Application of Integrals Test

Result

Application of Integrals Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

The area bounded by the curves $$y = x e ^ { x } , y = x e ^ { - x }$$ and the line $$x = 1 ,$$ is

A
$$\frac { 2 } { e }$$

B
$$1 - \frac { 2 } { e }$$

C
$$\frac { 1 } { e }$$

D
$$1 - \frac { 1 } { e }$$

Solution
Question 2
1 / -0

The area enclosed by the curve $$\left[x+3y\right]=\left[x-2\right]$$ where $$x\epsilon\left[3,4\right)$$ is (where $$[.]$$ denotes greatest integer function.)

A
$$\dfrac{2}{3}$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{1}{4}$$

D
$$1$$

Solution

$$\text { Given that: } \\$$
$$\qquad[x+3 y]=[x-2] \text { where } x \in[3,4) \\$$
$$x \in[3,4)$$

$$\therefore x-2 \in[1,2) \\$$
$$\therefore(x-2)=1$$

So,
$$[x+3 y]=1$$

that is for,
$$1 \leq x+3 y<2 \quad \& \quad 3 \leq x<4 \\$$
$$\frac{1-x}{3} \leq y<\frac{2-x}{3} \quad \& \quad 3 \leqslant x<4$$

$$\therefore \frac{1-x}{3} \leq y $$ & $$ y<\frac{2-x}{3} $$ & $$3 \leqslant x<4$$

$$\therefore x+3 y \geqslant 1 \quad \& \quad x+3 y<2$$
$$\& \quad 3 \leqslant x<4$$

The plot is as follows:

$$\therefore \text { Area }=\text { Area of } A B C D \text { - Area of } \triangle A B P \\$$
$$\text { - Area of } \triangle Q C D \\$$

= $$1\left(\frac{2}{3}\right)-\frac{1}{2}\left(\frac{1}{3}\right)-\frac{1}{2}\left(\frac{1}{3}\right) \\$$
$$=\frac{2}{3}-\frac{1}{3}$$

$$\text { doired Area } =\frac{1}{3} \text { sq. units }$$
Question 3
1 / -0

Area bounded by the parabola $${ x }^{ 2 }=36y$$ and its latus rectum is

A
116

B
616

C
216

D
126

Solution
Question 4
1 / -0

The area of the closed figure bounded by $$y = x , y = - x \&$$ the tangent to the curve $$y = \sqrt { x ^ { 2 } - 5 }$$ at the point $$( 3,2 )$$ is:

A
$$5$$

B
$$\frac { 15 } { 2 }$$

C
$$10$$

D
$$\frac { 35 } { 2 }$$

Solution
Question 5
1 / -0

Area of region bounded by $$[x]^2=[y]^2$$ if $$x\in [1,5]$$ where [.] represents the greatest integer function is-

A
$$10$$sq.units

B
$$8$$sq.units

C
$$6$$sq.units

D
$$5$$sq.units
Question 6
1 / -0

If $$\left[ \begin{array} { c c c } { 4 a ^ { 2 } } & { 4 a } & { 1 } \\ { 4 b ^ { 2 } } & { 4 b } & { 1 } \\ { 4 c ^ { 2 } } & { 4 c } & { 1 } \end{array} \right] \left[ \begin{array} { c } { f ( - 1 ) } \\ { f ( 1 ) } \\ { f ( 2 ) } \end{array} \right] = \left[ \begin{array} { c } { 3 a ^ { 2 } + 3 a } \\ { 3 b ^ { 2 } + 3 b } \\ { 3 c ^ { 2 } + 3 c } \end{array} \right]$$ $$f ( x )$$ is a quadratic function and its maximum value occurs at a point $$V. A$$ is a point of intersection of $$y = f ( x )$$ with $$x$$ -axis and point $$B$$ is such that chord $$AB$$ subtends a right angled at $$V .$$ Find The area enclosed by $$f ( x )$$ and chord $$A B .$$

A
$$\dfrac { 125 } { 3 }$$ sq unit

B
$$\dfrac { 115 } { 3 }$$ sq unit

C
$$\dfrac { 120 } { 3 }$$ sq unit

D
$$\dfrac { 130 } { 3 }$$ sq unit

Solution

Hence, Option (A) is the correct answer.
Question 7
1 / -0

If the area between the curves y=kx2 and x=ky2 is 1
then k is

A
1/\sqrt { 2 }

B
1/\sqrt { 3 }

C
1/\sqrt { 4 }

D
1
Question 8
1 / -0

The area between the parabola $${ y }^{ 2 }=4x$$, normal at one end of latusreetum and X-axis sin sq. units is

A
$$\frac { 1 }{ 3 } $$

B
$$\frac { 2 }{ 3 } $$

C
$$\frac { 10 }{ 3 } $$

D
$$\frac { 4 }{ 3 } $$

Solution
Question 9
1 / -0

Area bounded by Curve $${ y }^{ 2 }=4x,y$$ axis and line y=3 is :

A
7/4 Sq.unit

B
9/4 Sq. unit

C
5/4 Sq. unit

D
11/4 Sq. unit
Question 10
1 / -0

If $${ A }_{ n }$$ is the area bounded by y=x and y=$${ x }^{ n },n\epsilon N$$, then $${ A }_{ 2. }.{ A }_{ 3 }...{ A }_{ n }=$$

A
$$\dfrac { 1 }{ n\left( n+1 \right) } $$

B
$$\frac { 1 }{ { 2 }^{ n }n\left( n+1 \right) } $$

C
$$\frac { 1 }{ { 2 }^{ n-1 }n\left( n+1 \right) } $$

D
$$\frac { 1 }{ { 2 }^{ n-2 }n\left( n+1 \right) } $$

Solution