Free Objective Test 02 Practice Test - 11th and 12th 

Given $y=x^3-3x^2+6$
Differentiating y w.r.t. 'x', $\frac{dy}{dx}=3x^2-6x$
Putting $\frac{dy}{dx}=0$, we will get the values at which function is maximum or minimum
$\Rightarrow 3x^2-6x=0\Rightarrow x(3x-6)=0$
$\Rightarrow x=0,+2$
To distinguish values of x as the point of maximum or minimum, we need 2nd derivative of the function.
$\therefore \frac{d^{2}y}{d{x}^2}=6x-6;Now(\frac{d^{2}y}{d{x}^2}{)}_{x=0}=-6<0$
$\therefore Atx=0\Rightarrow \text{It is maximum}{\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=2}=6\left(2\right)-6=6>0\therefore Atx=2\Rightarrow \text{It is minimum}$
$Hencex=0isapointofmaximumandx=2isapointofminimumSo,maximumvalueofy=0^3-3.0+6=6minimumvalueofy=(2{)}^3-3(2{)}^2+6=2$

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)
=$2e^{4x}cos2x+sin2x4e^{4x}=2e^{4x}[cos2x+2sin2x]$

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 $\frac{dy}{dx}=\frac{dy}{du}.\frac{du}{dx}$
$\frac{dy}{dx}=\left(cosu\right).\left(2x\right)=2x.cos(x^2-4)$

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

$\frac{d{e}^x}{dx}=e^x$
$\frac{dy}{dx}$ for some common functions
$\begin{array}{cccc}y& \frac{dy}{dx}& y& \frac{dy}{dx}\\ x^n& n{x}^{n-1}& secx& secxtanx\\ sinx& cosx& cosecx& -cosecxcotx\\ cosx& -sinx& lnx& \frac{1}{x}\\ tanx& se{c}^{2}x& e^x& e^x\\ cotx& -cose{c}^{2}x& & \end{array}$

$y=e^{x}sinx$.
So $\frac{dy}{dx}=\frac{d}{dx}\left(e^{x}sinx\right)=e^x\frac{d}{dx}\left(sinx\right)+sinx\frac{d}{dx}\left(e^x\right)$
=$e^{x}cosx+e^{x}sinx=e^x(cosx+sinx)$.

Yes

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 = $x\frac{d\left(lnx\right)}{dx}+lnx\frac{d\left(x\right)}{dx}=x\frac{1}{x}+lnx.1=1+lnx$

$i=i_0{e}^{\frac{-t}{RC}}$
$i_{0}R,C$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
$\frac{di}{dt}=\frac{-i_o}{RC}{e}^{\frac{-10Rc}{RC}}=\frac{-i_o}{Rc{e}^{10}}$

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 

$A=\underset{0}{\overset{6}{\int }}{x}^{2}dx$

=$[\frac{x^3}{3}{]}_0^6=\frac{216-0}{3}=72$

The required area is
${\int }_0^{\infty }{e}^{-x}dx$
= $-e^{-x}{|}_0^{\infty }$

= $\frac{-1}{e^{\infty }}-[-\frac{1}{e^{\circ }}]$

= -0 + 1

= 1

$I={\int }_0^{10}se{c}^2(3x+6)dx$

Let $3x+6=u$

$\Rightarrow 3dx=du\Rightarrow dx=\frac{du}{3}$

$I=\int se{c}^{2}u\frac{du}{3}$

=$\frac{1}{3}tan\left(u\right)+c$
$I=\frac{1}{3}tan(3x+6{)}_0^{10}$
$\frac{1}{3}[tan36-tan6]$

Let $u=8x+6,\frac{du}{dx}=8\Rightarrow dx=\frac{du}{8}$

$t=7x+5,\frac{dt}{dx}=7\Rightarrow dx=\frac{dt}{7}$

$\int ydx=\int cosu\frac{du}{8}+\int cose{c}^{2}t\frac{dt}{7}+6\int secxtanxdx$

= $\frac{sinu}{8}+\frac{(-cott)}{7}+6secx+c$

= $\frac{sin(8x+6)}{8}-\frac{cot(7x+5)}{7}+6secx+c$

$y=e^{2x}+\frac{1}{x^2}$

Integration both side w.r.t. 'x',

$I=\int y.dx=\int ⟮e^{2x}+\frac{1}{x^2}⟯dx=\frac{e^{2x}}{2}+\frac{x^{-2+1}}{-2+1}+c[\because \int e^{ax}=\frac{e^{ax}}{a}+c]$

$I=\frac{e^{2x}}{2}-\frac{1}{x}+c$

$I={\int }_{\infty }^R\frac{GMm}{x^2}dx=GMm{\int }_{\infty }^R\frac{dx}{x^2}=GMm[-\frac{1}{x}{]}_{\infty }^R$

=$GMm|-\frac{1}{R}+\frac{1}{\infty }|=\frac{-GMm}{R}$

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).