Mathematics M.C.Q (1 Marks)

$x+2=4 y$     .......$(2)$

Solve $( 1)$ and $( 2)$

$x+2=x^{2}$

$\Rightarrow x^{2}-x-2=0$

$\Rightarrow(x-2)(x+1)=0 $

$\Rightarrow x=2,-1$

Area $ = \int\limits_{ - 1}^2 {\left( {\frac{{x + 2}}{4} - \frac{{{x^2}}}{4}} \right)dx} $

$=\frac{1}{4}\left[\frac{x^{2}}{2}+2 x-\frac{x^{3}}{3}\right]_{-1}^{2}$

$=\frac{1}{4}\left[\left(2+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)\right]$

$=\frac{1}{4}\left[\frac{10}{3}-\frac{3-12+2}{6}\right]=\frac{1}{4}\left[\frac{10}{3}+\frac{7}{6}\right]$

$=\frac{1}{4} \times \frac{27}{6}=\frac{9}{8}$

or $y=4 x-3$

Required Area $ = \int\limits_0^2 {\left( {{x^2} + 1} \right)} dx - \frac{1}{2}\left( {2 - \frac{3}{4}} \right) \cdot (5)$

$\left.=\frac{x^{3}}{3}+x\right]_{0}^{2}-\frac{25}{8}$

$=\frac{8}{3}-\frac{9}{8}=\frac{37}{24}$

$\frac{S(\lambda)}{S(4)}=\frac{2}{5} \Rightarrow \frac{\lambda^{3 / 2}}{4^{3 / 2}}=\frac{2}{5}$

$\Rightarrow \lambda=4\left(\frac{4}{25}\right)^{1 / 3}$

$x^{2}=y ; y=x+2$

$x^{2}=x+2$

$x^{2}-x-2=0$

$(x-2)(x-1)=0$

$x=2,-1$

Area $ = \int\limits_{ - 1}^2 {(x + 2) - } {x^2}dx = \frac{9}{2}$

$\lambda^{2} x^{2}=4 \lambda x$

$x=0$ and $x=\frac{4}{\lambda}$

Area $ = \int\limits_0^{4\lambda } {(\sqrt {4\lambda x}  - \lambda x)} dx = \frac{1}{9}$

$ \Rightarrow 2\sqrt \lambda   \times \left( {\frac{{{x^{3/2}}}}{{3/2}}} \right)_0^{4/\lambda } - \lambda \left( {\frac{{{x^2}}}{2}} \right)_0^{4/\lambda } = \frac{1}{9}$

$\frac{4}{3} \sqrt{\lambda} \times\left(2^{2}\right)^{3 / 4} \frac{x}{\lambda^{3 / 2}}-\frac{x}{2} \times \frac{16}{\lambda}=\frac{1}{9}$

$\Rightarrow \frac{32}{3 \lambda}-\frac{8}{\lambda}=\frac{1}{9}$

$\Rightarrow \frac{8}{3 \lambda}=\frac{1}{9}$

$\lambda=24$

From figure min area is $(-\sqrt{2}, \sqrt{2})$

$\Rightarrow x=k\left(k^{2} x^{4}\right) $

$\Rightarrow x=0$ or $x^{3}=\left(\frac{1}{k}\right)^{3}$

$ \Rightarrow x=\frac{1}{k}, 0$

Point of intersection are

$\left(\frac{1}{\mathrm{k}}, \frac{1}{\mathrm{k}}\right)$ and $(0,0)$

Area $\int\limits_0^{1/k} {\left( {\sqrt {\frac{x}{k}}  - k{x^\prime }} \right)dx}  = 1$

$ \Rightarrow {\left( {\frac{1}{{\sqrt k }}\frac{{{x^{3/2}}}}{{3/2}} - \frac{{k{x^2}}}{3}} \right)^{1/k}}$

$=1 \Rightarrow \frac{2}{3 k^{2}}=1$

$ \Rightarrow k^{2}=\frac{1}{3}$

$k=\frac{1}{\sqrt{3}}$

$\int\limits_0^3 {\left( {\left( {{x^2} + 2} \right) - (x + 1)} \right)} dx$

$ = \int\limits_0^3 {\left( {{x^2} - x + 1} \right)dx = \frac{{15}}{2}} $

$ = \int\limits_0^1 {\left( {{x^2} + 3x} \right)dx + } $ Area of

rectangle PQRS

$=\frac{11}{6}+8=\frac{59}{6}$

and $x-y-4=0$         ....$(2)$

solving $( 1)$ and $( 2)$

$(x-y)^{2}=2 x$

$\Rightarrow x^{2}-10 x+16=0$

$\Rightarrow x=8,2$ and $y=4,-2$

$A = \int\limits_{ - 2}^4 {\left( {y + 4 - \frac{{{y^2}}}{2}} \right)dy} $

$A=\left(\frac{y^{2}}{4}+4 y-\frac{y^{3}}{6}\right)_{-2}^{4}$

$A=\left(4+16-\frac{64}{6}\right)-\left(1-8+\frac{8}{6}\right)=18$

$\int\limits_0^1 {\left( {(x + 1) - {2^x}} \right)} dx$

$=\left(\frac{x^{2}}{2}+x-\frac{2^{x}}{\ln 2}\right)_{0}^{1}$

$=\left(\frac{1}{2}+1-\frac{2}{\ln 2}\right)-\left(0+0-\frac{1}{\ln 2}\right)$

$=\frac{3}{2}-\frac{1}{\ln 2}$

$A\int\limits_0^{3 - 2\sqrt 2 } {2\sqrt x dx + \frac{1}{2}(1 - (3 - 2\sqrt 2 ))(1 - (3 - 2\sqrt 2 ))} $

$=\frac{2\left[x^{3 / 2}\right]_{0}^{3-2 \sqrt{2}}}{3 / 2}+\frac{1}{2}(2 \sqrt{2}-2)(2 \sqrt{2}-2)$

$=\frac{8 \sqrt{2}}{3}+\left(-\frac{10}{3}\right)$

$a=\frac{8}{3}, b=-\frac{10}{3}$

$a-b=6$

$\Rightarrow(3 x-\pi)(6 x-\pi)=0$

$\Rightarrow \quad \alpha=\frac{\pi}{6}, \beta=\frac{\pi}{3}$

$\quad$ Also, $\operatorname{gof}(x)=\cos x$

$\therefore \quad$Req. area $ = \int\limits_{\pi /6}^{\pi /3} {\cos xdx = } \frac{{\sqrt 3  - 1}}{2}$

$y=\sqrt{x}$ is $(4,2)$

Therequiredarea

$=\int_{0}^{4} \sqrt{x} d x-\frac{1}{2} \times 2 \times 2$

$=\frac{16}{3}-2=\frac{10}{3}$

$y=x^{2}$ and $y=\frac{1}{x}$ is $(1,1)$

Area bounded by the curves is the region $\mathrm{ABCDA}$ is given as:

${\rm{ Area }} = \int_0^1 {{x^2}} dx + \int_1^t {\frac{1}{x}} dx$

$ = \left[ {\frac{{{x^3}}}{3}} \right]_0^1 + \left[ {\ln (x)_1^t} \right]$

$ = \frac{1}{3} + \ln (t)$

$\because$ area $=1$

$\Rightarrow \frac{1}{3}+\ln (t)=1 \quad \Rightarrow \ln (t)=\frac{2}{3} \Rightarrow t=e^{\frac{2}{3}}$

$ = \int\limits_0^1 {(1 + \sqrt x )dx}  + \int\limits_1^2 {(3 - x)dx - \int\limits_0^2 {\frac{{{x^2}}}{4}dx} } $

$\left. {\left. { = x]_0^1 + \left. {\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_0^1 + 3x} \right]_1^2 - \left. {\frac{{{x^2}}}{2}} \right]_{\kern 1pt} ^2 - \frac{{{x^3}}}{{12}}} \right]_0^2 = \frac{5}{2}$ sq.units

$x^{2}+3 x-4=0$

$ \Rightarrow $ $(x+4)(x-1)=0$

$x=-4, x=1$

when $x=1, y=\sqrt{3}$

$\therefore $ req. area $=$

$\left( {\int\limits_0^1 {\sqrt 3 } } \right.\left. { \cdot \sqrt x dx + \int\limits_1^2 {\sqrt {4 - {x^2}} }  \cdot dx} \right) \times 2$

$\left( {\sqrt 3 \left. {\left. {\left( {\frac{{{x^{3/2}}}}{{3/2}}} \right.} \right)_0^1 + \left( {\frac{x}{2}\sqrt {4 - {x^2}}  + 2{{\sin }^{ - 1}}\frac{x}{2}} \right)_1^2} \right)} \right. \times 2$

$ = \left( {\sqrt 3 \left( {\frac{2}{3}} \right) + \left\{ {2 \cdot \frac{\pi }{2} - \left( {\frac{{\sqrt 3 }}{2} + \frac{\pi }{3}} \right)} \right\}} \right) \times 2$

$ = \left( {\frac{2}{{\sqrt 3 }} - \frac{{\sqrt 3 }}{2} + \frac{{2\pi }}{3}} \right) \times 2$

$ = \left( {\frac{1}{{2\sqrt 3 }} + \frac{{2\pi }}{3}} \right) \times 2$

$ = \frac{1}{{\sqrt 3 }} + \frac{{4\pi }}{3}$

$=\pi-\sqrt{2} \cdot \frac{2}{3} 2 \sqrt{2}$

$=\pi-8 / 3$

$=\frac{1}{2} \times 2 \times 2+\left|\int_{3}^{4}\left(x^{2}-5 x+4\right) d x\right|$

$=2+\frac{7}{6}=\frac{19}{6}$ sq units

$y^{2}-2 x \quad \dots(i)$

and $y \geq 4 x-1$ represents a region to the left of the line

$y=4 x-1 \quad \ldots \ldots(ii)$

The point of intersection of the curves $(i)$ and $(ii)$ is

$(4 x-1)^{2}-2 x \Rightarrow 16 x^{2}+1-8 x=2 x$

$\Rightarrow 16 x^{2}-10 x+11=0 \Rightarrow x=\frac{1}{2}, \frac{1}{8}$

So, the points where these curves intersect are $\left(\frac{1}{2}, 1\right)$ and $\left(\frac{1}{8}, \frac{1}{2}\right)$

$\therefore$ Required area $ = \int\limits_{ - \frac{1}{2}}^1 {\left( {\frac{{y + 1}}{4} - \frac{{{y^2}}}{2}} \right)dy} $

$=\frac{1}{4}\left(\frac{y^{2}}{2}+y\right)_{-1 / 2}^{-1}-\frac{1}{6}\left(y^{3}\right)_{-1 / 2}^{1}$

$=\frac{1}{4}\left\{\left(\frac{1}{2}+1\right)-\left(\frac{1}{8}-\frac{1}{2}\right)\right\}-\frac{1}{6}\left(1+\frac{1}{8}\right)$

$=\frac{1}{4}\left\{\frac{3}{2}+\frac{3}{8}\right\}-\frac{1}{6}\left\{\frac{9}{8}\right\}$

$\begin{aligned}=\frac{1}{4} & \times \frac{15}{8}-\frac{3}{16} \\ &=\frac{9}{32} \text { sq units } \end{aligned}$

$y=2 x^{2}=0$

$y=3 x^{2}=1$

Point of intersection $(1,-2)$ and $(-1,-2)$

Area $ = 2\int\limits_0^1 {\left( {\left( {1 - 3{x^2}} \right) - \left( { - 2{x^2}} \right)dx} \right)} $

$ = 2\int\limits_0^1 {\left( {1 - {x^2}dx} \right) = 2\left( {x - \frac{{{x^3}}}{3}} \right)_0^1 = \frac{4}{3}} $

$A_{1}=2\left|\int_{1}^{0} 2 t^{2} d t\right|$

$A_{1}=4 \cdot \frac{1}{3}$

$\boxed{Area = \frac{\pi }{2} + \frac{4}{3}}$

$ = \left. {\frac{1}{2}\left[ {\frac{{{y^2}}}{2} + 4y} \right]_{ - 2}^4} \right| - \left| {\frac{1}{4}\left[ {\frac{{{y^3}}}{3}} \right]_{ - 2}^4} \right.$

$=\left|\frac{1}{2}[\{8+16\}-\{2-8\}]\right|-\left|\frac{1}{4}\left\{\frac{64}{3}+\frac{8}{3}\right\}\right|$

$=15-6=9$ sq units

when $x=\frac{\pi}{4}, y=1$

Equation of tangent at $P$ is

$y-1=\left(\sec ^{2} \frac{\pi}{4}\right)\left(x-\frac{\pi}{4}\right)$

or $y=2 x+1-\frac{\pi}{2}$         .....$(2)$

Area of shaded region

$=$ area of $\mathrm{OPMO}-a r(\Delta \mathrm{PLM})$

$=\int_{0}^{\frac{\pi}{4}} \tan x d x-\frac{1}{2}(\mathrm{OM}-\mathrm{OL}) \mathrm{PM}$

$ = \left[ {\left. {\log \sec x\frac{{\frac{\pi }{4}}}{0}} \right] - \frac{1}{2}\left\{ {\frac{\pi }{4} - \frac{{\pi  - 2}}{4}} \right\} \times 1} \right.$

$=\frac{1}{2}\left[\log 2-\frac{1}{2}\right]$ sq unit

${\rm{ and }}\quad 2y - x + 3 = 0\,\,\,\,\,\,\,\,\,\,.....\left( 2 \right)$

On solving both $y=-1,3$

Required area $ = \int\limits_0^3 {\left\{ {(2y + 3) - {y^2}} \right\}dy} $

$=y^{2}+3 y-\left.\frac{y^{3}}{3}\right|_{0} ^{3} $

${=9+9-9}$

${=9}$

$ = \frac{1}{3}\int\limits_{y = 1}^4 {{y^{1/2}}dy = \frac{1}{3} \times \frac{2}{3}} \left( {{y^{3/2}}} \right)_1^4$

$=\frac{2}{9}\left[\left(4^{1 / 2}\right)^{3}-\left(1^{1 / 2}\right)^{3}\right]=\frac{2}{9}[8-1]$

$=\frac{2}{9} \times 7=\frac{14}{9} \mathrm{sq.~units.}$

$ y=\ln (x) \text { and } y=\ln (3),$

$ \text { put } \ln (x)=\ln (3) $

$ \Rightarrow  \ln (x)-\ln (3)=0$

$ \Rightarrow  \ln (x)-\ln (3)=\ln (1) $

$ \Rightarrow  \frac{x}{3}=1, \Rightarrow x=3 $

Required area $ = \int\limits_0^3 {\ln } \left( 3 \right)dx - \int\limits_1^3 {\ln } \left( x \right)dx$

$=[x \ln (3)]_{0}^{3}-\left[x \ln (x)-x_{1}^{3}\right]=2$

Required area $ = 2\int\limits_0^{\pi /4} {(\cos x - \sin x)dx} $

$ = 2\left[ {\sin x + \cos x_0^{\pi /4}} \right]$

$=2\left[\frac{2}{\sqrt{2}}-1\right]$

$=(2 \sqrt{2}-2)$ sq. units

${{\rm{R}}_2} = \int\limits_{ - 2}^3 {xf(x)dx} $

$ = \int\limits_{ - 2}^3 {(1 - x)f(1 - x)dx} $

$\left[ {{\rm{Using}}\int\limits_a^b {f\left( x \right)dx}  = \int\limits_a^b {f\left( {a + b - x} \right)dx} } \right]$

$ \Rightarrow {{\rm{R}}_2} = \int\limits_{ - 2}^3 {(1 - x)f\left( x \right)dx} $

$(\because f(x)=f(1-x) \text { on }[-2,3])$

$\therefore {{\rm{R}}_2} + {R^2} = \int\limits_{ - 2}^3 {xf\left( x \right)dx + } \int\limits_{ - 2}^3 {(1 - x)f\left( x \right)dx} $

$ = \int\limits_{ - 2}^3 {f\left( x \right)dx = {R_1}} $

$\Rightarrow 2 \mathrm{R}_{2}=\mathrm{R}_{1}$

$\text { Area }= A _1+ A _2$

$\alpha \sqrt{2}=\int_0^2 2 \sqrt{ x } dx +\frac{1}{2} \times 3 \times 2 \sqrt{2}$

$\alpha \sqrt{2}=2\left[\frac{ x ^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^2+3 \sqrt{2}$

$\alpha \sqrt{2}=\frac{17 \sqrt{2}}{3}$

$\alpha=\frac{17}{3}$

$f^{\prime}(x)=x^2-2 x+\frac{5}{9}$

$f^{\prime}(x)=0 \text { at } x=\frac{1}{3} \text { in }[0,1]$

$A_R=\text { Area of Red region }$

$A_G=\text { Area of Green region }$

$A_R=\int_0^1 f(x) d x=\frac{1}{2}$

$\text { Total area }=1$

$\Rightarrow A_G=\frac{1}{2}$

$\int_0^1 f(x) d x=\frac{1}{2}$

$A_G=A_R$

$f(0)=\frac{17}{36}$

$f(1)=\frac{13}{36} \approx 0.36$

$f\left(\frac{1}{3}\right)=\frac{181}{324} \approx 0.558$

($A$) Correct when $\mathrm{h}=\frac{3}{4}$ but $\mathrm{h} \in\left[\frac{1}{4}, \frac{2}{3}\right]$

$\Rightarrow(A)$ is incorrect

($B$) Correct when $\mathrm{h}=\frac{1}{4}$

$\Rightarrow(B)$ is correct

($C$) When $\mathrm{h}=\frac{181}{324}, \mathrm{~A}_{\mathrm{R}}=\frac{1}{2}, \mathrm{~A}_{\mathrm{G}}<\frac{1}{2}$

$\mathrm{h}=\frac{13}{36}, \mathrm{~A}_{\mathrm{R}}<\frac{1}{2}, \mathrm{~A}_{\mathrm{G}}=\frac{1}{2}$

$\Rightarrow A_R=A_G$ for some $h \in\left(\frac{13}{36}, \frac{181}{324}\right)$

$\Rightarrow(C)$ is correct

($D$) Option ($D$) is remaining coloured part of option ($C$), hence option ($D$) is also correct.

$x^2+\frac{8 x}{3}+\frac{5}{12}-2=0$

$\begin{array}{l}12 x^2+32 x -19=0 \\ 12 x ^2+38 x -6 x -19=0 \\ 2 x (6 x +19)-1(6 x +19)=0 \\ (6 x +19)(2 x -1)=0 \\ x =\frac{1}{2} \\ \alpha=2 A _1+ A _2 \\ \alpha=2\left(\int_0^{1 / 2} x ^2+\frac{5}{12} dx +\frac{1}{2} \times \frac{1}{4} \times \frac{2}{3}\right) \\ \Rightarrow \alpha=2\left[\left(\frac{ x ^3}{3}+\frac{5 x }{12}\right)_0^{1 / 2}+\frac{1}{12}\right] \\ \Rightarrow \alpha=2\left[\frac{1}{24}+\frac{5}{24}+\frac{1}{12}\right] \\ \Rightarrow \alpha=2\left[\frac{1+5+2}{24}\right] \Rightarrow \alpha=2 \times \frac{8}{24} \Rightarrow 9 \alpha=9 \times \frac{8}{12} \\ \Rightarrow 9 \alpha=6\end{array}$

$P \left(\frac{3}{2}, \frac{1}{2}\right) ; Q (2,0) ; R \left(\frac{9}{4}, 0\right) ; S \left(\frac{9}{4}, \frac{3}{4}\right)$

Area $=\frac{1}{2}|| \begin{array}{ll}\frac{3}{2} & \frac{1}{2} \\ 2 & 0\end{array}|+| \begin{array}{ll}2 & 0 \\ \frac{9}{4} & 0\end{array}|+| \begin{array}{ll}\frac{9}{4} & 0 \\ \frac{9}{4} & \frac{3}{4}\end{array}|+| \begin{array}{ll}\frac{9}{4} & \frac{3}{4} \\ \frac{3}{2} & \frac{1}{2}\end{array}||$

$=\frac{1}{2}\left|(0-1)+(0-0)+\left(\frac{27}{16}-0\right)+\left(\frac{9}{8}-\frac{9}{8}\right)\right|=\frac{11}{32}$

$P =4(\alpha+\cos 2 \alpha)$

$\frac{ dP }{ d \alpha}=4(1-2 \sin 2 \alpha)=0$

$\sin 2 \alpha=\frac{1}{2}$

$2 \alpha=\frac{\pi}{6}, \frac{5 \pi}{6}$

$\frac{ d ^2 P }{ d \alpha^2}=-4 \cos 2 \alpha$

$\text { for maximum } \alpha=\frac{\pi}{12}$

$\text { Area }=(2 \alpha)(2 \cos 2 \alpha)$

$\quad=\frac{\pi}{6} \times 2 \times \frac{\sqrt{3}}{2}=\frac{\pi}{2 \sqrt{3}}$

$\begin{array}{l}f(x)=\left\{\begin{array}{cc}0 & x \leq 1 \\ e^{x-1}-e^{1-x} & x \geq 1\end{array}\right. \\ \& g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)\end{array}$

$\begin{array}{l}\text { solve } f(x) \& g(x) \Rightarrow x=1+\ell n \sqrt{3} \\ \text { So bounded area }=\int_0^1 \frac{1}{2}\left(e^{x-1}+e^{1-x}\right) d x+\int_1^{1+(n \sqrt{3}} \frac{1}{2}\left(e^{x-1}+e^{1-x}\right)-\left(e^{x-1}+e^{1-x}\right) d x \\ =\frac{1}{2}\left[e^{x-1}-e^{1-x}\right]_0^1+\left[-\frac{1}{2} e^{x-1}-\frac{3}{2} e^{1-x}\right]_1^{1+(n \sqrt{3}} \\ =\frac{1}{2}\left[e-\frac{1}{e}\right]+\left[\left(-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\right)+2\right]=2-\sqrt{3}+\frac{1}{2}\left(e-\frac{1}{e}\right)\end{array}$

Hence, required area $=  \int_1^4\left(\frac{8}{y}-\sqrt{ y }\right) dy$

$ =\left[8 \operatorname{lny}-\frac{2}{3} y ^{3 / 2}\right]_1^4=16 \ln 2-\frac{14}{3}$

$diff.\quad e^{-x} f^{\prime}(x)-e^{-x} f(x)=-(1-2 x) e^{-x}-2 e^{-x}+e^{-x} f(x)$

$f^{\prime}(x)-2 f(x)=-3+2 x$

$I.F.=e^{-2 x}$

Soln. $y \cdot e^{-2 x}=C+\int e^{-2 x}(2 x-3) d x$  $=C+2 \int e ^{-2 x } \cdot xdx -3 \cdot \int e ^{-2 x } dx$  $=C+2\left[\frac{e^{-2 x}}{-2} x-\frac{e^{-2 x}}{4}\right]+\frac{3}{2} e^{-2 x}$  $y=C \cdot e^{2 x}-x-\frac{1}{2}+\frac{3}{2}$

$\left.\frac{x^2}{2}\right|_0 ^1-\left.\frac{x^{n+1}}{n+1}\right|_0 ^1=\frac{3}{10}$

$\frac{1}{2}-\frac{1}{n+1}=\frac{3}{10}$

$\frac{1}{n+1}=\frac{1}{2}-\frac{3}{10}$

$n+1=5$

$n=4$

(image)

$\frac{1}{8}=\left|\frac{x^2}{2}-\frac{x^4}{4}\right|_0^\alpha$

(image)

$\frac{1}{2}=2 \alpha^2-\alpha^4$

$2 \alpha^4-4 \alpha^2+1=0$

$f(\alpha)=2 \alpha^4-4 \alpha^2+1$

$\text { Area } BCD \text { (under parabola) }=\int_{-3}^1 \sqrt{ x +3} dx =\frac{16}{3}$

$\text { Area of trapezium } ACDE =\frac{1}{2}(1+2)_5=\frac{15}{2}$

$\text { Required area }=\frac{15}{2}-\frac{16}{3}-\frac{2}{3}=\frac{3}{2}$

$\frac{d y}{d x}=\cos x-\sin x$

$y=|\cos x-\sin x|=\left[\begin{array}{cc}\cos x-\sin x \quad x \in[0, \pi / 4] \\ \sin x-\cos x \quad x \in[\pi / 4, \pi / 2]\end{array}\right.$ $Image$

required area $=\int_0^{\pi / 4}\left|(\sin x+\cos x)-(\cos x-\sin x) d d x+\int_{\pi / 4}^{\pi / 2}\right| 2 \cos x \mid d x$

$=\int_0^{\pi / 4}|2 \sin x| d x+\int_{\pi / 4}^{\pi / 2}|2 \cos x| d x$

$=2(-\cos x)_0^{\pi / 4}+2(\sin x)_{\pi / 4}^{\pi / 2} $ $Image$

$=2\left[-\frac{1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}}\right]$

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

$=2(2-\sqrt{2}) $

$=4-2 \sqrt{2} $

$=2 \sqrt{2}(\sqrt{2}-1)$

$-x^2 \leq 0 $

$e ^{- x ^2} \leq 1 $

$\int_0^1 e^{-x^2} d x \leq 1 $

$x^2 \leq x \Rightarrow-x^2 \geq-x \Rightarrow e^{-x^2} \geq e^{-x} $

$\Rightarrow I \geq \int_0^1 e^{-x} d x $

$\geq-\left(e^{-x}\right)_0^1 $

$\geq-\left(\frac{1}{e}-1\right) $

$I \geq 1-\frac{1}{e} \Rightarrow(B)$ is correct

Since If $I \geq 1-\frac{1}{e} \Rightarrow I>\frac{1}{e} \quad \Rightarrow(A)$ is correct

$1 < \frac{1}{\sqrt{2}} \times 1+\frac{1}{\sqrt{ e }} \times\left(1-\frac{1}{\sqrt{2}}\right)$

$=\int_{-1}^2(1-x) f(1-x) d x$

$=\int_{-1}^2(1-x) f(x) d x$

Hence $2 R_1=\int_{-1}^2 f(x) d x=R_2$.

$L: 2 x-3 y-1=0$

$L(0,0)=0-0-1 < 0$

$2 h-3 k-1 < 0$

$\left(2, \frac{3}{4}\right)$

$S: 4+\frac{9}{16}-6 < 0$

$L: 2 \cdot 2-3 \cdot \frac{3}{4}-1 > 0$

$\left(\frac{5}{2}, \frac{3}{4}\right)$

$S: \frac{25}{4}+\frac{9}{16} > 6$

$L: o$

$\left(\frac{1}{4},-\frac{1}{4}\right)$

$S < 0$

$L: \frac{1}{2}+\frac{3}{4}-1 > 0$

$\left(\frac{1}{8}, \frac{1}{4}\right)$

$S < 0$

$L: \frac{1}{4}-\frac{3}{4}-1 < 0$

$\Rightarrow\left[\frac{(x-1)^3}{3}\right]_0^b-\left[\frac{(x-1)^3}{3}\right]_b^1=\frac{1}{4}$

$\Rightarrow \frac{(b-1)^3}{3}+\frac{1}{3}-\left(0-\frac{(b-1)^3}{3}\right)=\frac{1}{4}$

$\Rightarrow \frac{2(b-1)^3}{3}=-\frac{1}{12}$

$\Rightarrow(b-1)^3=-\frac{1}{8} \Rightarrow b=\frac{1}{2}$

$2.$ $ -\frac{3}{4}$

$ \frac{1}{2}$

$ \int_0^{1 / 2}\left(4 x^3+3 x^2+2 x+1\right) d x<\text { area }<\int_0^{3 / 4}\left(4 x^3+3 x^2+2 x+1\right) d x $

$ {\left[x^4+x^3+x^2+x\right]_0^{1 / 2}<\text { area }<\left[x^4+x^3+x^2+x\right]_0^{3 / 4}} $

$ \frac{1}{16}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}<\text { area }<\frac{81}{256}+\frac{27}{64}+\frac{9}{16}+\frac{3}{4} $

$ \frac{15}{16}<\text { area }<\frac{525}{256} .$

$3.$ $\mathrm{f}^{\prime \prime}(\mathrm{x})=2[12 x+3]=0 \Rightarrow \mathrm{x}=-1 / 4 \text {. }$

$ =\int_0^{\pi / 4}\left(\sqrt{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}}-\sqrt{\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}}\right) d x=\int \frac{\left(1+\tan \frac{x}{2}\right)-\left(1-\tan \frac{x}{2}\right)}{\sqrt{1-\tan ^2 \frac{x}{2}}} d x $

$ =\int_0^{\pi / 4} \frac{2 \tan \frac{x}{2}}{\sqrt{1-\tan ^2 \frac{x}{2}}} d x=\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t \text { as } \tan \frac{x}{2}=t .$

(From the symmetry)

On solving, we get required area $ = \frac{1}{3}$ sq. unit.