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CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 14 - MCQExams.com
CBSE
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
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$$ \dfrac{2}{3} $$
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$$ \dfrac{4 \pi - \sqrt{3}}{8 \pi + \sqrt{3}} $$
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$$ \dfrac{4 \pi + \sqrt{3}}{8 \pi - \sqrt{3}} $$
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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.
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
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10/3
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27/10
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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
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$$\dfrac{2}{3}$$
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$$\dfrac{4}{3}$$
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$$\dfrac{8}{3}$$
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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
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$$K$$
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$$2K$$
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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
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$$ \dfrac{4-log27}{3} $$
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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
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$$ \dfrac{2}{3} $$
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$$ \dfrac{8}{3} $$
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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
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$$ log4 $$
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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
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$$ 2 \pi + \dfrac{8}{3} $$
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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
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$$2$$
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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
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$$ \dfrac{4-3 ln3}{2} $$
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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
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$$ \dfrac{\pi}{2} $$
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$$ \dfrac{\pi}{\sqrt{2}} $$
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$$ \dfrac{\pi}{2 \sqrt{2}} $$
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$$ \dfrac{\pi}{4} $$
If $$f(x)=x-1$$ 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
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$$ \dfrac{4}{3} (4 \sqrt{2} - 5)$$
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$$ \dfrac{4}{3} (4 \sqrt{2} - 3)$$
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$$ \dfrac{8}{3} (4 \sqrt{2} - 3)$$
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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
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$$6 \pi$$ sq units
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$$3 \pi$$ sq units
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$$12 \pi$$ sq units
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$$2 \pi$$ 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
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$$\dfrac{ \pi^2}{2} -1$$
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$$\dfrac{ \pi^2}{2} -2$$
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$$\dfrac{ \pi^2}{2} -4$$
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$$\dfrac{ \pi^2}{2}$$
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
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$$ \frac{\pi a^2}{2} $$
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$$ \frac{ a^2}{3} $$
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$$ \frac{2 a^2}{3} $$
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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
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$$ \overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^r. h(x) dx$$
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$$ \overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^{r+1}. h(x) dx$$
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$$2 \overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^r. h(x) dx$$
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$$\dfrac{1}{2} \ \overset{n}{\underset{r=0}{\Sigma}} \ \int^{x_{r+1}}_{x_r} \ (-1)^{r+1}. h(x) dx$$
The area enclosed by the asteroid $$ \bigg( \dfrac{x}{a} \bigg)^{2/3} + \bigg( \dfrac{y}{a} \bigg)^{2/3} =1$$ is
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$$ \dfrac{3}{4} a^2 \pi $$
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$$ \dfrac{3}{18} \pi a^2 $$
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$$ \dfrac{3}{8} \pi a^2 $$
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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
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$$ 4 \bigg[ \int^\alpha_0 \ e \ x \ ln x \ dx + \int^1_\alpha (1- \sqrt{1-(x-1)^2)} \ dx \bigg] $$
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$$ 4 \bigg[ \int^\alpha_0 \ e \ x \ ln x \ dx + \int^\alpha_1 (1- \sqrt{1-(x-1)^2)} \ dx \bigg] $$
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$$ 4 \bigg[ - \int^\alpha_0 \ e \ x \ ln x \ dx + \int^1_\alpha (1- \sqrt{1-(x-1)^2)} \ dx \bigg] $$
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$$ 2 \bigg[ \int^\alpha_0 \ e \ x \ ln x \ dx + \int^\alpha_1 (1- \sqrt{1-(x-1)^2)} \ dx \bigg] $$
Area of the region enclosed between the curves $$x=y^2 -1$$ and $$x= \mid y \mid \sqrt{1-y^2}$$ is
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$$1$$
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$$\dfrac{4}{3}$$
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$$\dfrac{2}{3}$$
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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
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$$ \dfrac{27}{5} $$
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$$ \dfrac{24}{5} $$
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$$ \dfrac{27}{6} $$
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$$ \dfrac{21}{5} $$
The area enclosed by the curves $$x = a \sin^3t$$ and $$y = a \cos^3 t$$ is equal to
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$$\displaystyle 12a^2 \int_0^{\pi/2} \cos^4 t \sin^2t \ dt$$
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$$\displaystyle 12a \int_0^{\pi/2} \cos^2 t \sin^4t \ dt$$
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$$\displaystyle 2 \int_{-a}^{a} (a^{2/3} - x^{2/3})^{3/2}dx$$
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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.
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If both Statement are correct and Statement 2 is the correct explanation of Statement 1
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If both Statement are correct and Statement 2 is not the correct explanation of Statement 1
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If Statement 1 is correct and Statement 2 is incorrect
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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.
The area enclosed by the ellipse $$\dfrac {x^{2}}{a^{2}} + \dfrac {y^{2}}{b^{2}} = 1$$ is equal to
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$$\pi^{2} ab$$
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$$\pi ab$$
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$$\pi a^{2}b$$
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$$\pi ab^{2}$$
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
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$$-\frac{1}{4}$$ sq. units
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$$-\frac{1}{2}$$ sq. units
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$$-\frac{3}{4}$$ sq. units
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$$-1$$ 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
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$$(2-\sqrt{3})+\dfrac{1}{2}(e-e^{-1})$$
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$$(2+\sqrt{3})+\dfrac{1}{2}(e-e^{-1})$$
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$$(2-\sqrt{3})+\dfrac{1}{2}(e+e^{-1})$$
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$$(2+\sqrt{3})+\dfrac{1}{2}(e+e^{-1})$$
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 $$f(x) = g(x)$$
$$e^{x-1} - e^{-(x-1)} = \dfrac{1}{2} (e^{x-1} + e^{1-x})$$
$$\dfrac{1}{2} e^{x-1} = \dfrac{3}{2} e^{1-x}$$
$$e^{2x}=3e^2$$
$$2x = 2 + \ell n3$$
$$x = 1 + \ell n \sqrt{3}$$
$$\displaystyle A = \int_0^1 g(x) dx + \int_1^{1+\ell n\sqrt{3}} (g(x) - f(x)) dx$$
$$\displaystyle = \dfrac{1}{2} \int_0^1 (e^{x-1} + e^{1-x})dx + \int_1^{1+\ell n \sqrt{3}} \dfrac{1}{2} (e^{x-1} + e^{1-x}) - (e^{x-1} - e^{1-x})dx$$
$$\displaystyle = \dfrac{1}{2} \int_1^{1+\ell n \sqrt{3}} (e^{x-1} + e^{1-x})dx - \int_1^{1+\ell n \sqrt{3}} (e^{x-1} - e^{1-x})dx$$
$$\displaystyle = \dfrac{1}{2} [e^{x-1} - e^{1-x} ]_0^{1+\ell n \sqrt{3}} - (e^{x-1} + e^{1-x})_1^{1+ \ell n \sqrt{3}}$$
$$= \displaystyle = \dfrac{1}{3} [ \sqrt{3} - \dfrac{1}{\sqrt{3}} - \dfrac{1}{e} + e] - [\sqrt{3} + \dfrac{1}{\sqrt{3}} - 1 - 1]$$
$$= 2 + \dfrac{1}{2} \left(e-\dfrac{1}{e}\right) - \dfrac{\sqrt{3}}{2} - \dfrac{3}{2\sqrt{3}}$$
$$= (2-\sqrt{3}) + \dfrac{e-\frac{1}{e}}{2}$$
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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers
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