MCQ 11 Mark
The area bounded by the curve $\text{y}=\log_{\text{e}}\text{x}$ and x-axis and the straight line x = e is:
AnswerCorrect option: B. $1\text{ sq. units}$
The point of intersection of the curve and the straight line is A(e, 1). Therefore, the area of the required region ABC,
$\text{A} = \int\limits^1_0(\text{x}_1-\text{x}_2)\text{dy}$ $(\text{where}, \text{x}_1 = \text{e}\text { and }\text{x}_2 = \text{e}_{\text{y}})$
$= \int\limits^1_0(\text{e}-\text{e}^{\text{y}})\text{dy}$
$=\big [\text{ey}-\text{e}^{\text{y}}\big]^1_0$
$=\big\{\text{e}(1)-\text{e}^{(1)}\big\} -\big \{\text{e}(0)-\text{e}^{(0)}\big\}$
$= \text{e}-\text{e}+1$
$= 1 \text{ square unit}$
View full question & answer→MCQ 21 Mark
Area bounded by the curve $y = x^3,$ the $x-$ axis and the ordinates $x = -2$ and $x = 1$ is :
- A
$-9$
- B
$\frac{-15}{4}$
- C
$\frac{15}{4}$
- ✓
$\frac{17}{4}$
AnswerCorrect option: D. $\frac{17}{4}$
$x = -2$ and $x = 1$ intersect the curve $y = x^3$ at $A(-2, -8)$ and $B(1, 1)$ respectively
If $P(x, y_1)$ lies on $OA , O(x, y_2)$ lies on curve $OB$
Then, $y_1 > 0 $
$\Rightarrow |y_1| = y_1$
$y_2 < 0$
$ \Rightarrow |y_2| = -y_2$
Area of curve bound by the two lines $=$ shaded are $\text{(OADO)}\ +$ shaded area $\text{(OCBO)}$
$= \int\limits^0_{-2}|\text{y}_2|\text{dx}+\int\limits^1_0|\text{y}_1|\text{dx}$
$= \int\limits^0_{-2}-\text{y}_2\text{dx}+\int\limits^1_0\text{y}_1\text{dx}$
$=\int\limits^0_{-2}-(\text{x}^3)\text{dx}+\int\limits^1_0\text{x}^3\text{dx}$
$= \Big[-\frac{\text{x}^4}{4}\Big]^0_{-2}+\Big[\frac{\text{x}^4}{4}\Big]^1_0$
$= 0-\Big(-\frac{16}{4}\Big)+\Big(\frac{1}{4}-0\Big)$
$= 4+\frac{1}{4}$
$=\frac{17}{4}\text{ sq. units}$
View full question & answer→MCQ 31 Mark
Area bounded by parabola $y^2 = x$ and straight line $2y = x$ is :
AnswerPoint of intersection is obtained by solving the equation of parabola $y^2 = x$ and equation of line $2y = x,$
we have
$y^2 = x$ and $2y = x$
$\Rightarrow y^2 = 2y$
$\Rightarrow y^2 - 2y = 0$
$\Rightarrow y = 0$ or $y = 2$
$\Rightarrow x = 0$ or $x = 4$
Thus $O(0, 0)$ and $A(4, 2)$ are the points of intersection of the curve and straight line.
Area bound by then
$\text{A}=\int\limits_0^4(\text{y}_1-\text{y}_2)\text{ dx}$
$\Big[$ Where, ${y}_1 =\sqrt{\text{x}}$ and ${y}_2=\frac{\text{x}}{2}\Big]$
$=\int\limits_0^4\Big(\sqrt{\text{x}}-\frac{\text{x}}{2}\Big)\text{dx}$
$=\Bigg[\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}-\frac{1}{2}\times\frac{\text{x}^2}{2}\Bigg]_0^4$
$=\Big[\frac{2}{3}\text{x}^\frac{3}{2}-\frac{\text{x}^2}{4}\Big]_0^4$
$=\frac{2}{3}4^\frac{3}{2}-\frac{1}{4}\times4^2-0$
$=\frac{2}{3}\times2^3-\frac{16}{4}$
$=\frac{16}{3}-4$
$=\frac{16-12}{3}$
$=\frac{4}{3}\text{ Sq units}$
View full question & answer→MCQ 41 Mark
The area bounded by the parabola $y^2 = 4ax$ and $x^2 = 4ay$ is :
- A
$\frac{8\text{a}^3}{3}$
- ✓
$\frac{16\text{a}^2}{3}$
- C
$\frac{32\text{a}^2}{3}$
- D
$\frac{64\text{a}^2}{3}$
AnswerCorrect option: B. $\frac{16\text{a}^2}{3}$
To find the point of intersection of the parabola substitute $\text{y} = \frac{\text{x}^{2}}{4\text{a}}$ in $y^2= 4ax$
We get,
$\frac{\text{x}^4}{16\text{a}^{2}}=4\text{ax}$
$\Rightarrow x^4 - 64a^3 x = 0$
$\Rightarrow x(x^3- 64a^3) = 0$
$\Rightarrow x = 0$ or $x = 4a$
$\Rightarrow y = 0$ or $y = 4a$
Therefore, the required area $\text{ABCD},$
$\text{A} =\int\limits^\text{4a}_0(\text{y}_1-\text{y}_2)\text{dx}$ $\Big($Where $, \text{ y}_1 = 2\sqrt{\text{ax}} $ and $\text{ y}_2=\frac{\text{x}^2}{\text{4a}}\Big)$
$= \int\limits^\text{4a}_0\Big(2\sqrt{\text{ax}}-\frac{\text{x}^2}{\text{4a}}\Big)\text{dx}$
$=\bigg[\frac{4\sqrt{\text{a}}}{3}\text{x}^\frac{3}{2}-\frac{\text{x}^{3}}{12\text{a}}\bigg]^\text{4a}_0$
$= \bigg[\frac{4\sqrt{\text{a}}}{3}(\text{4a}^\frac{3}{2})-\frac{(\text{4a})^3}{\text{12a}}\bigg]-\bigg[\frac{4\sqrt{\text{a}}}{3}(0)^\frac{3}{2}-\frac{(0)^3}{\text{12a}}\bigg]$
$= \bigg[\frac{4\sqrt{\text{a}}}{3}\text{8a}^\frac{3}{2}-\frac{64\text{a}^3}{12\text{a}}\bigg]-0$
$= \frac{32\text{a}^2}{3}-\frac{16\text{a}^3}{3}$
$= \frac{16\text{a}^2}{3}\text{ square units}$
View full question & answer→MCQ 51 Mark
Area lying between the curves $y^2 = 4x$ and $y = 2x$ is :
- A
$\frac{2}{3}$
- ✓
$\frac{1}{3}$
- C
$\frac{1}{4}$
- D
$\frac{3}{4}$
AnswerCorrect option: B. $\frac{1}{3}$
The points of intersection of the straight line and the parabola is obtained by solving the simultaneous equations,
$y^2 = 4x$ and $y = 2x$
$\Rightarrow (2x)^2 = 4x$
$\Rightarrow 4x^2 = 4x$
$\Rightarrow x(x - 1) = 0$
$\Rightarrow x = 0$ or $x = 1$
$\Rightarrow y = 0$ or $y = 2$
Thus, $O(0, 0)$ and $A(1, 2)$ are the points of intersection of the parabola and straight line shaded area is the required area.
Using the horizontal strip method, shaded area
$= \int\limits^2_0|\text{x}_2-\text{x}_1|\text{dy}$
$=\int\limits^2_0\Big[\Big(\frac{\text{y}}{2}\Big)-\Big(\frac{\text{y}^2}{4}\Big)\Big]\text{dy}$
$=\Big[\frac{1}{2}\Big(\frac{\text{y}^2}{2}\Big)-\frac{1}{4}\Big(\frac{\text{y}^3}{3}\Big)\Big]^2_0$
$=\frac{1}{4}(2)^2-\frac{1}{12}(2^3)-0$
$= 1 -\frac{8}{12}$
$= \frac{12-8}{12}$
$=\frac{1}{3}\text{ sq. units}$
View full question & answer→MCQ 61 Mark
The area bounded by the curve $y^2= 8x,$ the $x-$ axis and the lastus rectum is :
- ✓
$\frac{16}{3}$
- B
$\frac{23}{3}$
- C
$\frac{32}{3}$
- D
$\frac{16\sqrt{2}}{3}$
AnswerCorrect option: A. $\frac{16}{3}$
$y^2 = 8x$ represents a parabola opening side ways,
with vertex at $O(0, 0)$ and focus at $B(2, 0)$
Thus $AA\ ' $ represents the latus rectum of the parabola.
The points of intersection of the parabola and latus rectum are $A(2, 4)$ and $A\ '(2, -4)$
Area bound by curve $, x-$ axis and latus return is the area $\text{OABO},$
The approximating rectangle of with $= dx$ and length $= y$ has area $= y \ dx,$ and moves from $x = 0$ to
$\text{x} = 2\text{area}\text{ OABO}= \int\limits^2_0|\text{y}|\text{dx}$
$= \int\limits^2_0\text{y}\text{ dx} $
$\{\text{y}>0, \Rightarrow|\text{y}|=\text{y}\}$
$= \int\limits^2_0\sqrt{8\text{x}}\text{ dx}$
$= 2\sqrt{2}\int\limits^2_0\sqrt{\text{xdx}}$
$= 2\sqrt{2}\Bigg[\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}\Bigg]^2_0$
$=2 \sqrt{2}\times \frac{2}{3}\Big(2^\frac{3}{2}-0\Big)$
$= 4 \frac{\sqrt{2}}{3}\times2\sqrt{2}$
$=\frac{16}{3}\text{ sq. units}$
View full question & answer→MCQ 71 Mark
The area bounded by the curve $y = x^4 - 2x^3 + x^2 + 3$ with $x-$ axis and ordinates corresponding to the minima of $y$ is:
- A
$1$
- B
$\frac{91}{30}$
- C
$\frac{30}{9}$
- ✓
$4$
Answer
Clearly, from the figure the minimum value of $y$ is $3$ when $x=0$ or $1.$
Therefore, the required area $\text{ABCD},$
$\text{A} = \int\limits^1_0\text{y}\text{ dx}\ ($Where, $y = x^4 - 2x^3 + x^2 + 3)$
$= \int\limits^1_0(\text{x}^4-2\text{x}^3+\text{x}^2+3)\text{dx}$
$=\bigg[\frac{\text{x}^5}{5}-\frac{2\text{(x})^4}{4}+\frac{\text{x}^3}{3}+3\text{x}\bigg]^1_0$
$=\bigg[\frac{(1)^5}{5}-\frac{2(1)^4}{4}+\frac{(1)^3}{3}+3(1)\bigg]-\bigg[\frac{(0)^5}{5}-\frac{2(0)^4}{4}+\frac{(0)^3}{3}+3(0)\bigg]$
$=\big[\frac{1}{5}-\frac{1}{2}+\frac{1}{3}+3\big]- 0$
$=\frac{6-15+10+90}{3}$
$=\frac{91}{30}\text{ square units}$ View full question & answer→MCQ 81 Mark
The ratio of the areas between the curves $\text{y}=\cos\text{x}$ and $\text{y}=\cos2\text{x}$ and x-axis from x = 0 to x = 0 to $\text{x}=\frac{\pi}{3}$
- A
$1:2$
- B
$2:1$
- C
$\sqrt{3}:1$
- ✓
AnswerThe line $\text{x} = \pi3$ meets the curve $\text{y} = \cos\text {x}\text{ at}\text { B}\pi3,12$
Area between the curve y = cos x and x - axis from x = 0 and $\text{x} = 3\pi$ is,
$=\Big[2-\frac{1}{2}-\frac{1}{3}\Big]-\Big[-4-2+\frac{8}{3}\Big]$
$= 2 -\frac{1}{2}-\frac{1}{3}+4+2-\frac{8}{3}$
$=8-\frac{1}{2}-\frac{9}{3}$
$=5-\frac{1}{2}$
$=\frac{9}{2}\text{ square units}$
The line $\text{x}=\frac{\pi}{3}$ meets the curve y = cos 2x at $\text{B}'\pi3, -12$ Area between the curve y = cos 2x and x -axis from x = 0 and $\text{x}=\frac{\pi}{3}$ is,
$= \text{A}_2 = \int\limits^\frac{\pi}{4}_0\text{y}_2\text{ dx}-\int\limits^\frac{\pi}{3}_\frac{\pi}{4}\text{y}_2\text{dx}$ $\big[\text{where}, \text{y}_2 = \cos(2\text{x})\big]$
$=\int\limits^\frac{\pi}{4}_0\cos(2\text{x})\text{dx}-\int\limits^\frac{\pi}{3}_\frac{\pi}{4}\cos(2\text{x})\text{ dx}$
$=\Big[\frac{1}{2}\sin(2\text{x})\Big]^\frac{\pi}{4}_0-\Big[\frac{1}{2}\sin(2\text{x})\Big]^\frac{\pi}{3}_\frac{\pi}{4}$
$=\frac{1}{2}\Big[\sin\Big(\frac{\pi}{2}\Big)-\sin(0)\Big]-\frac{1}{2}\Big[\sin\Big(\frac{2\pi}{3}\Big)-\sin\Big]$
$= \frac{1}{2}-\frac{1}{2}\Big[\frac{\sqrt{3}}{2}-1\Big]$
$= \frac{1}{2}-\frac{\sqrt{3}}{4}+\frac{1}{2}$
$= 1-\frac{\sqrt{3}}{4}$
$=\frac{4-\sqrt{3}}{4}$
Therefore the retios will be
$\text{A}_1:\text{A}_2=\frac{\text{A}_1}{\text{A}_2}=\frac{\frac{\sqrt{3}}{2}}{\frac{4-\sqrt{3}}{4}}=\frac{2\sqrt{3}}{4-\sqrt{3}}$
View full question & answer→MCQ 91 Mark
The area bounded by the curve $\text{y}=\sin\text{x}$ between the ordinates $\text{x}=0,\text{x}=\pi$ and the x-axis is:
- ✓
$2\text{ sq. units}$
- B
$4\text{ sq. units}$
- C
$3\text{ sq. units}$
- D
$1\text{ sq. units}$
AnswerCorrect option: A. $2\text{ sq. units}$
$\text{A}=\int^\limits{\pi}_0\text{y}\text{ dx}$
$=\int^\limits{\pi}_0\sin(\text{x})\text{dx}$
$=\big[-\cos(\text{x})\big]^{\pi}_0$
$= -\cos(\pi)+\cos(0)$
$= 1 + 1$
$= 2 \text{ square units}$
View full question & answer→MCQ 101 Mark
The area bounded by the curve y = f(x), x-axis, and the ordinates x = 1 and $(\text{b}-1)\sin(3\text{b}+4)$ Then, f(x) is:
AnswerCorrect option: C. $\sin(3\text{x}+4)+3(\text{x}-1)\cos(3\text{x}+4)$
sin (3x + 4) + 3 (x - 1) cos (3x + 4)
y = fx
If A is the area bound by the curve, x-axis, x = 1 and x = b
$\Rightarrow \int\limits^\text{b}_1\text{f}(\text{x})\text{dx}=\big[\text{A}\big]^\text{b}_1 = (\text{b - 1})\sin(3\text{b} + 4)$ {given}
$\Rightarrow \text{f}(\text{x})=\frac{\text{d}}{\text{dx}}((\text{x} - 1)\sin(3\text{x}+ 4))$
$= \sin (3\text{x} + 4)\frac{\text{d}}{\text{dx}}(\text{x} - 1) + (\text{x} - 1)\frac{\text{d}}{\text{dx}}\sin(3\text{x} + 4)$
$= \sin (3\text{x} + 4) + 3(\text{x} - 1)\cos(3\text{x} + 4)$
View full question & answer→MCQ 111 Mark
Area lying first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the line $x = 0$ and $x = 2$, is :
- ✓
$\pi$
- B
$\frac{\pi}{2}$
- C
$\frac{\pi}{3}$
- D
$\frac{\pi}{4}$
Answer$x^2 + y^2 = 4$ represents a circle with centre at origion $O(0, 0)$ and radius $2$ units,
cutting the coordinate axis at $A, A\ ', B $ and $B\ ', x =2$
represents a straight line parallel to the $y-$ axis,
intersecting the circle at $A(2, 0)x = 0$ respresents the $y-$ axis
Area bounded by the circle and the two given lines in the first quadrant is the shaded area $\text{OBCAO}$
$\text{Area (OBCAO)}=\int\limits^2_0|\text{y}|\text{dx}$
$=\int\limits^2_0\sqrt{4-\text{x}^2}\text{dx}$
$=\bigg[\frac{1}{2}\text{x}\sqrt{4-\text{x}^2}+\frac{1}{2}\times4\sin^{-1}\Big(\frac{\text{x}}{2}\Big)\bigg]^2_0$
$= \frac{1}{2}\times2\sqrt{4-2^2}+\frac{1}{2}\times4\sin^{-1}\Big(\frac{2}{2}\Big)-0$
$= 0 + 2 \sin^{-1}(1)$
$= 2\times\frac{\pi}{2}$
$= \pi\text{ sq. units}$
View full question & answer→MCQ 121 Mark
The area of the region bounded by the parabola $y = x^2 + 1$ and the staight line $x + y = 3$ is given by :
- A
$\frac{45}{7}$
- B
$\frac{25}{4}$
- C
$\frac{\pi}{18}$
- ✓
$\frac{9}{2}$
AnswerCorrect option: D. $\frac{9}{2}$
To find the point of intersection of the parabola
$y = x^2 + 1$ and the line $x + y = 3$
substitute $y = 3 - x$ in $y = x^2 + 13 - x = x^2 + 1$
$\Rightarrow x^2 + x - 2 = 0$
$\Rightarrow (x - 1)(x + 2) = 0$
$\Rightarrow x = 1$ or $x = -2$
$\therefore y = 2$ or $y = 5$
So, we get the points of intersection $A (-2, 5)$ and $C (1, 2).$
Therefore, the required area $\text{ABC},$
$\text{A} = \int\limits^1_{-2}(\text{y}_1-\text{y}_2)\text{dx}$
$\big($Where $, \text{y}_1 = 3-\text{x }$ and $\text{ y}_2 = \text{x}^2+1\big)$
$=\int\limits^1_{-2}\big[(3-\text{x})-(\text{x}^2+1)\big]\text{dx}$
$=\int\limits^1_{-2}(3-\text{x}-\text{x}^2-1)\text{dx}$
$=\int\limits^1_{-2}\big(2-\text{x}-\text{x}^2\big)\text{dx}$
$= \Big[2\text{x}-\frac{\text{x}^2}{2}-\frac{\text{x}^3}{3}\Big]^1_{-2}$
$=\bigg[2(1)-\frac{(1)^2}{2}-\frac{(1)^3}{3}\bigg]-\bigg[2(-2)-\frac{(-2)^2}{2}-\frac{(-2)^3}{3}\bigg]$
$=\Big[2-\frac{1}{2}-\frac{1}{3}\Big]-\Big[-4-2+\frac{8}{3}\Big]$
$=2-\frac{1}{2}-\frac{1}{3}+4+2-\frac{8}{3}$
$=8-\frac{1}{2}-\frac{9}{3}$
$=5-\frac{1}{2}$
$=\frac{9}{2}\text{ square units}$
View full question & answer→MCQ 131 Mark
If the area above the $x-$ axis, bounded by the curve $y = 2kx$ and $x = 0$, and $x = 2$ is $\frac{3}{\log_{\text{e}}2},$ then the value of $k$ is:
- A
$\frac{1}{2}$
- ✓
$1$
- C
$-1$
- D
$2$
AnswerThe area bounded by the curves $y = 2^{kx}, x = 0$, and $x = 2$ is given by $\int\limits^2_02^\text{kx}\text{dx}.$ It is given that $\int\limits^2_02^{\text{kx}}\text{dx} = \frac{3}{\log_{\text{e}}(2)} $
$\Rightarrow\frac{1}{\text{k}}\bigg[\frac{2^\text{kx}}{\log_e(2)}\bigg]^2_0=\frac{3}{\log_{e}(2)}$
$\Rightarrow\frac{1}{\text{k}}\bigg[\frac{2^\text{k(2)}}{\log_e(2)}-\frac{2^{\text{k(0)}}}{\log_{e}(2)}\bigg]= \frac{3}{\log_{e}(2)}$
$\Rightarrow\frac{1}{\text{k}}\Big(\frac{2^{\text{2k}}}{\log_e(2)}-\frac{1}{\log_e(2)}\Big)= \frac{3}{\log_e(2)}$
$\Rightarrow\frac{1}{\text{k}}(2^{\text{2k}}-1)=3$
$\Rightarrow(2^{\text{2k}}-1)=3\text{k}$
$\Rightarrow2^{\text{2k}}-3\text{k}-1=0$
$\Rightarrow \text{k}= 1$
Clearly $, K = 1$ satisfies the equation.
Hence $, K = 1$
View full question & answer→MCQ 141 Mark
The area bounded by $y = 2 - x^2$ and $x + y = 0$ is :
AnswerCorrect option: B. $\frac{9}{2}\text{ sq. units}$
$($Image$)$
To find the points of intersection of $x + y = 0$ and $y = 2 - x^2$.
We put $x = -y$ in $y = 2 - x^2$,
We get $y = 2 - y^2$
$\Rightarrow y^2 + y - 2 = 0$
$\Rightarrow y - 1, y + 2 = 0$
$\Rightarrow y = 1, -2$
$\Rightarrow x = -1, 2$
Therefore, the points of intersection are $A(-1, 1)$ and $C(2, -2)$.
The area of the required region $\text{ABCD},$
$\text{A} = \int\limits^2_{-1}(\text{y}_1-\text{y}_{2})\text{dx}\ ($Where, $y_1 = 2 - x^2$ and $y_2 = -x)$
$=\int\limits^2_{-1}(2-\text{x}^{2}+\text{x})\text{dx}$
$ = \Big[2\text{x}-\frac{\text{x}^{3}}{3}+\frac{\text{x}^{2}}{2}\Big]^2_{-1}$
$= \bigg\{2(2)-\frac{(2)^{3}}{3}+\frac{(2)^{2}}{2}\bigg\}-\bigg\{2(-1)-\frac{(-1)^{3}}{3}+\frac{(-1)^{2}}{2}\bigg\}$
$= \Big(4-\frac{8}{3}+2\Big)-\Big(-2+\frac{1}{3}+\frac{1}{2}\Big)$
$=6-\frac{8}{3}+2-\frac{1}{3}-\frac{1}{2}$
$= 8 - \frac{9}{3}-\frac{1}{2}$
$= 5 -\frac{1}{2}$
$\frac{9}{2}\text{ sq. units}$
View full question & answer→MCQ 151 Mark
The area between x-axis and curve $\text{y}=\cos\text{x}$ when $0\leq\text{x}\leq2\pi$ is:
Answer
Required shaded area,
$\text{A} = \int\limits^\frac{\pi}{2}_0\cos\text{x}\text{ dx} + \int\limits^\frac{3\pi}{2}_\frac{\pi}{2}(-\cos\text{x})\text{dx} + \int\limits^{2\pi}_\frac{3\pi}{2}\cos\text{x}\text{ dx}$
$= \int\limits^\frac{\pi}{2}_0\cos\text{x}\text{ dx}-\int\limits^\frac{3\pi}{2}_\frac{\pi}{2}\cos\text{x}\text{ dx} + \int\limits^{2\pi}_\frac{3\pi}{2}\cos\text{x}\text{ dx}$
$=\Big[\sin\text{x}\Big]^{\frac{\pi}{2}}_0-\Big[\sin\text{x}\Big]^{\frac{3\pi}{2}}_\frac{\pi}{2}+\Big[\sin\text{x}\Big]^{2\pi}_\frac{3\pi}{2}$
$= \Big[\sin\text{x}\Big]^\frac{\pi}{2}_0-\Big[\sin\text{x}\Big]^\frac{3\pi}{2}_\frac{\pi}{2}+\Big[\sin\text{x}\Big]^{2\pi}_\frac{3\pi}{2}$
$=(1-0)-(-1-1)+\big[0-(-1)\big]$
$=1+2+1$
$=4\text{ sq. units}$
View full question & answer→MCQ 161 Mark
The area bounded by the curve $y = 4x - x^2$ and the $x-$ axis is :
- A
$\frac{30}{7}\text{ sq. units}$
- B
$\frac{31}{7}\text{ sq. units}$
- ✓
$\frac{32}{3}\text{ sq. units}$
- D
$\frac{34}{3}\text{ sq. units}$
AnswerCorrect option: C. $\frac{32}{3}\text{ sq. units}$
Point of intersection of parabola
$y = 4x - x^2$ with $x-$ axis is given by $y = 4x - x^2$ and $y = 0$
Equation of $x$ axis
$\Rightarrow 4x - x^2 = 0$
$\Rightarrow x = 0$ or $x = 4$
$\Rightarrow y = 0, y = 0$
Thus $0 (0, 0)$ and $B (4, 0)$ are points of intersection of parabola and $x -$ axis.
Required shaded area $= \int\limits^4_0(4\text{x}-\text{x}^2)\text{dx}$
$= \Big[2\text{ x}^2-\frac{\text{x}^3}{3}\Big]^4_0$
$= 2\times16-\frac{16}{3}-0$
$=\frac{96-64}{3}$
$=\frac{32}{3}\text{ sq. units}$
View full question & answer→MCQ 171 Mark
The area of the circle $x^2+ y^2 = 16$ enterior to the parabola $y^2 = 6x$ is :
- A
$\frac{4}{3}\big(4\pi-\sqrt{3}\big)$
- B
$\frac{4}{3}\big(4\pi+\sqrt{3}\big)$
- ✓
$\frac{4}{3}\big(8\pi-\sqrt{3}\big)$
- D
$\frac{4}{3}\big(8\pi+\sqrt{3}\big)$
AnswerCorrect option: C. $\frac{4}{3}\big(8\pi-\sqrt{3}\big)$
Points of intersection of the parabola and the circle is obtained by solving the simultaneous equation
$x^2 + y^2 = 16$ and $y^2 = 6x$
$\Rightarrow x^2 + 6x = 16$
$\Rightarrow x^2 = 6x -16 = 0$
$\Rightarrow (x + 8)(x - 2) = 0$
$\Rightarrow x = 2$ or $x = -8$
$\Rightarrow x = 2$ or $x = -8, x$ can not be $-8$ as in this case it will be the point outside circle.
$\therefore \text{x} = 2$
$\therefore$ when $ \text{x} = 2,\text{ y}=\pm\sqrt{6\times2}=\pm\sqrt{12}=\pm2\sqrt{3}$
$\therefore\text{B}(2, 2\sqrt{3})$ and $\text{ B}'(2,-2\sqrt{3})$ are points of intersection of the parabola and circle.
Required area
$=$ Area $\text{OB'C' A'CBO}$
$=$ area of circle $-$ area $\text{OBAB'O}$ Area of circle with radius $4$
$= \pi\times4^2$
$=16\pi$
Now,
Area $\text{OBAB'O}$
$= 2 \text{ areaOBAO}$
$= 2 \text{ area OBDO + area DBAD}$
$= 2\times\Bigg[\int\limits^2_0\sqrt{6\text{x}}\text{dx}+ \int\limits^4_2\sqrt{16-\text{x}^{2}}\Bigg]$
$= 2\times\Bigg\{\Bigg[\sqrt{6}\frac{\text{x}^\frac{3}{2}}{\frac{3}{2}}\Bigg]^2_0+ \Big[\frac{\text{x}}{2}\sqrt{16-\text{x}^2}+\frac{1}{2}\times16\sin^{-1}\Big(\frac{\text{x}}{4}\Big)\Big]^4_2\Bigg\}$
$= 2\times\Big\{\Big(\sqrt{6}\times\frac{2}{3}\times2^\frac{3}{2}-0\Big)+ \Big(\frac{1}{2}4\sqrt{16-(4)^2}\frac{1}{2}\times16\sin^{-1}\frac{4}{4}$
$-\frac{2}{2}\sqrt{16-2^2}-\frac{1}{2}\times16\sin^{-1}\frac{2}{4}\Big)\Big\}$
$= 2\times\bigg[\Big(\sqrt{6}\times\frac{2}{3}\times2\sqrt{2}\Big)+0+8\sin^{-1}(1)-\sqrt{12}-8\sin^{-1}\Big(\frac{1}{2}\Big)\bigg]$
$= 2\times \bigg[\frac{8\sqrt{3}}{3}+8\times\frac{\pi}{2}-2\sqrt{3}-8\frac{\pi}{6}\bigg]$
$= 2\bigg\{\frac{8\sqrt{3}-6\sqrt{3}}{3}+8\big(\frac{\pi}{2}-\frac{\pi}{2}\big)\bigg\}$
$= 2\bigg\{\frac{2\sqrt{3}}{3}+8\Big(\frac{2\pi}{6}\Big)\bigg\}$
$= \frac{4\sqrt{3}}{3}+\frac{16\pi}{3}$
View full question & answer→MCQ 181 Mark
If $A_n$ be the area bounded by the curve $y = (\tan x)^n$ and the lines $x = 0, y = 0 $ and $\text{x}=\frac{\pi}{4},$ then for $x > 2$
- ✓
$\text{A}_{\text{n}}+\text{A}_{\text{n}-2}=\frac{1}{\text{n}-1}$
- B
$\text{A}_{\text{n}}+\text{A}_{\text{n}-2} < \frac{1}{\text{n}-1}$
- C
$\text{A}_{\text{n}}-\text{A}_{\text{n}-2}=\frac{1}{\text{n}-1}$
- D
AnswerCorrect option: A. $\text{A}_{\text{n}}+\text{A}_{\text{n}-2}=\frac{1}{\text{n}-1}$
An $=$ Area bounded by the curve $\text{y}=\big\{\tan(\text{x})\big\}^\text{n}=\tan^\text{n}\text{(x)}$ and the lines $x = 0, y = 0,$ and $\text{x}=\frac{\pi}{4}.$
Therefore,
$\text{A}_\text{n}=\int\limits_\text{0}^{\frac{\pi}{4}}\tan^\text{n}\text{(x)}\text{dx}$
$\Rightarrow \text{A}_{\text{n}-2}=\int\limits_\text{0}^{\frac{\pi}{4}}\tan^{\text{n}-2}\text{(x)}\text{dx}$
Comsider, $\text{A}_\text{n}=\int\limits_\text{0}^{\frac{\pi}{4}}\tan^\text{n}\text{(x)}\text{dx}$
$\Rightarrow\ \text{A}_\text{n}=\int\limits_0^{\frac{\pi}{4}}\big\{\tan^{\text{n}-2}(\text{x})\big\}\big\{\tan^2(\text{x})\big\}\text{dx}$
$\Rightarrow\text{A}_\text{n}=\int\limits_0^\frac{\pi}{4}\big\{\tan^{\text{n}-2}(\text{x})\big\}\big\{\sec^2\text{(x)}-1\big\}\text{dx}$
$\Rightarrow\text{A}_\text{n}=\int\limits_0^\frac{\pi}{4}\big\{\tan^{\text{n}-2}\text{(x)} \sec^2(\text{x})-\tan^{\text{n}-2}(\text{x})\big\}\text{dx}$
$\Rightarrow\text{A}_\text{n}=\int\limits_0^{\frac{\pi}{4}}\big\{\tan^{\text{n}-2}\text{(x)}\sec^2(\text{x})\big\}\text{dx}-\int_0^\frac{\pi}{4}\tan^{\text{n}-2}\text{(x)}\text{ dx}$
$\Rightarrow\text{A}_\text{n}+\text{A}_{\text{n}-2}=\int\limits_0^\frac{\pi}{4}\tan^{\text{n}-2}(\text{x})\sec^2\text{(x) dx}$
Now, $\text{A}_\text{n}+\text{A}_{\text{n}+2}=\int\limits_0^\frac{\pi}{4}\tan^{\text{n}-2}\text{(x)}\sec^2(\text{x})\text{dx}$
Let $\text{u}=\tan\text{(x)}$
$\Rightarrow\text{du}=\sec^2\text{x }\text{dx}$
Also, when $x = 0, u = 0$ and when $\text{x}=\frac{\pi}{4},\text{u}=1$
Therefore,
$\text{A}_\text{n}+\text{A}_{\text{n}-2}=\int\limits_0^{\frac{\pi}{4}}\tan^{\text{n}-2}\text{(x)}\sec^2\text{(x) dx}$
$=\int\limits_0^1(\text{u}^{\text{n}-2})\text{du}$
$=\Big[\frac{\text{u}^{\text{n}-1}}{\text{n}-1}\Big]_0^1$
$=\Big[\frac{1}{\text{n}-1}-0\Big]=\frac{1}{\text{n}-1}$
View full question & answer→MCQ 191 Mark
The area of the region $($in square units$)$ bounded by the curve $x^2 = 4y,$ line $x = 2$ and $x-$ axis is :
- A
$1$
- ✓
$\frac{2}{3}$
- C
$\frac{4}{3}$
- D
$\frac{8}{3}$
AnswerCorrect option: B. $\frac{2}{3}$
$x^2 = 4y$ and $x= 2$
$\Rightarrow 4 = 4y$
$\Rightarrow y = 1$
$A(2, 1)$ is the point of intersection of curve and straight the
Area of shaded region $\text{OAB} = \int\limits^2_0\text{y}\text{ dx}$
$=\int\limits^2_0\frac{\text{x}^2}{4}\text{ dx}$
$=\Big[\frac{\text{x}^3}{12}\Big]^2_0$
$= \frac{2^3}{12}-0$
$= \frac{2}{3}\text{ square units}$
View full question & answer→MCQ 201 Mark
The area bounded by the y-axis, $\text{y}=\cos\text{x}$ and $\text{y}=\sin\text{x}$ when $0\leq\text{x}\leq\frac{\pi}{2}$ is:
- A
$2\big(\sqrt{2}-1\big)$
- ✓
$\sqrt{2}-1$
- C
$\sqrt{2}+1$
- D
$\sqrt{2}$
AnswerCorrect option: B. $\sqrt{2}-1$
Points of intersection is obtained by solving y = sinx and y = cos x
$\therefore\sin\text{x} = \cos \text{x}$
$\Rightarrow \text{x}=\frac{\pi}{4}$
Thus the two functions intesect at $\text{x}=\frac{\pi}{4}$
$\Rightarrow \text{y} = \sin \frac{\pi}{4} =\frac{1}{\sqrt{2}}$
Hence $\text{A}\Big(\frac{\pi}{4},\frac{1}{\sqrt{2}}\Big)$ is the point of intersection.
$\therefore$ Area bound by the curves and the y - axis when $0\leq\text{x}\leq\pi2$
$\text{A} = \int\limits^\frac{1}{\sqrt{2}}_0|\text{x}_1|\text{dy}+\int\limits^1_\frac{1}{\sqrt{2}}|\text{x}_2|\text{dy}$
$=\int\limits^\frac{1}{\sqrt{2}}_0\text{x}_1\text{ dy}+\int\limits^1_\frac{1}{\sqrt{2}}\text{x}_2\text{ dy}$
$= \int\limits^\frac{1}{\sqrt{2}}_0\sin^{-1}\text{y}\text{ dy}+\int\limits^1_\frac{1}{\sqrt{2}}\cos^{-1}\text{y}\text{ dy}$
$= \Big[\text{y}\sin^{-1}\text{y}+\sqrt{1-\text{y}^2}\Big]^\frac{1}{\sqrt{2}}_0+\Big[\text{y}\cos^{-1}\text{y}-\sqrt{1-\text{y}^2}\Big]^1_\frac{1}{\sqrt{2}}$
$= \Big[\frac{1}{\sqrt{2}}\sin^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{2}}-1\Big]\\+\bigg[1\times\cos^{-1}-0-\frac{1}{\sqrt{2}}\cos^{-1}\frac{1}{\sqrt{2}}+\sqrt{1-\frac{1}{2}}\bigg]$
$= \frac{2}{\sqrt{2}}-1$
$=\big(\sqrt{2}-1\big)\text{sq. units}$
View full question & answer→