MCQ Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 14

The area of the figure bounded by

$y=sinx, \ y=cosx$ is the first quadrant is

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$2(\sqrt{2-1})$
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$\sqrt{3}+1$
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$2(\sqrt{3}-1)$
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None of these

The areas of the figure into which the curve

$y^2=6x$ divides the circle

$x^2 + y^2 = 16$ are in the ratio

0%
$\dfrac{2}{3}$
0%
$\dfrac{4 \pi - \sqrt{3}}{8 \pi + \sqrt{3}}$
0%
$\dfrac{4 \pi + \sqrt{3}}{8 \pi - \sqrt{3}}$
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None of these

For

$a > 0$ , let the curves

$C_1 : y^2 = ax$ and

$C_2 : x^2 = ay$ intersect at origin

$O$ and a point

$P$ . Let the line

$x = b (0 < b < a)$ intersect the chord

$OP$ and the x-axis at points

$Q$ and

$R$ , rspectively. If the line

$x=b$ bisects the area bounded by the curves,

$C_1$ and

$C_2$ , and the area of

$\Delta OQR = \dfrac{1}{2}$ , then '

$a$ ' satisfies the equation:

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$x^6-12x^3 + 4 = 0$
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$x^6 + 6x^33 - 4 = 0$
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$x^6 - 12x^3 + 4 = 0$
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$x^6 - 6x^3 + 4 = 0$

Explanation

area

$\Delta OQR = \dfrac{1}{2}.....given$

For coordinates of

$p$

Curves :

$x^2 = ay, y^2 = ax$

$\Rightarrow y^2 = \left(\dfrac{x^2}{a} \right)^2 = ax$

$\dfrac{x^4}{a^2} = ax$

$x^4 = a^3x$

$x(x^3 - a^3) = 0$

$\Rightarrow x = 0, x=a....(i)$

$\Rightarrow y^2 = ax$

$\therefore y = 0, a....(from \ i)$

$\therefore$ equation of chord

$OP$

$O(0,0)\equiv(x_1,y_1) \ and \ P(a,a)\equiv(x_2,y_2)$

$y - y_1 = \dfrac{y_2-y_1}{x_2-x_1} (x-x_1)$

$y - 0 = \dfrac{a-0}{a-0} (x-o)$

$y = x$

$\therefore$ At point

$Q$

$x = b$ and

$y = b$

$\Rightarrow Q \equiv (b,b)$

$\therefore$ ar

$\Delta OQR = \dfrac{1}{2} \times b \times b = \dfrac{b^2}{2} = \dfrac{1}{2}$

$\Rightarrow b^2 = 1 \Rightarrow b=1$

Now, area bounded by the curves

$= \displaystyle \int_0^a \left(\sqrt{ax} - \dfrac{x^2}{a}\right)dx = \int_0^a a^{1/2} x^{1/2}dx - \int_0^a \dfrac{x^2}{a} dx$

$=a^{1/2}\left[\dfrac{x^{3/2}}{3/2}\right]_0^a-\dfrac{1}{a}\left[\dfrac{x^3}{3}\right]_0^a$

$=\dfrac{2}{3}a^{1/2}. a^{3/2} - \dfrac{1}{a} \dfrac{a^3}{3} = \dfrac{2}{3} a^2 - \dfrac{1}{3}a^2$

$= \dfrac{1}{3}a^2$

$\therefore \displaystyle \int_0^b \left(\sqrt{ax} - \dfrac{x^2}{a}\right)dx = \dfrac{1}{2} \times \dfrac{1}{3}a^2 = \dfrac{a^2}{6}$

$\dfrac{2}{3} a^{1/2} [x^{3/2}]_0^b - \dfrac{1}{3a} [x^3]_0^b = \dfrac{a^2}{6}$

$\Rightarrow \dfrac{2}{3}a^{1/2} b^{3/2} - \dfrac{1}{3a}b^3 = \dfrac{a^2}{6}$

Now after putting value of

$b$ .

we get,

$\dfrac{2}{3}a^{1/2} - \dfrac{1}{3a} = \dfrac{a^2}{6}$

$2 a^{3/2} -1 = \dfrac{3a^3}{6} = \dfrac{a^3}{2}$

$4a^{3/2} - 2 =a^3 \Rightarrow 4a^{3/2} = a^3 + 2$

Now squaring on both the side

$\Rightarrow a^6 + 4a^3 + 4 = 16a^3$

$\Rightarrow a^6 - 12a^3 + 4 = 0$

$\therefore$ a will satisfy

$x^6 - 12x^3 + 4 = 0$

$\therefore$

$\boxed{x^6 - 12x^3 + 4 = 0}....Answer$

Hence option

$'C'$ is the answer.

1477256_1697509_ans_2194c041b61a434db5fb2f0cc1847c9e.png

The area bounded by the curve

$y =x^4 -2x^3 + x^2 + 3$ the axis of abscissas and two oridnates corrsponding to the point of minimum of function y(x) is

0%
10/3
0%
27/10
0%
21/10
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None of these

The area of the region bounded by

$y^2=2x +1$ and

$x-y-1=0$ is

0%
$\dfrac{2}{3}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{8}{3}$
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$\dfrac{16}{3}$

The area bounded by the curves

$| x| +| y | > 1$ and

$x^2 + y^2 < 1$ is

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$2$ sq units
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$\pi$ sq units
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$( \pi - 2)$ sq units
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$( \pi +2) sq units$

Explanation

${\textbf{Step -1: Draw a figure and define formulas required to find the area.}}$

${\text{Area of circle=}}$

$\pi r^2$

$\Rightarrow r=1$

${\text{Area of triangle=}}$

$\dfrac{1}{2}\times b\times h$

${\text{But there are four triangles with same base and height}}$

$\Rightarrow b=1,h=1$

${\text{So, area of 4 triangles }}$

${ =4 \times \dfrac{1}{2} \times b \times h}$

${\textbf{ Step -2: Finding required area.}}$

${\text{Required area = Area of circle - Area of Triangle.}}$

$= \pi r^2-4 \times \dfrac{1}{2}\times b\times h$

$=\pi\times1^2-4\times\dfrac{1}{2}\times1\times1$

$=\pi-2\times1$

$=(\pi-2)$

${\text{sq.unit}}$

${\textbf{Hence , the area bounded by the curves is}}$

$\mathbf{(\pi-2)}$

${\textbf{sq.unit. So, option C is correct.}}$

The area of the region enclosed by the curve

$\mid y \mid =-(1- \mid x \mid)^2 +5$ , is

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$\dfrac{8}{3} (7 + 5 \sqrt{5})$ sq units
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$\dfrac{2}{3} (7 + 5 \sqrt{5})$ sq units
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$\dfrac{2}{3} (5 \sqrt{5}-7)$ sq units
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None of these

If the area bounded by the curve

$y=x^2 +1, y=x$ and the pair of lines

$x^2 + y^2 + 2xy - 4x -4y +3 =0$ is

$K$ units, then the area of the region bounded by the curve

$y=x^2 +1, \ y= \sqrt{x-1}$ and the pair of lines

$(x+y-1)(x+y-3)=0 ,$ is

0%
$K$
0%
$2K$
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$\dfrac{K}{2}$
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None of these

The area bounded by the curve

$y= \dfrac{3}{\mid x \mid}$ and

$y+ \mid 2-x \mid =2$ is

0%
$\dfrac{4-log27}{3}$
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$2-log3$
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$2+log3$
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None of these

The area bounded by the curves

$y=x^2 +2$ and

$y=2 \mid x \mid -cosx +x$ is

0%
$\dfrac{2}{3}$
0%
$\dfrac{8}{3}$
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$\dfrac{4}{3}$
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$\dfrac{1}{3}$

The area bounded by the curve

$f(x) = \mid \mid tanx + cotx \mid - \mid tanx - cotx \mid \mid$ between the lines

$x=0, \ x= \dfrac{\pi}{2}$ and the

$X-$ axis is

0%
$log4$
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$$ log \sqrt{2] $$
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$2log2$
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$\sqrt{2} log2$

The area bounded by the curve

$y^2 =4x$ and the circle

$x^2 + y^2 -2x -3=0$ is

0%
$2 \pi + \dfrac{8}{3}$
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$4 \pi + \dfrac{8}{3}$
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$\pi + \dfrac{8}{3}$
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$\pi - \dfrac{8}{3}$

The area of the region defined by

$1 \leq \mid x-2 \mid + \mid y+ 1 \mid \leq 2$ is

0%
$2$
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$4$
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$6$
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None of these

The area bounded by

$y=2- \mid 2-x \mid$ and

$y= \dfrac{3}{\mid x \mid}$ is

0%
$\dfrac{4-3 ln3}{2}$
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$\dfrac{19}{8} - 3 ln 2$
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$\dfrac{3}{2} + ln 3$
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$\dfrac{1}{2} + ln 3$

The area of the region defined by

$\mid \mid x \mid - \mid y \mid \mid \leq 1$ and

$x^2 + y^2 \leq 1$ in the

$xy$ plane is

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$\pi-2$
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$2 \pi$
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$3 \pi$
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$1$

If the length of latus rectum of ellipse

$E_1 : 4(x+y-1)^2 + 2(x-y+3)^2 =8$
and

$E_2: \dfrac{x^2}{p} + \dfrac{y^2}{p^2}=1, \ (0< p<1)$ are equal, then area of ellipse

$E_2,$ is

0%
$\dfrac{\pi}{2}$
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$\dfrac{\pi}{\sqrt{2}}$
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$\dfrac{\pi}{2 \sqrt{2}}$
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$\dfrac{\pi}{4}$

$f(x)=x-1$ and

$g(x) = \mid f( \mid x \mid ) - 2 \mid$ , then the area bounded by

$y=g(x)$ and the curve

$x^2 -4y+8=0$ is equal to

0%
$\dfrac{4}{3} (4 \sqrt{2} - 5)$
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$\dfrac{4}{3} (4 \sqrt{2} - 3)$
0%
$\dfrac{8}{3} (4 \sqrt{2} - 3)$
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$\dfrac{8}{3} (4 \sqrt{2} - 5)$

Area bounded by the ellipse

$\dfrac{x^2}{4} + \dfrac{y^2}{9} =1$ is equal to

0%
$6 \pi$ sq units
0%
$3 \pi$ sq units
0%
$12 \pi$ sq units
0%
$2 \pi$ sq units

Area bounded by

$y=f^{-1}(x)$ and tangent and normal drawn to it at the points with abscissae

$\pi$ and

$2 \pi$ , where

$f(x)=sin x- x$ is

0%
$\dfrac{ \pi^2}{2} -1$
0%
$\dfrac{ \pi^2}{2} -2$
0%
$\dfrac{ \pi^2}{2} -4$
0%
$\dfrac{ \pi^2}{2}$

A point

$P$ lying inside the curve

$y= \sqrt{2ax-x^2}$ is moving such that its shortest distance from the curve at any position is greater than its distance from

$X-$ axis. The point

$P$ enclose a region whose area is equal to

0%
$\frac{\pi a^2}{2}$
0%
$\frac{ a^2}{3}$
0%
$\frac{2 a^2}{3}$
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$\bigg( \frac{3 \pi -4}{6} \bigg) a^2$

Then, the absolute area enclosed by

$y=f(x)$ and

$y=g(x)$ is given by

0%
$\overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^r. h(x) dx$
0%
$\overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^{r+1}. h(x) dx$
0%
$2 \overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^r. h(x) dx$
0%
$\dfrac{1}{2} \ \overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^{r+1}. h(x) dx$

The area enclosed by the asteroid

$\bigg( \dfrac{x}{a} \bigg)^{2/3} + \bigg( \dfrac{y}{a} \bigg)^{2/3} =1$ is

0%
$\dfrac{3}{4} a^2 \pi$
0%
$\dfrac{3}{18} \pi a^2$
0%
$\dfrac{3}{8} \pi a^2$
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$\dfrac{3}{4} a \pi$

The area of the region bounded between the curves

$y=e \mid \mid x \mid \ ln \mid x \mid \mid, , \ x^2 + y^2 - 2( \mid x \mid + \mid y \mid) + 1 \geq 0$ and

$X-$ axis where

$\mid x \mid \leq 1,$ if

$\alpha$ is the

$x-$ coordinate of the point of intersection of curves in 1st quadrant, is

0%
$4 \bigg[ \int^\alpha_0 \ e \ x \ ln x \ dx + \int^1_\alpha (1- \sqrt{1-(x-1)^2)} \ dx \bigg]$
0%
$4 \bigg[ \int^\alpha_0 \ e \ x \ ln x \ dx + \int^\alpha_1 (1- \sqrt{1-(x-1)^2)} \ dx \bigg]$
0%
$4 \bigg[ - \int^\alpha_0 \ e \ x \ ln x \ dx + \int^1_\alpha (1- \sqrt{1-(x-1)^2)} \ dx \bigg]$
0%
$2 \bigg[ \int^\alpha_0 \ e \ x \ ln x \ dx + \int^\alpha_1 (1- \sqrt{1-(x-1)^2)} \ dx \bigg]$

Area of the region enclosed between the curves

$x=y^2 -1$ and

$x= \mid y \mid \sqrt{1-y^2}$ is

0%
$1$
0%
$\dfrac{4}{3}$
0%
$\dfrac{2}{3}$
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$2$

Area of the loop described as

$x=\dfrac{t}{3}(6-t), \ y=\dfrac{t^2}{8}(6-t)$ is

0%
$\dfrac{27}{5}$
0%
$\dfrac{24}{5}$
0%
$\dfrac{27}{6}$
0%
$\dfrac{21}{5}$

The area enclosed by the curves

$x = a \sin^3t$ and

$y = a \cos^3 t$ is equal to

0%
$\displaystyle 12a^2 \int_0^{\pi/2} \cos^4 t \sin^2t \ dt$
0%
$\displaystyle 12a \int_0^{\pi/2} \cos^2 t \sin^4t \ dt$
0%
$\displaystyle 2 \int_{-a}^{a} (a^{2/3} - x^{2/3})^{3/2}dx$
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$\displaystyle 4 \int_{0}^{a} (a^{2/3} - x^{2/3})^{3/2}dx$

Consider two regions

$R_1$ : Point

$P$ is nearer to

$(i, 0)$ than to

$x = -1$ .

$R_2$ : Point

$P$ is nearer to

$(0, 0)$ than to

$(8, 0)$ .
Statement 1 : Area of the region common to

$R_1$ and

$R_2$ is

$\dfrac{128}{3}$ sq. units.

Statement 2 : Area bounded by

$x = 4 \sqrt{y}$ and

$y = 4$ is

$\dfrac{32}{3}$ sq. units.

0%
If both Statement are correct and Statement 2 is the correct explanation of Statement 1
0%
If both Statement are correct and Statement 2 is not the correct explanation of Statement 1
0%
If Statement 1 is correct and Statement 2 is incorrect
0%
If Statement 1 is incorrect and Statement 2 is correct

Explanation

$R_1$ : points

$P(x, y)$ is nearer to

$(1, 0)$ then to

$x= -1$

$\Rightarrow \sqrt{(x-1)^2 +y^2} < |x+1|$

$\Rightarrow y^2 < 4x$

$\Rightarrow$ Point

$P$ lies inside parabola

$y^2 =4x$ .

$R_2$ : Poit

$P(x, y )$ us nearer to

$(0, 0)$ than to

$(8, 0)$

$\Rightarrow |x| < |x-8|$

$\Rightarrow x^2 < x^2 - 16x + 64$

$\Rightarrow x < 4$

$\Rightarrow$ Point

$P$ is towards left side of line

$x = 4$ .
The area of common region of

$R_1$ and

$R_2$ is the area bounded by

$x =4$ and

$y^2 = 4x$ .
This area is twice the area bounded by

$x = 4\sqrt{y}$ and

$y = 4$ .
Now, the area bounded by

$x = 4\sqrt{y}$ and

$y = 4$ is

$A = \displaystyle \int_0^4 \left(4 - \dfrac{x^2}{4}\right) dx = \left[4x - \dfrac{x^3}{12}\right]_0^4 = \left[ 16 - \dfrac{64}{12}\right] = \dfrac{32}{3}$ sq. units
Hence, the area bounded by

$R_1$ and

$R_2$ is

$\dfrac{64}{3}$ sq. units.
Thus, statement 1 is incorrect but statement 2 is correct.
1634344_1787868_ans_aeca2223fa3b41dfb22f926d5af0fa05.JPG

The area enclosed by the ellipse

$\dfrac {x^{2}}{a^{2}} + \dfrac {y^{2}}{b^{2}} = 1$ is equal to

0%
$\pi^{2} ab$
0%
$\pi ab$
0%
$\pi a^{2}b$
0%
$\pi ab^{2}$

The value of the parameter a such that the area bounded by

$y=a^2x^2+ax+1$ , coordinate axes and the line

$x=1$ attains its least value is equal to

0%
$-\frac{1}{4}$ sq. units
0%
$-\frac{1}{2}$ sq. units
0%
$-\frac{3}{4}$ sq. units
0%
$-1$ sq. units

Explanation

$\textbf{Step 1: Find the vertex of Parabola using}\boldsymbol{\left(\dfrac{-b}{2a},\dfrac{-D}{4a}\right) }\textbf{ and draw it's graph.}$

$\text{Given, }$

$y=a^2x^2+ax+1$

$\text{Vertex}=\left(\dfrac{-a}{2a^2},\dfrac{-(a^2-4(a^2)(1))}{4a^2}\right)$

$\text{So, its vertex }= \left(\dfrac{-1}{2a},\dfrac{3}{4}\right)$

$\textbf{Step 2: Find area under the curve using Integration.}$

$\int_{0}^{1}(a^2x^2+ax+1)dx$

$=[\dfrac{a^2x^3}{3}+\dfrac{ax^2}{2}+x]^1_0$

$=[\dfrac{-a^2}{3}-\dfrac{a}{2}+1]=f(a)$

$\textbf{Step 3: Find the Least value of Quadratic, using }\boldsymbol{\dfrac{-b}{2a}.}$

$f(a)=-\dfrac{a^2}{3}-\dfrac{a}{2}+1$

$\text{minimum at }\dfrac{-b}{2a}=-\left[\dfrac{\dfrac{1}{2}}{\dfrac{-2}{3}}\right]$

$a=\dfrac{-3}{4}$ sq. units.

$\textbf{Hence, Option C is correct.}$

Let the functions

$f:R\rightarrow R$ and

$𝑔:R\rightarrow R$ be defined by

$f(x)=e^{ x-1 }-e^{ -|x-1| }$ and

$g(x)=\dfrac{1}{2}(e^{x-1}+e^{1-x})$ . Then the area of the region in the first quadrant bounded by the curves

$𝑦=𝑓(𝑥), 𝑦=𝑔(𝑥)$ and

$x=0$ is

0%
$(2-\sqrt{3})+\dfrac{1}{2}(e-e^{-1})$
0%
$(2+\sqrt{3})+\dfrac{1}{2}(e-e^{-1})$
0%
$(2-\sqrt{3})+\dfrac{1}{2}(e+e^{-1})$
0%
$(2+\sqrt{3})+\dfrac{1}{2}(e+e^{-1})$

Explanation

$f(x) = 0$

$; x < 1$

$= e^{x-1} - e^{-(x-1)}$

$; x \ge 1$

while

$g(x) \ge 1$

so they will intresect in the region

$x > 1$

solve

$f(x) = g(x)$

$e^{x-1} - e^{-(x-1)} = \dfrac{1}{2} (e^{x-1} + e^{1-x})$

$\dfrac{1}{2} e^{x-1} = \dfrac{3}{2} e^{1-x}$

$e^{2x}=3e^2$

$2x = 2 + \ell n3$

$x = 1 + \ell n \sqrt{3}$

$\displaystyle A = \int_0^1 g(x) dx + \int_1^{1+\ell n\sqrt{3}} (g(x) - f(x)) dx$

$\displaystyle = \dfrac{1}{2} \int_0^1 (e^{x-1} + e^{1-x})dx + \int_1^{1+\ell n \sqrt{3}} \dfrac{1}{2} (e^{x-1} + e^{1-x}) - (e^{x-1} - e^{1-x})dx$

$\displaystyle = \dfrac{1}{2} \int_1^{1+\ell n \sqrt{3}} (e^{x-1} + e^{1-x})dx - \int_1^{1+\ell n \sqrt{3}} (e^{x-1} - e^{1-x})dx$

$\displaystyle = \dfrac{1}{2} [e^{x-1} - e^{1-x} ]_0^{1+\ell n \sqrt{3}} - (e^{x-1} + e^{1-x})_1^{1+ \ell n \sqrt{3}}$

$= \displaystyle = \dfrac{1}{3} [ \sqrt{3} - \dfrac{1}{\sqrt{3}} - \dfrac{1}{e} + e] - [\sqrt{3} + \dfrac{1}{\sqrt{3}} - 1 - 1]$

$= 2 + \dfrac{1}{2} \left(e-\dfrac{1}{e}\right) - \dfrac{\sqrt{3}}{2} - \dfrac{3}{2\sqrt{3}}$

$= (2-\sqrt{3}) + \dfrac{e-\frac{1}{e}}{2}$

1993865_1950151_ans_b0eb0d07d68c4246aa7c02bbb4e3f140.png

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 14 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 14 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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