2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
Let P(x) be a real polynomial of degree 3 which vanishes at x = $-$3. Let P(x) have local minima at x = 1, local maxima at x = $-$1 and $\int\limits_{ - 1}^1 {P(x)dx} $ = 18, then the sum of all the coefficients of the polynomial P(x) is equal to _________.
Correct Answer: 8
Explanation:
P'(x) = a(x + 1)(x $-$ 1)
$ \therefore $ P(x) = ${{a{x^3}} \over 3}$ $-$ ax + C
P($-$3) = 0 (given)
$ \Rightarrow $ a($-$9 + 3) + C = 0
$ \Rightarrow $ 6a = C ..... (i)
Also, $\int\limits_{ - 1}^1 {P(x)dx} = 18 $
$\Rightarrow \int\limits_{ - 1}^1 {\left( {a\left( {{{{x^3}} \over 3} - x} \right) + C} \right)} dx = 18$
$ \Rightarrow 0 + 2C = 18 \Rightarrow C = 9$
from (i)
$a = {3 \over 2}$
$ \therefore $ $P(x) = {{{x^3}} \over 2} - {3 \over 2}x + 9$
Sum of co-efficient
= ${1 \over 2} - {3 \over 2} + 9$ = $-$1 + 9 = 8
$ \therefore $ P(x) = ${{a{x^3}} \over 3}$ $-$ ax + C
P($-$3) = 0 (given)
$ \Rightarrow $ a($-$9 + 3) + C = 0
$ \Rightarrow $ 6a = C ..... (i)
Also, $\int\limits_{ - 1}^1 {P(x)dx} = 18 $
$\Rightarrow \int\limits_{ - 1}^1 {\left( {a\left( {{{{x^3}} \over 3} - x} \right) + C} \right)} dx = 18$
$ \Rightarrow 0 + 2C = 18 \Rightarrow C = 9$
from (i)
$a = {3 \over 2}$
$ \therefore $ $P(x) = {{{x^3}} \over 2} - {3 \over 2}x + 9$
Sum of co-efficient
= ${1 \over 2} - {3 \over 2} + 9$ = $-$1 + 9 = 8
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
Let f(x) and g(x) be two functions satisfying f(x2) + g(4 $-$ x) = 4x3 and g(4 $-$ x) + g(x) = 0, then the value of $\int\limits_{ - 4}^4 {f{{(x)}^2}dx} $ is
Correct Answer: 512
Explanation:
$I = 2\int\limits_0^4 {f({x^2})dx} $ ............(1)
$ \Rightarrow I = 2\int\limits_0^4 {f({{(4 - x)}^2})dx} $ ..............(2)
Adding equation (1) & (2)
$2I = 2\int\limits_0^4 {\left[ {f{{(x)}^2} + f{{(4 - x)}^2}} \right]} \,dx$ ............(3)
Now using $f{(x^2)} + g(4 - x) = 4{x^3}$ ............. (4)
$x \to 4 - x$
$f({(4 - x)^2}) + g(x) = 4{(4 - x)^3}$ ..............(5)
Adding equation (4) & (5)
$f({x^2}) + f(4 - {x^2}) + g(x) + g(4 - x) = 4({x^3} + {(4 - x)^3}]$
$ \Rightarrow f({x^2}) + f(4 - {x^2}) = 4({x^3} + {(4 - x)^3}]$
Now, $I = 4\int\limits_0^4 {\left( {{x^3} + {{(4 - x)}^3}} \right)dx = 512} $
$ \Rightarrow I = 2\int\limits_0^4 {f({{(4 - x)}^2})dx} $ ..............(2)
Adding equation (1) & (2)
$2I = 2\int\limits_0^4 {\left[ {f{{(x)}^2} + f{{(4 - x)}^2}} \right]} \,dx$ ............(3)
Now using $f{(x^2)} + g(4 - x) = 4{x^3}$ ............. (4)
$x \to 4 - x$
$f({(4 - x)^2}) + g(x) = 4{(4 - x)^3}$ ..............(5)
Adding equation (4) & (5)
$f({x^2}) + f(4 - {x^2}) + g(x) + g(4 - x) = 4({x^3} + {(4 - x)^3}]$
$ \Rightarrow f({x^2}) + f(4 - {x^2}) = 4({x^3} + {(4 - x)^3}]$
Now, $I = 4\int\limits_0^4 {\left( {{x^3} + {{(4 - x)}^3}} \right)dx = 512} $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
Let ${I_n} = \int_1^e {{x^{19}}{{(\log |x|)}^n}} dx$, where n$\in$N. If (20)I10 = $\alpha$I9 + $\beta$I8, for natural numbers $\alpha$ and $\beta$, then $\alpha$ $-$ $\beta$ equals to ___________.
Correct Answer: 1
Explanation:
${I_n} = 2\int\limits_1^e {{x^{19}}{{(\ln x)}^n}\,.\,dx} $
$ = {(\ln x)^n}\,.\,\left. {{{{x^{20}}} \over {20}}} \right|_1^e -\int\limits_1^e {n{{{{(\ln x)}^{n - 1}}} \over x}{{{x^{20}}} \over {20}}dx} $
${I_n} = {{{e^{20}}} \over {20}} - {n \over {20}}({I_{n - 1}})$
$20{I_n} = {e^{20}} - n\,{I_{n - 1}}$
Putting n = 10, we get
$20{I_{10}} = ({e^{20}} - 10{I_9})$ ...... (1)
Putting n = 9, we get
$20{I_9} = {e^{20}} - 9{I_8}$ ....... (2)
Subtracting (2) from (1), we get
$20{I_{10}} = 10{I_9} + 9{I_8}$
By comparing with (20)I10 = $\alpha$I9 + $\beta$I8, we get
$\alpha$ = 10, $\beta$ = 9 $ \Rightarrow $ $\alpha$ $-$ $\beta$ = 1
$ = {(\ln x)^n}\,.\,\left. {{{{x^{20}}} \over {20}}} \right|_1^e -\int\limits_1^e {n{{{{(\ln x)}^{n - 1}}} \over x}{{{x^{20}}} \over {20}}dx} $
${I_n} = {{{e^{20}}} \over {20}} - {n \over {20}}({I_{n - 1}})$
$20{I_n} = {e^{20}} - n\,{I_{n - 1}}$
Putting n = 10, we get
$20{I_{10}} = ({e^{20}} - 10{I_9})$ ...... (1)
Putting n = 9, we get
$20{I_9} = {e^{20}} - 9{I_8}$ ....... (2)
Subtracting (2) from (1), we get
$20{I_{10}} = 10{I_9} + 9{I_8}$
By comparing with (20)I10 = $\alpha$I9 + $\beta$I8, we get
$\alpha$ = 10, $\beta$ = 9 $ \Rightarrow $ $\alpha$ $-$ $\beta$ = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
If [ . ] represents the greatest integer function, then the value of
$\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$ is ____________.
$\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$ is ____________.
Correct Answer: 1
Explanation:
$\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]} dx$
$ = \int\limits_0^1 {[ - \cos x]dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {[1 - \cos x]dx} $
= $\int\limits_0^1 {\left[ { - \cos x} \right]} dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {1dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {\left[ { - \cos x} \right]} dx$
When 0 $ \le $ x $ \le $ 1 then -1 $ \le $ -cosx $ \le $ 0
$ \therefore $ ${\left[ { - \cos x} \right]}$ = -1
When 1 $ \le $ x $ \le $ ${\sqrt {{\pi \over 2}} }$ = 1.24 then -0.33 $ \le $ -cosx $ \le $ 0
$ \therefore $ ${\left[ { - \cos x} \right]}$ = -1 (Integer value present in the left side of -0.33)
$ = - \int\limits_0^1 {dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {dx - \int\limits_1^{\sqrt {{\pi \over 2}} } {dx} } } $
$ = - (x)_0^1 = - 1$
$ \therefore $ $\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$ = 1
$ = \int\limits_0^1 {[ - \cos x]dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {[1 - \cos x]dx} $
= $\int\limits_0^1 {\left[ { - \cos x} \right]} dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {1dx} + \int\limits_1^{\sqrt {{\pi \over 2}} } {\left[ { - \cos x} \right]} dx$
When 0 $ \le $ x $ \le $ 1 then -1 $ \le $ -cosx $ \le $ 0
$ \therefore $ ${\left[ { - \cos x} \right]}$ = -1
When 1 $ \le $ x $ \le $ ${\sqrt {{\pi \over 2}} }$ = 1.24 then -0.33 $ \le $ -cosx $ \le $ 0
$ \therefore $ ${\left[ { - \cos x} \right]}$ = -1 (Integer value present in the left side of -0.33)
$ = - \int\limits_0^1 {dx + \int\limits_1^{\sqrt {{\pi \over 2}} } {dx - \int\limits_1^{\sqrt {{\pi \over 2}} } {dx} } } $
$ = - (x)_0^1 = - 1$
$ \therefore $ $\left| {\int\limits_0^{\sqrt {{\pi \over 2}} } {\left[ {[{x^2}] - \cos x} \right]dx} } \right|$ = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
Let f : R $ \to $ R be a continuous function such that f(x) + f(x + 1) = 2, for all x$\in$R.
If ${I_1} = \int\limits_0^8 {f(x)dx} $ and ${I_2} = \int\limits_{ - 1}^3 {f(x)dx} $, then the value of I1 + 2I2 is equal to ____________.
If ${I_1} = \int\limits_0^8 {f(x)dx} $ and ${I_2} = \int\limits_{ - 1}^3 {f(x)dx} $, then the value of I1 + 2I2 is equal to ____________.
Correct Answer: 16
Explanation:
$f(x) + f(x + 1) = 2$ .... (i)
$x \to (x + 1)$
$f(x + 1) + f(x + 2) = 2$ .... (ii)
by (i) & (ii)
$f(x) - f(x + 2) = 0$
$f(x + 2) = f(x)$
$ \therefore $ f(x) is periodic with T = 2
${I_1} = \int_0^{2 \times 4} {f(x)dx} = 4\int_0^2 {f(x)dx} $
${I_2} = \int_{ - 1}^3 {f(x)dx} = \int_0^4 {f(x + 1)dx} = \int_0^4 {(2 - f(x))dx} $
$ \Rightarrow $ ${I_2} = 8 - 2\int_0^2 {f(x)dx} $
$ \Rightarrow $ ${I_2} = 8 - $${{{I_1}} \over 2}$
$ \Rightarrow $ ${I_1} + 2{I_2} = 16$
$x \to (x + 1)$
$f(x + 1) + f(x + 2) = 2$ .... (ii)
by (i) & (ii)
$f(x) - f(x + 2) = 0$
$f(x + 2) = f(x)$
$ \therefore $ f(x) is periodic with T = 2
${I_1} = \int_0^{2 \times 4} {f(x)dx} = 4\int_0^2 {f(x)dx} $
${I_2} = \int_{ - 1}^3 {f(x)dx} = \int_0^4 {f(x + 1)dx} = \int_0^4 {(2 - f(x))dx} $
$ \Rightarrow $ ${I_2} = 8 - 2\int_0^2 {f(x)dx} $
$ \Rightarrow $ ${I_2} = 8 - $${{{I_1}} \over 2}$
$ \Rightarrow $ ${I_1} + 2{I_2} = 16$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
Let f : (0, 2) $ \to $ R be defined as f(x) = log2$\left( {1 + \tan \left( {{{\pi x} \over 4}} \right)} \right)$. Then, $\mathop {\lim }\limits_{n \to \infty } {2 \over n}\left( {f\left( {{1 \over n}} \right) + f\left( {{2 \over n}} \right) + ... + f(1)} \right)$ is equal to ___________.
Correct Answer: 1
Explanation:
$E = 2\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} f\left( {{r \over n}} \right)$
$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {{\pi x} \over 4}} \right)dx} $ ..... (i)
replacing x $ \to $ 1 $-$ x
$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {\pi \over 4}(1 - x)} \right)dx} $
$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan \left( {{\pi \over 4} - {\pi \over 4}x} \right)} \right)dx} $
$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + {{1 - \tan {\pi \over 4}x} \over {1 + \tan {\pi \over 4}x}}} \right)dx} $
$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( { {2 \over {1 + \tan {{\pi x} \over 4}}}} \right)dx} $
$E = {2 \over {\ln 2}}\int_0^1 {\left( {\ln 2 - \ln \left( {1 + \tan {{\pi x} \over 4}} \right)} \right)dx} $ ..... (ii)
equation (i) + (ii)
2E = 2 $ \Rightarrow $ E = 1
$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {{\pi x} \over 4}} \right)dx} $ ..... (i)
replacing x $ \to $ 1 $-$ x
$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {\pi \over 4}(1 - x)} \right)dx} $
$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan \left( {{\pi \over 4} - {\pi \over 4}x} \right)} \right)dx} $
$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + {{1 - \tan {\pi \over 4}x} \over {1 + \tan {\pi \over 4}x}}} \right)dx} $
$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( { {2 \over {1 + \tan {{\pi x} \over 4}}}} \right)dx} $
$E = {2 \over {\ln 2}}\int_0^1 {\left( {\ln 2 - \ln \left( {1 + \tan {{\pi x} \over 4}} \right)} \right)dx} $ ..... (ii)
equation (i) + (ii)
2E = 2 $ \Rightarrow $ E = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
If the normal to the curve y(x) = $\int\limits_0^x {(2{t^2} - 15t + 10)dt} $ at a point (a, b) is parallel to the line x + 3y = $-$5, a > 1, then the value of | a + 6b | is equal to ___________.
Correct Answer: 406
Explanation:
Normal to the curve at point P(a, b) is parallel to the line x + 3y = $-$5.
mnormal = $ - {1 \over 3}$
$ \therefore $ mtangent = 3 = ${{dy} \over {dx}}$
Given y(x) = $\int\limits_0^x {(2{t^2} - 15t + 10)dt} $
$ \Rightarrow $ y'(x) = (2x2 $-$ 15x + 10)
at point P(a, b)
3 = (2a2 $-$ 15a + 10)
$ \Rightarrow $ 2a2 $-$ 15a + 7 = 0
$ \Rightarrow $ 2a2 $-$ 14a $-$ a + 7 = 0
$ \Rightarrow $ 2a(a $-$ 7) $-$ 1 (a $-$ 7) = 0
a = ${1 \over 2}$ or 7,
given a > 1 $ \therefore $ a = 7
As P(a, b) lies on curve
$ \therefore $ $b = \int_0^a {(2{t^2} - 15t + 10)dt} $
$b = \int_0^7 {(2{t^2} - 15t + 10)dt} $
$6b = - 413$
$ \therefore $ $|a + 6b|\, = 406$
mnormal = $ - {1 \over 3}$
$ \therefore $ mtangent = 3 = ${{dy} \over {dx}}$
Given y(x) = $\int\limits_0^x {(2{t^2} - 15t + 10)dt} $
$ \Rightarrow $ y'(x) = (2x2 $-$ 15x + 10)
at point P(a, b)
3 = (2a2 $-$ 15a + 10)
$ \Rightarrow $ 2a2 $-$ 15a + 7 = 0
$ \Rightarrow $ 2a2 $-$ 14a $-$ a + 7 = 0
$ \Rightarrow $ 2a(a $-$ 7) $-$ 1 (a $-$ 7) = 0
a = ${1 \over 2}$ or 7,
given a > 1 $ \therefore $ a = 7
As P(a, b) lies on curve
$ \therefore $ $b = \int_0^a {(2{t^2} - 15t + 10)dt} $
$b = \int_0^7 {(2{t^2} - 15t + 10)dt} $
$6b = - 413$
$ \therefore $ $|a + 6b|\, = 406$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
If ${I_{m,n}} = \int\limits_0^1 {{x^{m - 1}}{{(1 - x)}^{n - 1}}dx} $, for m, $n \ge 1$, and
$\int\limits_0^1 {{{{x^{m - 1}} + {x^{n - 1}}} \over {{{(1 + x)}^{m + 1}}}}} dx = \alpha {I_{m,n}}\alpha \in R$, then $\alpha$ equals ___________.
$\int\limits_0^1 {{{{x^{m - 1}} + {x^{n - 1}}} \over {{{(1 + x)}^{m + 1}}}}} dx = \alpha {I_{m,n}}\alpha \in R$, then $\alpha$ equals ___________.
Correct Answer: 1
Explanation:
${I_{m,n}} = \int\limits_0^1 {{x^{m - 1}}} .{(1 - x)^{n - 1}}dx$
Put $x = {1 \over {y + 1}} \Rightarrow dx = {{ - 1} \over {{{(y + 1)}^2}}}dy$
$1 - x = {y \over {y + 1}}$
$ \therefore $ ${I_{m,n}} = \int\limits_\infty ^0 {{{{y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}( - 1)dy = } \int\limits_0^\infty {{{{y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $ .... (i)
Similarly, ${I_{m,n}} = \int\limits_0^1 {{x^{n - 1}}.{{(1 - x)}^{m - 1}}dx} $
$ \Rightarrow {I_{m,n}} = \int\limits_0^\infty {{{{y^{m - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $ .... (ii)
From (i) & (ii)
$2{I_{m,n}} = \int\limits_0^\infty {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $
$ \Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_0^\infty {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} = {I_1} + {I_2}$
Put $y = {1 \over z}$ in I2
$dy = - {1 \over {{z^2}}}dz$
$ \Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_1^0 {{{{z^{m + 1}} + {z^{n - 1}}} \over {{{(z + 1)}^{m + n}}}}( - dz)} $
$ \Rightarrow {I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} \Rightarrow \alpha = 1$
Put $x = {1 \over {y + 1}} \Rightarrow dx = {{ - 1} \over {{{(y + 1)}^2}}}dy$
$1 - x = {y \over {y + 1}}$
$ \therefore $ ${I_{m,n}} = \int\limits_\infty ^0 {{{{y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}( - 1)dy = } \int\limits_0^\infty {{{{y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $ .... (i)
Similarly, ${I_{m,n}} = \int\limits_0^1 {{x^{n - 1}}.{{(1 - x)}^{m - 1}}dx} $
$ \Rightarrow {I_{m,n}} = \int\limits_0^\infty {{{{y^{m - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $ .... (ii)
From (i) & (ii)
$2{I_{m,n}} = \int\limits_0^\infty {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $
$ \Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_0^\infty {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} = {I_1} + {I_2}$
Put $y = {1 \over z}$ in I2
$dy = - {1 \over {{z^2}}}dz$
$ \Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_1^0 {{{{z^{m + 1}} + {z^{n - 1}}} \over {{{(z + 1)}^{m + n}}}}( - dz)} $
$ \Rightarrow {I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} \Rightarrow \alpha = 1$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
The value of the integral $\int\limits_0^\pi {|{{\sin }\,}2x|dx} $ is ___________.
Correct Answer: 2
Explanation:
$\begin{aligned} & \text { Let } I=\int_0^\pi|\sin 2 x| d x
\\\\ & =2 \int_0^{\pi / 2}|\sin 2 x| d x \quad[\because \sin 2 x \text { is periodic function }]
\\\\ & =2 \int_0^{\pi / 2} \sin 2 x \,d x[\sin 2 x \text { is positive in range }(0, \pi / 2)]
\\\\ & =2\left[\frac{-\cos 2 x}{2}\right]_0^{\pi / 2} \\\\ & =-[\cos \pi-\cos 0]=-(-1-1)=2 \end{aligned}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
The value of $\int\limits_{ - 2}^2 {|3{x^2} - 3x - 6|dx} $ is ___________.
Correct Answer: 19
Explanation:
x2 – x – 2 = (x - 2)(x + 1)
$\int\limits_{ - 2}^2 {3|{x^2} - x - 2|dx} $
$ = 3\int\limits_{ - 2}^2 {|{x^2} - x - 2|dx} $
$ = 3\left[ {\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - x - 2} \right)dx} + \int\limits_{ - 1}^2 { - \left( {{x^2} - x - 2} \right)dx} } \right]$
$ = 3\left[ {\left. {\left( {{{{x^3}} \over 3} - {{{x^2}} \over 2} - 2x} \right)} \right|_{ - 2}^{ - 1} - \left( {{{{x^3}} \over 3} - {{{x^2}} \over 2} - 2x} \right)_{ - 1}^2} \right]$
$ = 3\left[ {7 - {2 \over 3}} \right]$
= 19
$\int\limits_{ - 2}^2 {3|{x^2} - x - 2|dx} $
$ = 3\int\limits_{ - 2}^2 {|{x^2} - x - 2|dx} $
$ = 3\left[ {\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - x - 2} \right)dx} + \int\limits_{ - 1}^2 { - \left( {{x^2} - x - 2} \right)dx} } \right]$
$ = 3\left[ {\left. {\left( {{{{x^3}} \over 3} - {{{x^2}} \over 2} - 2x} \right)} \right|_{ - 2}^{ - 1} - \left( {{{{x^3}} \over 3} - {{{x^2}} \over 2} - 2x} \right)_{ - 1}^2} \right]$
$ = 3\left[ {7 - {2 \over 3}} \right]$
= 19
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
If $\int\limits_{ - a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx = 22$, (a > 2) and [x] denotes the greatest integer $ \le $ x, then$\int\limits_{ - a}^a {\left( {x + \left[ x \right]} \right)} dx$ is equal to _________.
Correct Answer: 3
Explanation:
$\int\limits_{ - a}^a {\left( {\left| x \right| + \left| {x - 2} \right|} \right)} dx = 22$
$ \Rightarrow $ $\int\limits_{ - a}^0 {( - 2x + 2)dx} + \int\limits_0^2 {(x + 2 - x)dx} + \int\limits_2^a {(2x - 2)dx} = 22$
$ \Rightarrow $ ${x^2} - 2x|_0^{ - a} + 2x|_0^2 + {x^2} - 2x|_2^a = 22$
$ \Rightarrow $ ${a^2} + 2a + 4 + {a^2} - 2a - (4 - 4) = 22$
$ \Rightarrow $ $2{a^2} = 18 \Rightarrow a = 3$
$\int\limits_3^{ - 3} {(x + [x])dx} = - \left( {\int\limits_{ - 3}^3 {(x + [x])dx} } \right) = - \left( {\int\limits_{ - 3}^3 {[x]dx} } \right)$
= -($\int\limits_{ - 3}^{ - 2} {\left[ x \right]dx} + \int\limits_{ - 2}^{ - 1} {\left[ x \right]dx} + \int\limits_{ - 1}^0 {\left[ x \right]dx} $
+ $\int\limits_0^1 {\left[ x \right]dx} + \int\limits_1^2 {\left[ x \right]dx} + \int\limits_2^3 {\left[ x \right]dx} $)
$ = - ( - 3 - 2 - 1 + 0 + 1 + 2) = 3$
$ \Rightarrow $ $\int\limits_{ - a}^0 {( - 2x + 2)dx} + \int\limits_0^2 {(x + 2 - x)dx} + \int\limits_2^a {(2x - 2)dx} = 22$
$ \Rightarrow $ ${x^2} - 2x|_0^{ - a} + 2x|_0^2 + {x^2} - 2x|_2^a = 22$
$ \Rightarrow $ ${a^2} + 2a + 4 + {a^2} - 2a - (4 - 4) = 22$
$ \Rightarrow $ $2{a^2} = 18 \Rightarrow a = 3$
$\int\limits_3^{ - 3} {(x + [x])dx} = - \left( {\int\limits_{ - 3}^3 {(x + [x])dx} } \right) = - \left( {\int\limits_{ - 3}^3 {[x]dx} } \right)$
= -($\int\limits_{ - 3}^{ - 2} {\left[ x \right]dx} + \int\limits_{ - 2}^{ - 1} {\left[ x \right]dx} + \int\limits_{ - 1}^0 {\left[ x \right]dx} $
+ $\int\limits_0^1 {\left[ x \right]dx} + \int\limits_1^2 {\left[ x \right]dx} + \int\limits_2^3 {\left[ x \right]dx} $)
$ = - ( - 3 - 2 - 1 + 0 + 1 + 2) = 3$
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Evening Slot
Let {x} and [x] denote the fractional part of x and
the greatest integer $ \le $ x respectively of a real
number x. If $\int_0^n {\left\{ x \right\}dx} ,\int_0^n {\left[ x \right]dx} $ and 10(n2 – n),
$\left( {n \in N,n > 1} \right)$ are three consecutive terms of a G.P., then n is equal to_____.
the greatest integer $ \le $ x respectively of a real
number x. If $\int_0^n {\left\{ x \right\}dx} ,\int_0^n {\left[ x \right]dx} $ and 10(n2 – n),
$\left( {n \in N,n > 1} \right)$ are three consecutive terms of a G.P., then n is equal to_____.
Correct Answer: 21
Explanation:
$\int\limits_0^n {\left\{ x \right\}} dx = n\int\limits_0^1 x dx = n\left( {{{{x^2}} \over 2}} \right)_0^1 = {n \over 2}$
[As period of {x} = 1]
$\int\limits_0^n {\left[ x \right]} dx = \int\limits_0^1 0 dx + \int\limits_1^2 1 dx + ... + \int\limits_{n - 1}^n {\left( {n - 1} \right)} dx$
= 1 + 2 + 3 + ....+ (n - 1)
= ${{n\left( {n - 1} \right)} \over 2}$
As ${n \over 2}$, ${{n\left( {n - 1} \right)} \over 2}$, 10(n2 – n) are in GP.
$ \therefore $ ${\left[ {{{n\left( {n - 1} \right)} \over 2}} \right]^2} = {n \over 2} \times 10\left( {{n^2} - n} \right)$
$ \Rightarrow $ n2 = 21n
$ \Rightarrow $ n = 21
[As period of {x} = 1]
$\int\limits_0^n {\left[ x \right]} dx = \int\limits_0^1 0 dx + \int\limits_1^2 1 dx + ... + \int\limits_{n - 1}^n {\left( {n - 1} \right)} dx$
= 1 + 2 + 3 + ....+ (n - 1)
= ${{n\left( {n - 1} \right)} \over 2}$
As ${n \over 2}$, ${{n\left( {n - 1} \right)} \over 2}$, 10(n2 – n) are in GP.
$ \therefore $ ${\left[ {{{n\left( {n - 1} \right)} \over 2}} \right]^2} = {n \over 2} \times 10\left( {{n^2} - n} \right)$
$ \Rightarrow $ n2 = 21n
$ \Rightarrow $ n = 21
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Evening Slot
Let [t] denote the greatest integer less than or
equal to t.
Then the value of $\int\limits_1^2 {\left| {2x - \left[ {3x} \right]} \right|dx} $ is ______.
Then the value of $\int\limits_1^2 {\left| {2x - \left[ {3x} \right]} \right|dx} $ is ______.
Correct Answer: 1
Explanation:
$\int\limits_1^2 {\left| {2x - \left[ {3x} \right]} \right|dx} $
$ = \int\limits_1^2 {\left| {2x - \left( {3x - \left\{ {3x} \right\}} \right)} \right|dx} $
$ = \int\limits_1^2 {\left| {\left\{ {3x} \right\} - x} \right|dx} $
We know, 0 $ \le $ $\left\{ {3x} \right\} < 1$ and x > 1
$\therefore \left\{ {3x} \right\} - x < 0$
So, $\left| {\left\{ {3x} \right\} - 2} \right| = - \left[ {\left\{ {3x} \right\} - x} \right]$
$ = \int\limits_1^2 {\left( {x - \left\{ {3x} \right\}} \right)dx} $
$ = \left[ {{{{x^2}} \over 2}} \right]_1^2 - \int\limits_{3 \times {1 \over 3}}^{6 \times {1 \over 3}} {\left\{ {3x} \right\}dx} $ [Period of {x} = 1, so period of {3x} = ${1 \over 3}$]
$ = \left( {2 - {1 \over 2}} \right) - \left( {6 - 3} \right)\int\limits_0^{{1 \over 3}} {3xdx} $
$ = {3 \over 2} - 3 \times 3\left[ {{{{x^2}} \over 2}} \right]_0^{{1 \over 3}}$
$ = {3 \over 2} - {9 \over 2}\left[ {{1 \over 9} - 0} \right]$
$ = {3 \over 2} - {1 \over 2}$ = 1
$ = \int\limits_1^2 {\left| {2x - \left( {3x - \left\{ {3x} \right\}} \right)} \right|dx} $
$ = \int\limits_1^2 {\left| {\left\{ {3x} \right\} - x} \right|dx} $
We know, 0 $ \le $ $\left\{ {3x} \right\} < 1$ and x > 1
$\therefore \left\{ {3x} \right\} - x < 0$
So, $\left| {\left\{ {3x} \right\} - 2} \right| = - \left[ {\left\{ {3x} \right\} - x} \right]$
$ = \int\limits_1^2 {\left( {x - \left\{ {3x} \right\}} \right)dx} $
$ = \left[ {{{{x^2}} \over 2}} \right]_1^2 - \int\limits_{3 \times {1 \over 3}}^{6 \times {1 \over 3}} {\left\{ {3x} \right\}dx} $ [Period of {x} = 1, so period of {3x} = ${1 \over 3}$]
$ = \left( {2 - {1 \over 2}} \right) - \left( {6 - 3} \right)\int\limits_0^{{1 \over 3}} {3xdx} $
$ = {3 \over 2} - 3 \times 3\left[ {{{{x^2}} \over 2}} \right]_0^{{1 \over 3}}$
$ = {3 \over 2} - {9 \over 2}\left[ {{1 \over 9} - 0} \right]$
$ = {3 \over 2} - {1 \over 2}$ = 1
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Morning Slot
The integral $\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx} $
is equal to______.
is equal to______.
Correct Answer: 1.50
Explanation:
$\int\limits_0^2 {\left| {\left| {x - 1} \right| - x} \right|dx} $
= $\int\limits_0^1 {\left| { - \left( {x - 1} \right) - x} \right|} dx$ + $ + \int\limits_1^2 {\left| {\left( {x - 1} \right) - x} \right|} dx$
= $\int\limits_0^1 {\left| {1 - x - x} \right|} dx + \int\limits_1^2 {dx} $
= $\int\limits_0^1 {\left| {1 - 2x} \right|} dx + \int\limits_1^2 {dx} $
= $\int\limits_0^{{1 \over 2}} {\left( {1 - 2x} \right)} dx + \int\limits_{{1 \over 2}}^1 {\left( {2x - 1} \right)dx} + \int\limits_1^2 {dx} $
= $\left[ {x - {x^2}} \right]_0^{{1 \over 2}} + \left[ {x - {x^2}} \right]_{{1 \over 2}}^1 + \left( {2 - 1} \right)$
= $\left( {{1 \over 2} - {1 \over 4}} \right) + \left[ {\left( {1 - 1} \right) - \left( {{1 \over 4} - {1 \over 2}} \right)} \right] + 1$
= ${3 \over 2}$ = 1.5
= $\int\limits_0^1 {\left| { - \left( {x - 1} \right) - x} \right|} dx$ + $ + \int\limits_1^2 {\left| {\left( {x - 1} \right) - x} \right|} dx$
= $\int\limits_0^1 {\left| {1 - x - x} \right|} dx + \int\limits_1^2 {dx} $
= $\int\limits_0^1 {\left| {1 - 2x} \right|} dx + \int\limits_1^2 {dx} $
= $\int\limits_0^{{1 \over 2}} {\left( {1 - 2x} \right)} dx + \int\limits_{{1 \over 2}}^1 {\left( {2x - 1} \right)dx} + \int\limits_1^2 {dx} $
= $\left[ {x - {x^2}} \right]_0^{{1 \over 2}} + \left[ {x - {x^2}} \right]_{{1 \over 2}}^1 + \left( {2 - 1} \right)$
= $\left( {{1 \over 2} - {1 \over 4}} \right) + \left[ {\left( {1 - 1} \right) - \left( {{1 \over 4} - {1 \over 2}} \right)} \right] + 1$
= ${3 \over 2}$ = 1.5
2012
JEE Mains
MSQ
AIEEE 2012
If $g\left( x \right) = \int\limits_0^x {\cos 4t\,dt,} $ then $g\left( {x + \pi } \right)$ equals
A.
${{g\left( x \right)} \over {8\left( \pi \right)}}$
B.
$g\left( x \right) + g\left( \pi \right)$
C.
$g\left( x \right) - g\left( \pi \right)$
D.
$g\left( x \right) . g\left( \pi \right)$
2007
JEE Advanced
Numerical
IIT-JEE 2007
Match the integrals in Column $I$ with the values in Column $II$ and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the $ORS$.
Column $I$
(A) $\int\limits_{ - 1}^1 {{{dx} \over {1 + {x^2}}}} $
(B) $\int\limits_0^1 {{{dx} \over {\sqrt {1 - {x^2}} }}} $
(C) $\int\limits_2^3 {{{dx} \over {1 - {x^2}}}} $
(D) $\int\limits_1^2 {{{dx} \over {x\sqrt {{x^2} - 1} }}} $
Column $II$
(p) ${1 \over 2}\log \left( {{2 \over 3}} \right)$
(q) $2\log \left( {{2 \over 3}} \right)$
(r) ${{\pi \over 3}}$
(s) ${{\pi \over 2}}$
Correct Answer: $$\left( A \right) - \left( s \right),\left( B \right) - \left( s \right),\left( C \right) - \left( p \right),\left( D \right) - \left( r \right)$$
2005
JEE Advanced
Numerical
IIT-JEE 2005
Evaluate $\,\int\limits_0^\pi {{e^{\left| {\cos x} \right|}}} \left( {2\sin \left( {{1 \over 2}\cos x} \right) + 3\cos \left( {{1 \over 2}\cos x} \right)} \right)\sin x\,\,dx$
Correct Answer: $${{24} \over 5}\left[ {e\cos \left( {{1 \over 2}} \right) + {1 \over 2}e\sin \left( {{1 \over 2}} \right) - 1} \right]$$
2004
JEE Advanced
Numerical
IIT-JEE 2004
If $y\left( x \right) = \int\limits_{{x^2}/16}^{{x^2}} {{{\cos x\cos \sqrt \theta } \over {1 + {{\sin }^2}\sqrt \theta }}d\theta ,} $ then find ${{dy} \over {dx}}$ at $x = \pi $
Correct Answer: $$2\pi $$
2004
JEE Advanced
Numerical
IIT-JEE 2004
Find the value of $\int\limits_{ - \pi /3}^{\pi /3} {{{\pi + 4{x^3}} \over {2 - \cos \left( {\left| x \right| + {\pi \over 3}} \right)}}dx} $
Correct Answer: $${{4\pi } \over {\sqrt 3 }}\left[ {{{\tan }^{ - 1}}3 - {\pi \over 4}} \right]$$
2003
JEE Advanced
Numerical
IIT-JEE 2003
If $f$ is an even function then prove that
$\int\limits_0^{\pi /2} {f\left( {\cos 2x} \right)\cos x\,dx = \sqrt 2 } \int\limits_0^{\pi /4} {f\left( {\sin 2x} \right)\cos x\,dx.} $
$\int\limits_0^{\pi /2} {f\left( {\cos 2x} \right)\cos x\,dx = \sqrt 2 } \int\limits_0^{\pi /4} {f\left( {\sin 2x} \right)\cos x\,dx.} $
Correct Answer: Solve it.
2000
JEE Advanced
Numerical
IIT-JEE 2000
For $x>0,$ let $f\left( x \right) = \int\limits_e^x {{{\ln t} \over {1 + t}}dt.} $ Find the function
$f\left( x \right) + f\left( {{1 \over x}} \right)$ and show that $f\left( e \right) + f\left( {{1 \over e}} \right) = {1 \over 2}.$
Here, $\ln t = {\log _e}t$.
$f\left( x \right) + f\left( {{1 \over x}} \right)$ and show that $f\left( e \right) + f\left( {{1 \over e}} \right) = {1 \over 2}.$
Here, $\ln t = {\log _e}t$.
Correct Answer: Solve it.
1999
JEE Advanced
Numerical
IIT-JEE 1999
Integrate $\int\limits_0^\pi {{{{e^{\cos x}}} \over {{e^{\cos x}} + {e^{ - \cos x}}}}\,dx.} $
Correct Answer: $${\pi \over 2}$$
1998
JEE Advanced
Numerical
IIT-JEE 1998
Prove that $\int_0^1 {{{\tan }^{ - 1}}} \,\left( {{1 \over {1 - x + {x^2}}}} \right)dx = 2\int_0^1 {{{\tan }^{ - 1}}} \,x\,dx.$
Hence or otherwise, evaluate the integral
$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx.} $
Hence or otherwise, evaluate the integral
$\int_0^1 {{{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)dx.} $
Correct Answer: $$\log 2$$
1997
JEE Advanced
Numerical
IIT-JEE 1997
Determine the value of $\int_\pi ^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} \,dx.$
Correct Answer: $${\pi ^2}$$
1995
JEE Advanced
Numerical
IIT-JEE 1995
Let ${I_m} = \int\limits_0^\pi {{{1 - \cos mx} \over {1 - \cos x}}} dx.$ Use mathematical induction to prove that ${I_m} = m\,\pi ,m = 0,1,2,........$
Correct Answer: Solve it.
1995
JEE Advanced
Numerical
IIT-JEE 1995
Evaluate the definite integral :
$$\int\limits_{ - 1/\sqrt 3 }^{1/\sqrt 3 } {\left( {{{{x^4}} \over {1 - {x^4}}}} \right){{\cos }^{ - 1}}\left( {{{2x} \over {1 + {x^2}}}} \right)} dx$$
Correct Answer: $${\pi \over {12}}\left[ {\pi + 3{{\log }_e}\left( {2 + \sqrt 3 } \right) - 4\sqrt 3 } \right]$$
1994
JEE Advanced
Numerical
IIT-JEE 1994
Show that $\int\limits_0^{n\pi + v} {\left| {\sin x} \right|dx = 2n + 1 - \cos \,v} $ where $n$ is a positive integer and $\,0 \le v < \pi .$
Correct Answer: $$2n + 1 - \cos \gamma $$
1993
JEE Advanced
Numerical
IIT-JEE 1993
Evaluate $\int_2^3 {{{2{x^5} + {x^4} - 2{x^3} + 2{x^2} + 1} \over {\left( {{x^2} + 1} \right)\left( {{x^4} - 1} \right)}}} dx.$
Correct Answer: $${1 \over 6}\log 6 - {1 \over {10}}$$
1992
JEE Advanced
Numerical
IIT-JEE 1992
Determine a positive integer $n \le 5,$ such that
$$\int\limits_0^1 {{e^x}{{\left( {x - 1} \right)}^n}dx = 16 - 6e} $$
Correct Answer: $$n = 3$$
1991
JEE Advanced
Numerical
IIT-JEE 1991
Evaluate $\,\int\limits_0^\pi {{{x\,\sin \,2x\,\sin \left( {{\pi \over 2}\cos x} \right)} \over {2x - \pi }}dx} $
Correct Answer: $${8 \over {{\pi ^2}}}$$
1990
JEE Advanced
Numerical
IIT-JEE 1990
Prove that for any positive integer $k$,
${{\sin 2kx} \over {\sin x}} = 2\left[ {\cos x + \cos 3x + ......... + \cos \left( {2k - 1} \right)x} \right]$
Hence prove that $\int\limits_0^{\pi /2} {\sin 2kx\,\cot \,x\,dx = {\pi \over 2}} $
${{\sin 2kx} \over {\sin x}} = 2\left[ {\cos x + \cos 3x + ......... + \cos \left( {2k - 1} \right)x} \right]$
Hence prove that $\int\limits_0^{\pi /2} {\sin 2kx\,\cot \,x\,dx = {\pi \over 2}} $
Correct Answer: Solve it.
1990
JEE Advanced
Numerical
IIT-JEE 1990
Show that $\int\limits_0^{\pi /2} {f\left( {\sin 2x} \right)\sin x\,dx = \sqrt 2 } \int\limits_0^{\pi /4} {f\left( {\cos 2x} \right)\cos x\,dx} $
Correct Answer: Solve it.
1989
JEE Advanced
Numerical
IIT-JEE 1989
If $f$ and $g$ are continuous function on $\left[ {0,a} \right]$ satisfying
$f\left( x \right) = f\left( {a - x} \right)$ and $g\left( x \right) + g\left( {a - x} \right) = 2,$
then show that $\int\limits_0^a {f\left( x \right)g\left( x \right)dx = \int\limits_0^a {f\left( x \right)dx} } $
$f\left( x \right) = f\left( {a - x} \right)$ and $g\left( x \right) + g\left( {a - x} \right) = 2,$
then show that $\int\limits_0^a {f\left( x \right)g\left( x \right)dx = \int\limits_0^a {f\left( x \right)dx} } $
Correct Answer: Solve it.
1988
JEE Advanced
Numerical
IIT-JEE 1988
Evaluate $\int\limits_0^1 {\log \left[ {\sqrt {1 - x} + \sqrt {1 + x} } \right]dx} $
Correct Answer: $${1 \over 2}\left[ {\log 2 + {\pi \over 2} - 1} \right]$$
1986
JEE Advanced
Numerical
IIT-JEE 1986
Evaluate : $\int\limits_0^\pi {{{x\,dx} \over {1 + \cos \,\alpha \,\sin x}},0 < \alpha < \pi } $
Correct Answer: $${{\pi \alpha } \over {\sin x}}$$
1985
JEE Advanced
Numerical
IIT-JEE 1985
Evaluate the following : $\,\,\int\limits_0^{\pi /2} {{{x\sin x\cos x} \over {{{\cos }^4}x + {{\sin }^4}x}}} dx$
Correct Answer: $${{{\pi ^2}} \over {16}}$$
1984
JEE Advanced
Numerical
IIT-JEE 1984
Given a function $f(x)$ such that
(i) it is integrable over every interval on the real line and
(ii) $f(t+x)=f(x),$ for every $x$ and a real $t$, then show that
the integral $\int\limits_a^{a + 1} {f\,\,\left( x \right)} \,dx$ is independent of a.
(i) it is integrable over every interval on the real line and
(ii) $f(t+x)=f(x),$ for every $x$ and a real $t$, then show that
the integral $\int\limits_a^{a + 1} {f\,\,\left( x \right)} \,dx$ is independent of a.
Correct Answer: Solve it.
1984
JEE Advanced
Numerical
IIT-JEE 1984
Evaluate the following $\int\limits_0^{{1 \over 2}} {{{x{{\sin }^{ - 1}}x} \over {\sqrt {1 - {x^2}} }}dx} $
Correct Answer: $${{6 - \pi \sqrt 3 } \over {12}}$$
1983
JEE Advanced
Numerical
IIT-JEE 1983
Evaluate : $\int\limits_0^{\pi /4} {{{\sin x + \cos x} \over {9 + 16\sin 2x}}dx} $
Correct Answer: $${1 \over {20}}\log 3$$
1982
JEE Advanced
Numerical
IIT-JEE 1982
Show that $\int\limits_0^\pi {xf\left( {\sin x} \right)dx} = {\pi \over 2}\int\limits_0^\pi {f\left( {\sin x} \right)dx.} $
Correct Answer: Solve it.
1982
JEE Advanced
Numerical
IIT-JEE 1982
Find the value of $\int\limits_{ - 1}^{3/2} {\left| {x\sin \,\pi \,x} \right|\,dx} $
Correct Answer: $${3 \over \pi } + {1 \over {{\pi ^2}}}$$
1981
JEE Advanced
Numerical
IIT-JEE 1981
Show that : $\mathop {\lim }\limits_{n \to \infty } \left( {{1 \over {n + 1}} + {1 \over {n + 2}} + .... + {1 \over {6n}}} \right) = \log 6$
Correct Answer: Solve it.
2025
JEE Advanced
Numerical
JEE Advanced 2025 Paper 2 Online
If
$ \alpha=\int\limits_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x $
then the value of $\sqrt{7} \tan \left(\frac{2 \alpha \sqrt{7}}{\pi}\right)$ is _________.
(Here, the inverse trigonometric function $\tan ^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.)
Correct Answer: 21
Explanation:
$ \begin{aligned} & \alpha=\int_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{2 x^2-3 x+2} d x ........(i) \\ & \text { Let } x=\frac{1}{t} \\ & \qquad d x=-\frac{1}{t^2} d t \\ & \alpha=\int_2^{\frac{1}{2}} \frac{\tan ^{-1}\left(\frac{1}{t}\right)}{\frac{2}{t^2}-\frac{3}{t}+2}\left(\frac{-1}{t^2}\right) d t \end{aligned} $
$ \begin{aligned} & \alpha=\int_2^{\frac{1}{2}} \frac{\tan ^{-1}\left(\frac{1}{t}\right)}{\frac{2}{t^2}-\frac{3}{t}+2}\left(\frac{-1}{t^2}\right) d t \\ & \alpha=\int_{\frac{1}{2}}^2 \frac{\cot ^{-1} t}{2 t^2-3 t+2} d t .......(ii) \end{aligned} $
Now by (i) + (ii)
$ \begin{aligned} & 2 \alpha=\int_{\frac{1}{2}}^2 \frac{\frac{\pi}{2}}{2 x^2-3 x+2} d x \\ & \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2 \frac{d x}{x^2-\frac{3 x}{2}+1} \\ & \alpha=\frac{\pi}{8} \int_{\frac{1}{2}}^2 \frac{d x}{\left(x-\frac{3}{4}\right)^2+\frac{7}{16}} \\ & \alpha=\frac{\pi}{8 \times \frac{\sqrt{7}}{4}}\left[\tan ^{-1}\left(\frac{x-\frac{3}{4}}{\frac{\sqrt{7}}{4}}\right)\right]_{\frac{1}{2}}^2 \end{aligned} $
$\begin{aligned} & \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{4 x-3}{\sqrt{7}}\right]_{\frac{1}{2}}^2 \\ & \alpha=\frac{\pi}{2 \sqrt{7}}\left[\tan ^{-1} \frac{5}{\sqrt{7}}-\tan ^{-1}\left(-\frac{1}{\sqrt{7}}\right)\right] \\ & \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1} \frac{\left(\frac{5}{\sqrt{7}}+\frac{1}{\sqrt{7}}\right)}{1-\frac{5}{7}} \\ & \alpha=\frac{\pi}{2 \sqrt{7}} \tan ^{-1}(3 \sqrt{7}) \end{aligned}$
Now $\sqrt{7} \tan \left(\frac{2 \sqrt{7} \alpha}{\pi}\right)$
$ \begin{aligned} & \sqrt{7} \times \tan \left(\tan ^{-1}(3 \sqrt{7})\right) \\ & \sqrt{7} \times 3 \sqrt{7} \\ & =21 \end{aligned} $
2024
JEE Advanced
Numerical
JEE Advanced 2024 Paper 2 Online
The value of $2 \int\limits_0^{\frac{\pi}{2}} f(x) g(x) d x-\int\limits_0^{\frac{\pi}{2}} g(x) d x$ is ____________.
Correct Answer: 0
Explanation:
$\mathrm{I}=2 \int_\limits0^{\frac{\pi}{2}} \underbrace{\sin ^2 \mathrm{x} \cdot \sqrt{\frac{\pi x}{2}-\mathrm{x}^2}}_{\mathrm{i}_1}-\int_\limits0^{\frac{\pi}{2}} \mathrm{~g}(\mathrm{x}) \mathrm{dx}$
$\text { Let } I_1=\int_\limits0^{\frac{\pi}{2}} \sin ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} \quad \text { (making perfect square) }$
apply kings
$I_1=\int_\limits0^{\frac{\pi}{2}} \cos ^2 x \sqrt{\left(\frac{\pi}{4}\right)^2-\left(\frac{\pi}{2}-x\right)^2}$
add both
$2 I_1=\int_\limits0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2}$
i.e. $2 I_1=\int_\limits0^{\frac{\pi}{2}} g(x)$
Now $I=2 I_1-\int_\limits0^{\frac{\pi}{2}} g(x)=0$
2024
JEE Advanced
Numerical
JEE Advanced 2024 Paper 2 Online
The value of $\frac{16}{\pi^3} \int\limits_0^{\frac{\pi}{2}} f(x) g(x) d x$ is ______.
Correct Answer: 0.25
Explanation:
Now $I_1=\int_\limits0^{\frac{\pi}{2}} f(x) \cdot g(x) d x=\frac{1}{2} \int_\limits0^{\frac{\pi}{2}} g(x) d x$
i.e. $\frac{1}{2} \int_0^{\frac{\pi}{2}} \sqrt{\left(\frac{\pi}{4}\right)^2-\left(x-\frac{\pi}{4}\right)^2} d x$
Using $\int \sqrt{a^2-x^2}=\frac{1}{2}\left(x \sqrt{a^2-x^2}+a^2 \sin ^{-1}\left(\frac{x}{a}\right)\right)+C$
$\begin{aligned} & \Rightarrow \frac{1}{2}\left[\frac{\left(x-\frac{\pi}{4}\right)}{2} \sqrt{\frac{\pi x}{2}-x^2}+\frac{\frac{\pi^2}{16}}{2} \sin ^{-1}\left(\frac{x-\frac{\pi}{4}}{\frac{\pi}{4}}\right)\right]_0^{\pi / 2} \\ & \Rightarrow \frac{1}{2}\left[\left(0+\frac{\pi^3}{64}\right)-\left(0+\left(\frac{-\pi^3}{64}\right)\right)\right] \\ & \Rightarrow \frac{1}{2} \times \frac{\pi^3}{32} \end{aligned}$
Now $\frac{16}{\pi^3} \times \frac{\pi^3}{64}=\frac{1}{4}=0.25$
2023
JEE Advanced
Numerical
JEE Advanced 2023 Paper 2 Online
For $x \in \mathbb{R}$, let $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the minimum value of the function $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x)=\int\limits_0^{x \tan ^{-1} x} \frac{e^{(t-\cos t)}}{1+t^{2023}} d t$ is :
Correct Answer: 0
Explanation:
$\begin{aligned} f(x) & =\int\limits_0^{x \tan ^{-1} x} \frac{e^{t-\cos t}}{1+t^{2023}} d t \\\\ f^{\prime}(x) & =\frac{e^{x \tan ^{-1} x-\cos \left(x \tan ^{-1} x\right)}}{1+\left(x \tan ^{-1} x\right)^{2023}}\left(\frac{x}{1-x^2}+\tan ^{-1} x\right)\end{aligned}$
For max/min put $f^{\prime}(x)=0$
$ \begin{aligned} & \Rightarrow \frac{x}{1+x^2}+\tan ^{-1} x=0 \\\\ & \Rightarrow x=0 \end{aligned} $
$\therefore f(x)$ has minimum at $x=0$
And $f(x)_{\min }=f(0)=0$
For max/min put $f^{\prime}(x)=0$
$ \begin{aligned} & \Rightarrow \frac{x}{1+x^2}+\tan ^{-1} x=0 \\\\ & \Rightarrow x=0 \end{aligned} $
$\therefore f(x)$ has minimum at $x=0$
And $f(x)_{\min }=f(0)=0$
2022
JEE Advanced
Numerical
JEE Advanced 2022 Paper 2 Online
The greatest integer less than or equal to
$ \int_{1}^{2} \log _{2}\left(x^{3}+1\right) d x+\int_{1}^{\log _{2} 9}\left(2^{x}-1\right)^{\frac{1}{3}} d x $
is ___________.
$ \int_{1}^{2} \log _{2}\left(x^{3}+1\right) d x+\int_{1}^{\log _{2} 9}\left(2^{x}-1\right)^{\frac{1}{3}} d x $
is ___________.
Correct Answer: 5
Explanation:
Let $I = \int_1^2 {{{\log }_2}({x^3} + 1)dx + \int_1^{\log _2^9} {{{({2^x} - 1)}^{{1 \over 3}}}dx} } $
Let ${I_1} = \int_1^2 {{{\log }_2}({x^3} + 1)dx} $
and ${I_2} = \int_1^{\log _2^9} {{{({2^x} - 1)}^{{1 \over 3}}}dx} $
Let ${2^x} - 1 = {y^3}$
$ \Rightarrow {2^x}\,.\,\log _2^2 = 3{y^2}\,.\,{{dy} \over {dx}}$
$ \Rightarrow dx = {{3{y^2}} \over {{2^x}}}dy$
$ \Rightarrow dx = {{3{y^2}dy} \over {{y^3} + 1}}$
When $x = 1$ then ${y^3} = {2^1} - 1 = 1 \Rightarrow y = 1$
When $x = \log _2^9$ then ${y^3} = {2^{\log _2^9}} - 1 \Rightarrow {y^3} = 8 \Rightarrow y = 2$
$\therefore$ ${I_2} = \int_1^2 {y\,.\,{{3{y^2}} \over {{y^3} + 1}}dy} $
$ = \int_1^2 {{{3{y^3}dy} \over {{y^3} + 1}}} $
Let $y = x$
$ \Rightarrow dy = dx$
$\therefore$ ${I_2} = \int_1^2 {{{3{x^3}dx} \over {{x^3} + 1}}} $
Now, ${I_1} = \int_1^2 {{{\log }_2}({x^3} + 1)dx} $
$ = \left[ {{{\log }_2}({x^3} + 1)\,.\,x} \right]_1^2 - \int_1^2 {{1 \over {{x^3} + 1}}\,.\,{{3{x^2}} \over {\log _2^2}}\,.\,x\,dx} $
$ = \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_2^2 - \int_1^2 {{{3{x^3}dx} \over {{x^3} + 1}}} $
$\therefore$ $I = {I_1} + {I_2}$
$ = \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_1^2 - \int_1^2 {{{3{x^3}dx} \over {({x^3} + 1)}} + \int_1^2 {{{3{x^3}dx} \over {({x^3} + 1)}}} } $
$ = \left[ {x\,.\,{{\log }_2}({x^3} + 1)} \right]_1^2$
$ = 2\log _2^9 - 1\,.\,\log _2^2$
$ = 2\log _2^9 - 1$
$ = 4\log _2^3 - 1$
$ = 4 \times 1.58 - 1$
$ = 6.32 - 1$
$ = 5.32$
$\therefore$ Greatest integer value of
$I = [5.32] = 5$
Other Method :-
Let $f(x) = {\log _2}({x^3} + 1) = y$
$ \Rightarrow {x^3} + 1 = {2^y}$
$ \Rightarrow {x^3} = {2^y} - 1$
$ \Rightarrow x = {\left( {{2^y} - 1} \right)^{{1 \over 3}}}$
$\therefore$ ${f^{ - 1}}(x) = {\left( {{2^x} - 1} \right)^{{1 \over 3}}}$
And $f(1) = \log _2^{(1 + 1)} = 1 = f(a)$ (Assume)
$f(2) = \log _2^{(8 + 1)} = \log _2^9 = f(b)$
$\therefore$ $I = \int_a^b {f(x)dx + \int_{f(a)}^{f(b)} {{f^{ - 1}}(x)dx} } $
Let ${f^{ - 1}}(x) = t$
$ \Rightarrow x = f(t)$
$ \Rightarrow dx = f'(t)dt$
When $x = f(a)$ then $f(a) = f(t) \Rightarrow t = a$
When $x = f(b)$ then $f(b) = f(t) \Rightarrow t = b$
$\therefore$ $I = \int_a^b {f(t)dt + \int_a^b {t\,.\,f'(t)dt} } $
$ = \int_a^b {\left( {f(t) + t\,.\,f'(t)} \right)dt} $
$ = \left[ {t\,.\,f(t)} \right]_a^b$
$ = b\,.\,f(b) - a\,.\,f(a)$
Here $b = 2$, $f(b) = \log _2^9$
and $a = 1$, $f(a) = 1$
$\therefore$ $I = 2\log _2^9 - 1\,.\,1$
$ = 4\,.\,\log _2^3 - 1$
$ = 4 \times 1.58 - 1$
$ = 6.32 - 1$
$ = 5.32$
$\therefore$ $[I] = 5$
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 2 Online
Let ${g_i}:\left[ {{\pi \over 8},{{3\pi } \over 8}} \right] \to R,i = 1,2$, and $f:\left[ {{\pi \over 8},{{3\pi } \over 8}} \right] \to R$ be functions such that ${g_1}(x) = 1,{g_2}(x) = |4x - \pi |$ and $f(x) = {\sin ^2}x$, for all $x \in \left[ {{\pi \over 8},{{3\pi } \over 8}} \right]$. Define ${S_i} = \int\limits_{{\pi \over 8}}^{{{3\pi } \over 8}} {f(x).{g_i}(x)dx} $, i = 1, 2
The value of ${{16{S_1}} \over \pi }$ is _____________.
The value of ${{16{S_1}} \over \pi }$ is _____________.
Correct Answer: 2.00
Explanation:
${S_1} = \int_{\pi /8}^{3\pi /8} {{{\sin }^2}x.1\,dx = \int_{\pi /8}^{3\pi /8} {\left( {{{1 - \cos 2x} \over 2}} \right)dx} } $
$ = \left( {{x \over 2} - {{\sin 2x} \over 4}} \right)_{\pi /8}^{3\pi /8} = {\pi \over 8}$
$\therefore$ ${{16{S_1}} \over \pi } = {{16} \over \pi } \times {\pi \over 8} = 2$
$ = \left( {{x \over 2} - {{\sin 2x} \over 4}} \right)_{\pi /8}^{3\pi /8} = {\pi \over 8}$
$\therefore$ ${{16{S_1}} \over \pi } = {{16} \over \pi } \times {\pi \over 8} = 2$
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 2 Online
Let ${g_i}:\left[ {{\pi \over 8},{{3\pi } \over 8}} \right] \to R,i = 1,2$, and $f:\left[ {{\pi \over 8},{{3\pi } \over 8}} \right] \to R$ be functions such that ${g_1}(x) = 1,{g_2}(x) = |4x - \pi |$ and $f(x) = {\sin ^2}x$, for all $x \in \left[ {{\pi \over 8},{{3\pi } \over 8}} \right]$. Define ${S_i} = \int\limits_{{\pi \over 8}}^{{{3\pi } \over 8}} {f(x).{g_i}(x)dx} $, i = 1, 2
The value of ${{48{S_2}} \over {{\pi ^2}}}$ is ___________.
The value of ${{48{S_2}} \over {{\pi ^2}}}$ is ___________.
Correct Answer: 1.50
Explanation:
${S_2} = \int\limits_{\pi /8}^{3\pi /8} {{{\sin }^2}x\left| {4x - \pi } \right|dx} $ .... (i)
${S_2} = \int_{\pi /8}^{3\pi /8} {{{\sin }^2}\left( {{{3\pi } \over 8} + {\pi \over 8} - x} \right)\left| {4\left( {{{3\pi } \over 8} + {\pi \over 8} - x} \right) - \pi } \right|dx} $
$[\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx]} } $
$ \Rightarrow {S_2} = \int_{\pi /8}^{3\pi /8} {{{\cos }^2}x\left| {\pi - 4x} \right|dx} $ .... (ii)
Adding Eqs. (i) and (ii), we get
$2{S_2} = \int_{\pi /8}^{3\pi /8} {\left| {4x - \pi } \right|dx} $
From figure,
${A_1} = {1 \over 2} \times {\pi \over 8} \times {\pi \over 2} = {{{\pi ^2}} \over {32}} = {A_2}$
$\therefore$ $2{S_2} = 2{A_1} = {{{\pi ^2}} \over {16}} \Rightarrow {S_2} = {{{\pi ^2}} \over {32}}$
Hence, ${{48{S_2}} \over {{\pi ^2}}} = {{48} \over {32}} = {3 \over 2} = 1.5$
${S_2} = \int_{\pi /8}^{3\pi /8} {{{\sin }^2}\left( {{{3\pi } \over 8} + {\pi \over 8} - x} \right)\left| {4\left( {{{3\pi } \over 8} + {\pi \over 8} - x} \right) - \pi } \right|dx} $
$[\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx]} } $
$ \Rightarrow {S_2} = \int_{\pi /8}^{3\pi /8} {{{\cos }^2}x\left| {\pi - 4x} \right|dx} $ .... (ii)
Adding Eqs. (i) and (ii), we get
$2{S_2} = \int_{\pi /8}^{3\pi /8} {\left| {4x - \pi } \right|dx} $
From figure,
${A_1} = {1 \over 2} \times {\pi \over 8} \times {\pi \over 2} = {{{\pi ^2}} \over {32}} = {A_2}$
$\therefore$ $2{S_2} = 2{A_1} = {{{\pi ^2}} \over {16}} \Rightarrow {S_2} = {{{\pi ^2}} \over {32}}$
Hence, ${{48{S_2}} \over {{\pi ^2}}} = {{48} \over {32}} = {3 \over 2} = 1.5$
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 2 Online
For any real number x, let [ x ] denote the largest integer less than or equal to x. If $I = \int\limits_0^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx} $, then the value of 9I is __________.
Correct Answer: 182
Explanation:
Given, $I = \int_0^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]d} x$ .... (i)
${{10x} \over {x + 1}} = 1 \Rightarrow 10x = x + 1 \Rightarrow x = {1 \over 9}$
${{10x} \over {x + 1}} = 4 \Rightarrow 10x = 4x + 4 \Rightarrow x = {2 \over 3}$
${{10x} \over {x + 1}} = 9 \Rightarrow 10x = 9x + 9 \Rightarrow x = 9$
From Eq. (i), we get
$I = \int_0^{1/9} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_{1/9}^{2/3} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_{2/3}^9 {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_9^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx} } } } $
$ = \int_0^{1/9} {0.dx + \int_{1/9}^{2/3} {1.dx + \int_{2/3}^9 {2.dx + \int_9^{10} {3.dx} } } } $
$ = 0 + [x]_{1/9}^{2/3} + [2x]_{2/3}^9 + [3x]_9^{10}$
$ = {2 \over 3} - {1 \over 9} + 18 - {4 \over 3} + 30 - 27 = 21 - {2 \over 3} - {1 \over 9} = {{182} \over 9}$
Hence, 9I = 182
${{10x} \over {x + 1}} = 1 \Rightarrow 10x = x + 1 \Rightarrow x = {1 \over 9}$
${{10x} \over {x + 1}} = 4 \Rightarrow 10x = 4x + 4 \Rightarrow x = {2 \over 3}$
${{10x} \over {x + 1}} = 9 \Rightarrow 10x = 9x + 9 \Rightarrow x = 9$
From Eq. (i), we get
$I = \int_0^{1/9} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_{1/9}^{2/3} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_{2/3}^9 {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_9^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx} } } } $
$ = \int_0^{1/9} {0.dx + \int_{1/9}^{2/3} {1.dx + \int_{2/3}^9 {2.dx + \int_9^{10} {3.dx} } } } $
$ = 0 + [x]_{1/9}^{2/3} + [2x]_{2/3}^9 + [3x]_9^{10}$
$ = {2 \over 3} - {1 \over 9} + 18 - {4 \over 3} + 30 - 27 = 21 - {2 \over 3} - {1 \over 9} = {{182} \over 9}$
Hence, 9I = 182