Let S be the area of the region bouded by $y= e^{-x^2}$, y = 0, x = 0 and x = 1. Then

(A) $S \geqslant \frac{1}{e}$

(B) $S \geqslant 1-\frac{1}{e}$

(C) $S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$

(D) $S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

Solution: Area S $\geqslant$ area of rectangle 0 ABC $=1 \times \frac{1}{e}$ $S \geqslant \frac{1}{e}$

Also, $e^{-x^2} \geqslant e^{-x}$ for all $x \in[0,1]$ because $x^2 \leqslant x$ for $x \in[0,1]$

$\therefore \quad \int_0^1 e^{-x^2} d x \geqslant \int_0^1 e^{-x} d x=1-\frac{1}{e}$

$\therefore$ Option (B) is correct. By the figure,

$S \leqslant \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

Now,

$\left(1-\frac{1}{e}\right)-\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$

$=\frac{3}{4}-\frac{1}{e}-\frac{1}{4 \sqrt{e}}=\frac{1}{4 e}(3 e-4-\sqrt{e})$

We know that 2<e<3

$\therefore \quad 3 e-4-\sqrt{e}>3 \times 2-4-\sqrt{3}=2-\sqrt{3}>0$

$\therefore \quad 1-\frac{1}{e}>\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$

Since $S \geqslant 1-\frac{1}{e}, S>\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$

$\therefore$ Option (C) is Wrong

Let $f:\left[\frac{1}{2}, 1\right] \rightarrow[0, \infty)$ be a now constant differentiable function such that

$f^{\prime}(x)<2 f(x)$ and $f\left(\frac{1}{2}\right)=1$

Then the value of $\int_{1 / 2}^1 f(x) d x$ lies in the interval

(A) $(2 e-1,2 e)$

(B) $(e-1,2 e-1)$

(C) $\left(\frac{e-1}{2}, e-1\right)$

(D) $\left(0, \frac{e-1}{2}\right)$

$f^{\prime}(x)<2 f(x)$

$\Rightarrow f^{\prime}(x)-2 f(x)<0 \quad \forall x \in\left[\frac{1}{2}, 1\right]$

$\Rightarrow e^{-2 x} \left[f^{\prime}(x)-2 f(x)\right]<0 \ \Rightarrow$

$\Rightarrow \frac{d}{d x}\left[e^{-2 x} f(x)\right]<0 \quad \forall x \in\left[\frac{1}{2}, 1\right] \ \Rightarrow$

$\Rightarrow e^{-2 x} f(x) \text { is a decreasing function } \ in \left[\frac{1}{2}, 1\right] \therefore$

$\therefore \quad e^{-2 x} f(x)<e^{-2\left(\frac{1}{2}\right)} f\left(\frac{1}{2}\right) \quad \forall x>\frac{1}{2}$

$= \frac{1}{e}$

$\Rightarrow \quad f(x)<e^{2 x-1} \quad$ for $x \in\left(\frac{1}{2}, 1\right)$

$\begin{aligned} \Rightarrow \quad \int_{1 / 2}^1 f(x) d x<\int_{1 / 2}^1 e^{2 x-1} d x & =\left.\frac{e^{2 x-1}}{2}\right|_{1 / 2} ^1 \ & =\frac{e}{2}-\frac{1}{2}=\frac{e-}{2}\end{aligned}$

$\therefore \quad \int_{1 / 2}^1 f(x) d x<\frac{e-1}{2}$

$Also, since f(x)>0, \int_{1 / 2}^1 f(x) d x>0$

Let $f(x)$ and $g(x)$ be non constant differentiable functions on $R$ such that

$f^{\prime}(x)=e^{f(x)-g(x))} g^{\prime}(x) \quad \forall x \in \mathbb{R}$ and

$f(1)=g(2)=1$. Then

(A) $f(2)<1-\ln 2$

(B) $f(2)>1-\ln 2$

(C) $g(1) >1-\ln 2$

(D) $g(1)<1-\ln 2$

$f^{\prime}(x)=e^{f(x)-g(x)} \cdot g^{\prime}(x)$

$\Rightarrow \quad \underbrace{e^{-f(x)} f^{\prime}(x)}_{\frac{d}{d x}\left[e^{-f(x)}\right]}=$

$\underbrace{e^{-g(x)} g^{\prime}(x)}_{\frac{d}{d x} [e^{-g(x)}]} \quad \forall x$

$\Rightarrow \quad e^{-f(x)}=e^{-g(x)}+C$, for some constant C.

$\therefore \quad e^{-f(1)}=e^{-g(1)}+c \Rightarrow e^{-8(1)}+c=e^{-1}-(i)$

$e^{-f(2)}=e^{-g(2)}+c=e^{-1}+c \quad-$ (ii)

from (i) & (ii)

$e^{-f(2)}-\frac{1}{e}=\frac{1}{e}-e^{-g(1)}$

$\Rightarrow \quad e^{-f(2)}+e^{-g(1)}=\frac{2}{e}$

$\therefore \quad e^{-f(2)}<\frac{2}{e} \quad$ & $e^{-g(1)}<\frac{2}{e}$

$\Rightarrow \quad e^{f(2)}>\frac{e}{2} \quad$ & $e^{g(1)}>\frac{e}{2}$

$\Rightarrow \quad f(2)>\ln (\frac{e}{2})=1-\ln 2 \quad$ & $\quad g(1)>1-\ln 2$

Solution

$g\left(\frac{1}{2}\right)=\int_0^1 t^{-1 / 2}(1-t)^{-1 / 2} d t$

$=\int_0^1 \frac{1}{\sqrt{t(1-t)}} d t$

$=\int_0^1 \frac{1}{\sqrt{\frac{1}{4}-\left(t-\frac{1}{2}\right)^2}} d t$

$=\left.\sin ^{-1}\left(\frac{t-\frac{1}{2}}{1 / 2}\right)\right|_0 ^1$

$=\sin ^{-1}(1)-\sin ^{-1}(-1)$

$=\quad \frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi$

$\therefore \quad g\left(\frac{1}{2}\right)=\pi$

$g(a)=\int_0^1 t^{-a}(1-t)^{a-1} d t$

$\therefore \quad g(1-a)=\int_0^1 t^{-(1-a)}(1-t)^{(1-a)-1} d t$

$=\int_0^1 t^{a-1}(1-t)^{-a} d t$

$\left(\because \int_a^b f(x) d x\right.$ $=\int_a^b f(a+b-x) d x)$

$= g (a)$

$\therefore g(a)=g(1-a) \quad \forall a \in(0,1)$

Differentiating, We get

$g^{\prime}(a)=-g^{\prime}(1-a)$

Putting $a=\frac{1}{2}$, $g^{\prime}\left(\frac{1}{2}\right)=-g^{\prime}\left(\frac{1}{2}\right)$

$\Rightarrow g^{\prime}\left(\frac{1}{2}\right)=0$

Another way:

$g(a)=\int_0^1 t^{-a}(1-t)^{a-1} d t$

$g^{\prime}(a)= \int_0^1 \frac{\partial}{\partial a}\left[t^{-a}(1-t)^{a-1}\right] dt$

$=\int_0^1\left[-t^{-a} \ln t(1-t)^{a-1}+t^{-a}(1-t)^{a-1} \ln(1-t)\right] dt$

$\text {0 at a} =\frac{1}{2}$

$\therefore \quad g^{\prime}\left(\frac{1}{2}\right)=0$

Find the value of $\int_0^1 4x^3 \frac{d^2}{d x^2}\left[\left(1-x^2\right)^5\right] d x$

Solution Direct way: First calculate $\frac{d^2}{d x^2}\left[\left(1-x^2\right)^5\right]$ and the integrate. But this involves let of computations.

Smarter way: Use integration by parts formula.

$\int f(x) g^{\prime}(x) d x=f(x) g(x)-\int f^{\prime}(x) g(x) d x$ or,

$\int f(x) g(x) d x=f(x) \int g(x) d x-\int f^{\prime}(x) \int g(x) d x d x$

$\int_0^1 4 x^3 f^{\prime \prime}(x) d x$ $=\left.4 x^3 f^{\prime}(x)\right|_0 ^1-\int_0^1 12 x^2 f^{\prime}(x) d x$

$=\left.4 x^3 f^{\prime}(x)\right|_0 ^1-\left[\left.12 x^2 f(x)\right|_0 ^1-\int_0^1 24 x f(x) d x\right]$

$f^{\prime}(1)=0$

$\therefore \int_0^1 4 x^3 f^{\prime \prime}(x) d x=24 \int_0^1 x f(x) d x$

$=24 \int_0^1 x\left(1-x^2\right)^5 d x$

$\quad$Put

$\quad 1-x^2 =y$

$\quad -2 x d x =d y$

$=12 \int_0^1 y^5 d y$

$=\left.2 y^6\right|_0 ^1=2$

$\therefore \quad I=2$

If $I_n$ $=\int_{-\pi}^\pi \frac{\sin (n x)}{\left(1+\pi^x\right) \sin x} d x$ for $x=0,1,2, \cdots$,

Then,

(A) $\quad I_n=I_{n+2} \quad \forall n$

(B) $\quad \sum_{m=1}^{10} I_{2 m+1}=10 \pi$

(C) $\quad \sum_{m=1}^{10} I_{2 m}=0$

(D) $\quad I_n=I_{n+1} \quad \forall n$

Solution $I_n=\int_0^\pi\left[\frac{\sin (n x)}{\left(1+\pi^x\right) \sin x}+\frac{\sin (-n x)}{\left(1+\pi^{-x}\right) \sin (-x)}\right] d x$

because

$\int_{-a}^a f(x) d x=\int_0^a[f(x)+f(-x)] d x$

$\therefore I_n=\int_0^\pi \frac{\sin (n x)}{\sin x} d x$

$I_0 = 0$, $I_1=\int_0^\pi \frac{\sin x}{\sin x} d x=\pi$

$I_0 \neq I_1 \Rightarrow(D)$ is wrong.

Now, $I_{n+2}-I_n=\int_0^\pi \frac{\sin (n+2) x-\sin (n x)}{\sin x} d x$ $=\int_0^\pi \frac{2 \cos (n+1) x \cdot \sin x}{\sin x} d x$

$= \left| 2 \frac{\sin (n+1)^x}{n+1} \right| _ 0 ^{\pi}=0 \therefore \quad I_ {n+2}=I_n \quad \forall n$

Since, $\begin{aligned} I_0=0, \quad I_{2 m}=0 \quad \forall m\end{aligned}$ $\begin{aligned}I_1=\pi, \quad I_{2 m+1}=\pi \quad \forall m\end{aligned}$

The total number of distinct $x \in[0,1]$ for which $\int_0^x \frac{t^2}{1+t^4} d t=2 x-1$ is

Solution Let $f(x)=\int_0^x \frac{t^2}{1+t^4} d t-(2 x-1)$ for $x \in[0,1]$

clearly, f is continuous and Differentiable function

Also,

$f(0)=0-(-1)=1>0$

$f(1)=\int_0^1 \frac{t^2}{1+t^4} d t-1$

For $0<t<1, \quad \frac{t^2}{1+t^4}<1$

$\Rightarrow \quad \int_0^1 \frac{t^2}{1+t^4} d t<1$

$\Rightarrow \quad f(1)<0$

By the intermediate value theorem, there exists at least one $x \in(0,1)$ for which $f(x)=0$.

Also, $f^{\prime}(x)=\frac{x^2}{1+x^4}-2<1-2=-1<0$

$\Rightarrow f$ is strictly decreasing in $(0,1)$.

$\Rightarrow f$ can have at most one zero in $(0,1)$

$\therefore$ The number of $x$ for which $f(x)=0$ is one.