If the area (in sq. units) bounded by the parabola $y^{2}=4\lambda x$ and the line $y=\lambda x,\ \lambda>0$, is $\dfrac{1}{9}$, then $\lambda$ is equal to:
If the area of the region $\left\{(x, y):-1 \leq x \leq 1,0 \leq y \leq \mathrm{a}+\mathrm{e}^{|x|}-\mathrm{e}^{-x}, \mathrm{a}>0\right\}$ is $\frac{\mathrm{e}^2+8 \mathrm{e}+1}{\mathrm{e}}$, then the value of $a$ is
A line passing through the point $A(-2,0)$ touches the parabola $P: y^2=x-2$ at the point $B$ in the first quadrant. The area of the region bounded by the line $\overline{AB}$, parabola $P$ and the $x$-axis is:
$ \text{The area enclosed by the curves } y=\log_{e}(x+e^{2}),; x=\log_{e}!\left(\dfrac{2}{y}\right) \text{ and } x=\log_{e}2,\ \text{above the line } y=1,\ \text{is:} $
Let $g(x)=\cos x^{2}$, $f(x)=\sqrt{x}$ and $\alpha,\beta\ (\alpha<\beta)$ be the roots of the quadratic equation $18x^{2}-9\pi x+\pi^{2}=0$. Then the area (in sq. units) bounded by the curve $y=(g\circ f)(x)$ and the lines $x=\alpha$, $x=\beta$ and $y=0$ is :
Area (in sq. units) of the region outside
$\frac{|x|}{2} + \frac{|y|}{3} = 1$
and inside the ellipse
$\frac{x^2}{4}$ + $\frac{y^2}{9} = 1$ is
\[
2x - y + 2z = 2
\]
\[
x - 2y + \lambda z = -4
\]
\[
x + \lambda y + z = 4
\]
has no solution. Then the set $S$ :
If the area of the region $\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0<\mathrm{a}<1\right\}$ is $\left(\log _{\mathrm{e}} 2\right)-\frac{1}{7}$ then the value of $7 \mathrm{a}-3$ is equal to
Consider a region R = {(x, y) $ \in $ R : x2 $ \le $ y $ \le $ 2x}.
if a line y = $\alpha $ divides the area of region R intotwo equal parts, then which of the following istrue?
Let $S(\alpha)={(x,y):, y^{2}\le x,\ 0\le x\le \alpha}$ and $A(\alpha)$ be the area of the region $S(\alpha)$. If for a $\lambda$, $0<\lambda<4$, $A(\lambda):A(4)=2:5$, then $\lambda$ equals:
Let $\Delta$ be the area of the region $\{(x,y)\in\mathbb{R}^{2}:\ x^{2}+y^{2}\le 21,\ y^{2}\le 4x,\ x\ge 1\}$.
Then $\dfrac{1}{2}\Big(\Delta-21\sin^{-1}\!\dfrac{2}{\sqrt{7}}\Big)$ is equal to:
Let
$$A=\{(x,y)\in\mathbb{R}^{2}:\ y\ge 0,\ 2x\le y\le \sqrt{4-(x-1)^{2}}\}$$
and
$$B=\{(x,y)\in\mathbb{R}\times\mathbb{R}:\ 0\le y\le \min\{2x,\ \sqrt{4-(x-1)^{2}}\}\}.$$
Then the ratio of the area of $A$ to the area of $B$ is
Let $q$ be the maximum integral value of $p$ in $[0,10]$ for which the roots of the equation
$x^{2}-px+\dfrac{5}{4}p=0$ are rational. Then the area of the region
$\left\{(x,y): 0\le y\le (x-q)^{2},\ 0\le x\le q\right\}$ is:
One of the points of intersection of the curves
$y=1+3x-2x^2$ and $y=\dfrac{1}{x}$
is $\left(\dfrac{1}{2},\,2\right)$.
Let the area of the region enclosed by these curves be
$\dfrac{1}{24}\big(l\sqrt{5}+m\big)-n\ln(1+\sqrt{5})$,
where $l,m,n\in\mathbb{N}$.
Then $l+m+n$ is equal to: