Application of Integration · Mathematics · JEE Advanced
MCQ (Single Correct Answer)
$$\left\{ {\matrix{ {(x,y):0 \le x \le {9 \over 4},} & {0 \le y \le 1,} & {x \ge 3y,} & {x + y \ge 2} \cr } } \right\}$$ is
f(x) = ex $$-$$ 1 $$-$$ e$$-$$|x $$-$$ 1|
and g(x) = $${1 \over 2}$$(ex $$-$$ 1 + e1 $$-$$ x).
The the area of the region in the first quadrant bounded by the curves y = f(x), y = g(x) and x = 0 is
{(x, y) : xy $$ \le $$ 8, 1 $$ \le $$ y $$ \le $$ x2} is
$$\left\{ {\left( {x,y} \right) \in {R^2}:y \ge \sqrt {\left| {x + 3} \right|} ,5y \le x + 9 \le 15} \right\}$$
is equal to
$$y = {\left( {1 - x} \right)^2},y = 0,$$ and $$x=0$$ into two parts $${R_1}\left( {0 \le x \le b} \right)$$ and
$${R_2}\left( {b \le x \le 1} \right)$$ such that $${R_1} - {R_2} = {1 \over 4}.$$ Then $$b$$ equals
$$f\left( x \right) = f\left( {1 - x} \right)$$ for all $$x \in \left[ { - 1,2} \right]$$
Let $${R_1} = \int\limits_{ - 1}^2 {xf\left( x \right)dx,} $$ and $${R_2}$$ be the area of the region bounded by $$y=f(x),$$ $$x=-1,$$ $$x=2,$$ and the $$x$$-axis. Then
Consider the polynomial
$$f\left( x \right) = 1 + 2x + 3{x^2} + 4{x^3}.$$
Let $$s$$ be the sum of all distinct real roots of $$f(x)$$ and let $$t = \left| s \right|.$$
The area bounded by the curve $$y=f(x)$$ and the lines $$x=0,$$ $$y=0$$ and $$x=t,$$ lies in the interval
Let $$f$$ be a non-negative function defined on the interval $$[0,1]$$.
If $$\int\limits_0^x {\sqrt {1 - {{(f'(t))}^2}dt} = \int\limits_0^x {f(t)dt,0 \le x \le 1} } $$, and $$f(0) = 0$$, then
and $$y = \sqrt {{{1 - \sin x} \over {\cos x}}} $$ bounded by the lines $$x=0$$ and $$x = {\pi \over 4}$$ is
The area of the region bounded by the curve $$y=f(x),$$ the
$$x$$-axis, and the lines $$x=a$$ and $$x=b$$, where $$ - \infty < a < b < - 2,$$ is :
$$y = {\left( {x - 1} \right)^2}$$ and the line $$y=1/4$$ is
$$x = a{y^2}\left( {a > 0} \right)$$ is $$1$$ sq. unit, then the value of $$a$$ is
$$x$$-axis in the 1st quadrant is
$${x^2} - f'\left( x \right) = 0$$ are
Numerical
Let the function $f:[1, \infty) \rightarrow \mathbb{R}$ be defined by
$$ f(t)=\left\{\begin{array}{cc} (-1)^{n+1} 2, & \text { if } t=2 n-1, n \in \mathbb{N}, \\ \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1), & \text { if } 2 n-1 < t < 2 n+1, n \in \mathbb{N} . \end{array}\right. $$
Define $g(x)=\int_1^x f(t) d t, x \in(1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x)=0$ in the interval $(1,8]$ and $\beta=\lim \limits_{x \rightarrow l+} \frac{g(x)}{x-1}$.
Then the value of $\alpha+\beta$ is equal to _______.
$$ f(x)= \begin{cases}n(1-2 n x) & \text { if } 0 \leq x \leq \frac{1}{2 n} \\\\ 2 n(2 n x-1) & \text { if } \frac{1}{2 n} \leq x \leq \frac{3}{4 n} \\\\ 4 n(1-n x) & \text { if } \frac{3}{4 n} \leq x \leq \frac{1}{n} \\\\ \frac{n}{n-1}(n x-1) & \text { if } \frac{1}{n} \leq x \leq 1\end{cases} $$
If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is 4 , then the maximum value of the function $f$ is :
$$ f(x)=x^{2}+\frac{5}{12} \quad \text { and } \quad g(x)= \begin{cases}2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4} \\ 0, & |x|>\frac{3}{4}\end{cases} $$
If $\alpha$ is the area of the region
$$ \left\{(x, y) \in \mathbb{R} \times \mathbb{R}:|x| \leq \frac{3}{4}, 0 \leq y \leq \min \{f(x), g(x)\}\right\}, $$
then the value of $9 \alpha$ is
The value of $$2{m_1} + 3{n_1} + {m_1}{n_1}$$ is ___________.
The value of $$6{m_2} + 4{n_2} + 8{m_2}{n_2}$$ is ___________.
MCQ (More than One Correct Answer)
$$f(x) = 1 - 2x + \int_0^x {{e^{x - t}}f(t)dt} $$ for all x $$ \in $$ [0, $$\infty $$). Then, which of the following statement(s) is (are) TRUE?
$$F\left( 1 \right) = 0,F\left( 3 \right) = - 4$$ and $$F\left( x \right) < 0$$ for all $$x \in \left( {{1 \over 2},3} \right).$$ Let $$f\left( x \right) = xF\left( x \right)$$ for all $$x \in R.$$
If $$\int_1^3 {{x^2}F'\left( x \right)dx = - 12} $$ and $$\int_1^3 {{x^3}F''\left( x \right)dx = 40,} $$ then the correct expression(s) is (are)
by $$\,f\left( x \right) = \ln x + \int\limits_0^x {\sqrt {1 + \sin t\,} dt.} $$ then which of the following
statement(s) is (are) true?
Subjective
Column $$I$$
(A) $$\int\limits_0^{\pi /2} {{{\left( {\sin x} \right)}^{\cos x}}\left( {\cos x\cot x - \log {{\left( {\sin x} \right)}^{\sin x}}} \right)dx} $$
(B) Area bounded by $$ - 4{y^2} = x$$ and $$x - 1 = - 5{y^2}$$
(C) Cosine of the angle of intersection of curves $$y = {3^{x - 1}}\log x$$ and $$y = {x^x} - 1$$ is
(D) Let $${{dy} \over {dx}} = {6 \over {x + y}}$$ where $$y(0)=0$$ then value of $$y$$ when $$x+y=6$$ is
Column $$II$$
(p) $$1$$
(q) $$0$$
(r) $$6\ln 2$$
(s) $${4 \over 3}$$
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 angle at $$V$$. Find the area enclosed by $$f(x)$$ and chord $$AB$$.
the region bounded by the $$y$$-axis and the curve $$x{e^{ay}} = \sin $$ by,
$${{jr} \over b} \le y \le {{\left( {j + 1} \right)\pi } \over b}.$$ Show that $${S_0},{S_1},{S_2},\,....,\,{S_n}$$ are in
geometric progression. Also, find their sum for $$a=-1$$ and $$b = \pi .$$
Find the area of the region in the third quadrant bounded by the curves $$x = - 2{y^2}$$ and $$y=f(x)$$ lying on the left of the line $$8x+1=0.$$
Determine the area of the region bounded by the curves
$$y = f\left( x \right),$$ $$x$$-axes, $$x=0$$ and $$x=1.$$
lines $$x=0,$$ $$y=0,$$ and $$x = {\pi \over 4}.$$ Prove that for $$n > 2,$$
$${A_n} + {A_{n - 2}} = {1 \over {n - 1}}$$ and deduce $${1 \over {2n + 2}} < {A_n} < {1 \over {2n - 2}}.$$
bounded by the parabolas $$y = 4x - {x^2}$$ and $$y = {x^2} - x?$$
$$y = {2 \over {1 + {x^2}}}.$$ Find the area.
$$x = {1 \over 2},x = 2,y = \ln \,x$$ and $$y = {2^x}.$$ Find the area of this region.
intersects the curve $${y^2} + \int\limits_0^x {f\left( t \right)dt = 2!} $$
$$\tan x,$$ tangent drawn to $$C$$ at $$x = {\pi \over 4}$$ and the $$x$$-axis.
the ordinates at $$x=2$$ and $$x=4$$. If the ordinate at $$x=a$$ divides the area into two equal parts, find $$a$$.
hyperbola $${x^2} - {y^2} = 1$$. Show that the area bounded by this hyperbola and the lines joining its centre to the points corresponding to $${t_1}$$ and $$-{t_1}$$ is $${t_1}$$.