosu$ we can write $\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}$
$\frac{dy}{dx}=\left(cosu\right).\left(2x\right)=2x.cos(x^2-4)$

Answer: Option C. -> 2e4x[cos2x+2sin2x]
:
C
Here $u=e^{4x}v=si{n}^{2x}$, Differentiating both sides w.r.t. 'x'
$\frac{dy}{dx}=\frac{d}{dx}\left[e^{4x}sin2x\right]$
=$e^{4x}\frac{d}{dx}\left[sin2x\right]+sin2x\frac{d}{dx}\left[e^{4x}\right]$(Using product rule)
=$2e^{4x}cos2x+sin2x4e^{4x}=2e^{4x}[cos2x+2sin2x]$

Answer: Option A. -> −GMmR
:
A
$I={\int }_{\infty }^R\frac{GMm}{x^2}dx=GMm{\int }_{\infty }^R\frac{dx}{x^2}=GMm[-\frac{1}{x}{]}_{\infty }^R$
=$GMm|-\frac{1}{R}+\frac{1}{\infty }|=\frac{-GMm}{R}$
This expression represents the gravitational potential energy which is obtained by integrating $GMm/x^2$ between the limits infinity $\left(\infty \right)$ and radius of the earth (R).

Answer: Option A. -> 1 + lnx
:
A
Here u=x, v=ln x
Differentiating both sides w.r.t. 'x', $\frac{dy}{dx}=\frac{d\left[xlnx\right]}{dx}$
Using product rule = $x\frac{d\left(lnx\right)}{dx}+lnx\frac{d\left(x\right)}{dx}=x\frac{1}{x}+lnx.1=1+lnx$

Answer: Option B. -> 72 unit2
:
B
The area can be divided into strips by drawing ordinates between x = 0 and x = 6 at a regular interval of dx. Consider the strip between the ordinates at x and x + dx. The height of this strip is $y=x^2$. The area of this strip is dA = y dx= $x^{2}dx$.
The total area of the shaded part is obtained by summing up these strip-areas with x varying from 0 to 6. Thus
$A=\underset{0}{\overset{6}{\int }}{x}^{2}dx$
=$[\frac{x^3}{3}{]}_0^6=\frac{216-0}{3}=72$

Answer: Option A. -> x-(i); y-(iii); z-(iv)
:
A
$i=i_0{e}^{\frac{-t}{RC}}$
$i_{0}R,C$are constants
To find the rate of change of i
i.e.,
$\frac{di}{dt}$
$\Rightarrow \frac{di}{dt}=\frac{-i_o}{RC}{e}^{\frac{-t}{RC}}$
(x) at t=0
$\frac{di}{dt}=\frac{-i_o}{RC}$
(y) at t =RC
$\frac{di}{dt}=\frac{-i_o}{RC}{e}^{-1}=\frac{-i_o}{RCe}$
(z) at t = 10RC
$\frac{di}{dt}=\frac{-i_o}{RC}{e}^{\frac{-10Rc}{RC}}=\frac{-i_o}{Rc{e}^{10}}$

Answer: Option D. -> None of these
:
D
$I={\int }_0^{10}se{c}^2(3x+6)dx$
Let $3x+6=u$
$\Rightarrow 3dx=du\Rightarrow dx=\frac{du}{3}$
$I=\int se{c}^{2}u\frac{du}{3}$
=$\frac{1}{3}tan\left(u\right)+c$
$I=\frac{1}{3}tan(3x+6{)}_0^{10}$
$\frac{1}{3}[tan36-tan6]$

Answer: Option C. -> sin(8x+6)8−cot(7x+5)7+6sec x + c
:
C
Let $u=8x+6,\frac{du}{dx}=8\Rightarrow dx=\frac{du}{8}$
$t=7x+5,\frac{dt}{dx}=7\Rightarrow dx=\frac{dt}{7}$
$\int ydx=\int cosu\frac{du}{8}+\int cose{c}^{2}t\frac{dt}{7}+6\int secxtanxdx$
= $\frac{sinu}{8}+\frac{(-cott)}{7}+6secx+c$
= $\frac{sin(8x+6)}{8}-\frac{cot(7x+5)}{7}+6secx+c$