Definite Integration
315 Questions
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
The value of $\int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$ is :
A.
$2\pi $
B.
${\pi \over 2}$
C.
$4\pi $
D.
${\pi \over 4}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
The value of $\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}dx} } $, where [ x ] is the greatest integer $ \le $ x, is :
A.
100e
B.
100(e $-$ 1)
C.
100(1 + e)
D.
100(1 $-$ e)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
If ${I_n} = \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^n}x\,dx} $, then :
A.
${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$ are in A.P.
B.
I2 + I4, I3 + I5, I4 + I6 are in A.P.
C.
${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$ are in G.P.
D.
I2 + I4, (I3 + I5)2, I4 + I6 are in G.P.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ........ + {n \over {{{(2n + 1)}^2}}}} \right]$ is equal to :
A.
${{1 \over 2}}$
B.
${{1 \over 3}}$
C.
1
D.
${{1 \over 4}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
The value of $\int\limits_{ - 1}^1 {{x^2}{e^{[{x^3}]}}} dx$, where [ t ] denotes the greatest integer $ \le $ t, is :
A.
${{e + 1} \over 3}$
B.
${{e - 1} \over {3e}}$
C.
${1 \over {3e}}$
D.
${{e + 1} \over {3e}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
The value of the integral, $\int\limits_1^3 {[{x^2} - 2x - 2]dx} $, where [x] denotes the greatest integer less than or equal to x, is :
A.
$-$ 5
B.
$ - \sqrt 2 - \sqrt 3 + 1$
C.
$-$ 4
D.
$ - \sqrt 2 - \sqrt 3 - 1$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
Let f(x) be a differentiable function defined on [0, 2] such that f'(x) = f'(2 $-$ x) for all x$ \in $ (0, 2), f(0) = 1 and f(2) = e2. Then the value of $\int\limits_0^2 {f(x)} dx$ is :
A.
1 + e2
B.
2(1 + e2)
C.
1 $-$ e2
D.
2(1 $-$ e2)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Evening Shift
Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'(x) $ \ne $ 0 for all x $ \in $ R. If $\left| {\matrix{
{f(x)} & {f'(x)} \cr
{f'(x)} & {f''(x)} \cr
} } \right|$ = 0, for all x$ \in $R, then the value of f(1) lies in the interval :
A.
(0, 3)
B.
(9, 12)
C.
(3, 6)
D.
(6, 9)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {\left( {\sin \sqrt t } \right)dt} } \over {{x^3}}}$ is equal to :
A.
${1 \over {15}}$
B.
0
C.
${2 \over 3}$
D.
${3 \over 2}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
Let [t] denote the greatest integer $\le$ t. Then the value of
$8.\int\limits_{ - {1 \over 2}}^1 {([2x] + |x|)dx} $ is ___________.
$8.\int\limits_{ - {1 \over 2}}^1 {([2x] + |x|)dx} $ is ___________.
Correct Answer: 5
Explanation:
$I = \,\int\limits_{ - {1 \over 2}}^1 {([2x] + |x|)dx} $
$ = \int\limits_{ - 1/2}^1 {[2x]\,dx + \int\limits_{ - 1/2}^1 {|x|\,} dx} $
$ = 0 + \int\limits_{ - 1/2}^0 {( - x)\,} dx + \int\limits_0^1 {x\,} dx$
$ = \left( { - {{{x^2}} \over 2}} \right)_{ - 1/2}^0 + \left( {{{{x^2}} \over 2}} \right)_0^1$
$ = \left( {0 + {1 \over 8}} \right) + {1 \over 2}$
$ = {5 \over 8}$
$ \therefore $ 8I = 5
$ = \int\limits_{ - 1/2}^1 {[2x]\,dx + \int\limits_{ - 1/2}^1 {|x|\,} dx} $
$ = 0 + \int\limits_{ - 1/2}^0 {( - x)\,} dx + \int\limits_0^1 {x\,} dx$
$ = \left( { - {{{x^2}} \over 2}} \right)_{ - 1/2}^0 + \left( {{{{x^2}} \over 2}} \right)_0^1$
$ = \left( {0 + {1 \over 8}} \right) + {1 \over 2}$
$ = {5 \over 8}$
$ \therefore $ 8I = 5
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
If $x\phi (x) = \int\limits_5^x {(3{t^2} - 2\phi '(t))dt} $, x > $-$2, and $\phi$(0) = 4, then $\phi$(2) is __________.
Correct Answer: 4
Explanation:
$x\phi (x) = \int\limits_5^x {3{t^2} - 2\phi '(t)dt} $
$x\phi (x) = {x^3} - 125 - 2[\phi (x) - \phi (5)]$
$x\phi (x) = {x^3} - 125 - 2\phi (x) - 2\phi (5)$
$\phi (0) = 4 \Rightarrow \phi (5) = {{133} \over 2}$
$\phi (x) = {{{x^3} + 8} \over {x + 2}}$
$\phi (2) = 4$
$x\phi (x) = {x^3} - 125 - 2[\phi (x) - \phi (5)]$
$x\phi (x) = {x^3} - 125 - 2\phi (x) - 2\phi (5)$
$\phi (0) = 4 \Rightarrow \phi (5) = {{133} \over 2}$
$\phi (x) = {{{x^3} + 8} \over {x + 2}}$
$\phi (2) = 4$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
If $\int_0^\pi {({{\sin }^3}x){e^{ - {{\sin }^2}x}}dx = \alpha - {\beta \over e}\int_0^1 {\sqrt t {e^t}dt} } $, then $\alpha$ + $\beta$ is equal to ____________.
Correct Answer: 5
Explanation:
$I = 2\int_0^{\pi /2} {{{\sin }^3}x{e^{ - {{\sin }^2}x}}dx} $
$ = 2\int_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \int\limits_0^{\pi /2} {\mathop {\cos x}\limits_I \,\underbrace {{e^{ - {{\sin }^2}x}}( - \sin 2x)}_{II}dx} $$ = 2\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \left[ {\cos x{e^{ - {{\sin }^2}x}}} \right]_0^{\pi /2} + \int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}} dx$
$ = 3\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} - 1$
$ = {3 \over 2}\int\limits_{ - 1}^0 {{{{e^\alpha }d\alpha } \over {\sqrt {1 + \alpha } }} - 1} $ (Put $-$sin2x = t)
$ = {3 \over {2e}}\int\limits_0^1 {{{{e^x}} \over {\sqrt x }}dx - 1} $ (put 1 + $\alpha$ = x)
$ = {3 \over {2e}}\int\limits_0^1 {\mathop {{e^x}}\limits_{II} } \mathop {{1 \over {\sqrt x }}}\limits_{II} dx - 1$
$ = 2 - {3 \over e}\int\limits_0^1 {{e^x}\sqrt x } dx$
Hence, $\alpha$ + $\beta$ = 5
$ = 2\int_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \int\limits_0^{\pi /2} {\mathop {\cos x}\limits_I \,\underbrace {{e^{ - {{\sin }^2}x}}( - \sin 2x)}_{II}dx} $$ = 2\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} + \left[ {\cos x{e^{ - {{\sin }^2}x}}} \right]_0^{\pi /2} + \int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}} dx$
$ = 3\int\limits_0^{\pi /2} {\sin x{e^{ - {{\sin }^2}x}}dx} - 1$
$ = {3 \over 2}\int\limits_{ - 1}^0 {{{{e^\alpha }d\alpha } \over {\sqrt {1 + \alpha } }} - 1} $ (Put $-$sin2x = t)
$ = {3 \over {2e}}\int\limits_0^1 {{{{e^x}} \over {\sqrt x }}dx - 1} $ (put 1 + $\alpha$ = x)
$ = {3 \over {2e}}\int\limits_0^1 {\mathop {{e^x}}\limits_{II} } \mathop {{1 \over {\sqrt x }}}\limits_{II} dx - 1$
$ = 2 - {3 \over e}\int\limits_0^1 {{e^x}\sqrt x } dx$
Hence, $\alpha$ + $\beta$ = 5
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
Let the domain of the function
$f(x) = {\log _4}\left( {{{\log }_5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right)} \right)$ be (a, b). Then the value of the integral $\int\limits_a^b {{{{{\sin }^3}x} \over {({{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dx$ is equal to _____________.
$f(x) = {\log _4}\left( {{{\log }_5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right)} \right)$ be (a, b). Then the value of the integral $\int\limits_a^b {{{{{\sin }^3}x} \over {({{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dx$ is equal to _____________.
Correct Answer: 1
Explanation:
For domain
${\log _5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right) > 0$
${\log _3}(18x - {x^2} - 77) > 1$
$18x - {x^2} - 77 > 3$
${x^2} - 18x + 80 < 0$
$x \in (8,10)$
$\Rightarrow$ a = 8 and b = 10
$I = \int\limits_a^b {{{{{\sin }^3}x} \over {{{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dx$
$I = \int\limits_a^b {{{{{\sin }^3}x(a + b - x)} \over {{{\sin }^3}x + {{\sin }^3}(a + b - x)}}} $
$2I = (b - a) \Rightarrow I = {{b - a} \over 2}$ ($\because$ a = 8 and b = 10)
$I = {{10 - 8} \over 2} = 1$
${\log _5}\left( {{{\log }_3}(18x - {x^2} - 77)} \right) > 0$
${\log _3}(18x - {x^2} - 77) > 1$
$18x - {x^2} - 77 > 3$
${x^2} - 18x + 80 < 0$
$x \in (8,10)$
$\Rightarrow$ a = 8 and b = 10
$I = \int\limits_a^b {{{{{\sin }^3}x} \over {{{\sin }^3}x + {{\sin }^3}(a + b - x)}}} dx$
$I = \int\limits_a^b {{{{{\sin }^3}x(a + b - x)} \over {{{\sin }^3}x + {{\sin }^3}(a + b - x)}}} $
$2I = (b - a) \Rightarrow I = {{b - a} \over 2}$ ($\because$ a = 8 and b = 10)
$I = {{10 - 8} \over 2} = 1$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
Let $F:[3,5] \to R$ be a twice differentiable function on (3, 5) such that
$F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $. If $F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$, then $\alpha$ + $\beta$ is equal to _______________.
$F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $. If $F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$, then $\alpha$ + $\beta$ is equal to _______________.
Correct Answer: 16
Explanation:
$F(3) = 0$
${e^x}F(x) = \int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $
${e^x}F(x) + {e^x}F'(x) = 3{x^2} + 2x + 4F'(x)$
$({e^x} - 4){{dy} \over {dx}} + {e^x}y = (3{x^2} + 2x)$
${{dy} \over {dx}} + {{{e^x}} \over {({e^x} - 4)}}y = {{(3{x^2} + 2x)} \over {({e^x} - 4)}}$
$y{e^{\int {{{{e^x}} \over {({e^x} - 4)}}dx} }} = \int {{{(3{x^2} + 2x)} \over {({e^x} - 4)}}{e^{\int {{{{e^x}} \over {{e^x} - 4}}dx} }}dx} $
$y.({e^x} - 4) = \int {(3{x^2} + 2x)dx + c} $
$y({e^x} - 4) = {x^3} + {x^2} + c$
Put x = 3 $\Rightarrow$ c = $-$36
$F(x) = {{({x^3} + {x^2} - 36)} \over {({e^x} - 4)}}$
$F'(x) = {{(3{x^2} + 2x)({e^x} - 4) - ({x^3} + {x^2} - 36){e^x}} \over {{{({e^x} - 4)}^2}}}$
Now, put value of x = 4 we will get $\alpha$ = 12 & $\beta$ = 4
${e^x}F(x) = \int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $
${e^x}F(x) + {e^x}F'(x) = 3{x^2} + 2x + 4F'(x)$
$({e^x} - 4){{dy} \over {dx}} + {e^x}y = (3{x^2} + 2x)$
${{dy} \over {dx}} + {{{e^x}} \over {({e^x} - 4)}}y = {{(3{x^2} + 2x)} \over {({e^x} - 4)}}$
$y{e^{\int {{{{e^x}} \over {({e^x} - 4)}}dx} }} = \int {{{(3{x^2} + 2x)} \over {({e^x} - 4)}}{e^{\int {{{{e^x}} \over {{e^x} - 4}}dx} }}dx} $
$y.({e^x} - 4) = \int {(3{x^2} + 2x)dx + c} $
$y({e^x} - 4) = {x^3} + {x^2} + c$
Put x = 3 $\Rightarrow$ c = $-$36
$F(x) = {{({x^3} + {x^2} - 36)} \over {({e^x} - 4)}}$
$F'(x) = {{(3{x^2} + 2x)({e^x} - 4) - ({x^3} + {x^2} - 36){e^x}} \over {{{({e^x} - 4)}^2}}}$
Now, put value of x = 4 we will get $\alpha$ = 12 & $\beta$ = 4
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
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
The integral $\int\limits_1^2 {{e^x}.{x^x}\left( {2 + {{\log }_e}x} \right)} dx$ equals :
A.
e(4e + 1)
B.
e(2e – 1)
C.
e(4e – 1)
D.
4e2 – 1
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
If I1 = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{100}}} dx$ and
I2 = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$ such
that I2 = $\alpha $I1 then $\alpha $ equals to :
I2 = $\int\limits_0^1 {{{\left( {1 - {x^{50}}} \right)}^{101}}} dx$ such
that I2 = $\alpha $I1 then $\alpha $ equals to :
A.
${{5051} \over {5050}}$
B.
${{5050} \over {5051}}$
C.
${{5050} \over {5049}}$
D.
${{5049} \over {5050}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$
A.
is equal to 0
B.
is equal to ${1 \over 2}$
C.
does not exist
D.
is equal to $ - {1 \over 2}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
The value of $\int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{1 \over {1 + {e^{\sin x}}}}dx} $ is:
A.
$\pi $
B.
${{3\pi \over 2}}$
C.
${{\pi \over 2}}$
D.
${{\pi \over 4}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
The integral
$\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} $
is equal to:
$\int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{\tan }^3}x.{{\sin }^2}3x\left( {2{{\sec }^2}x.{{\sin }^2}3x + 3\tan x.\sin 6x} \right)dx} $
is equal to:
A.
$ - {1 \over {9}}$
B.
$ - {1 \over {18}}$
C.
$ {7 \over {18}}$
D.
${9 \over 2}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
Let $f(x) = \left| {x - 2} \right|$ and g(x) = f(f(x)), $x \in \left[ {0,4} \right]$. Then
$\int\limits_0^3 {\left( {g(x) - f(x)} \right)} dx$ is equal to:
$\int\limits_0^3 {\left( {g(x) - f(x)} \right)} dx$ is equal to:
A.
1
B.
0
C.
${1 \over 2}$
D.
${3 \over 2}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
If the value of the integral
$\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx$
is ${k \over 6}$, then k is equal to :
$\int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx$
is ${k \over 6}$, then k is equal to :
A.
$2\sqrt 3 + \pi $
B.
$3\sqrt 2 - \pi $
C.
$3\sqrt 2 + \pi $
D.
$2\sqrt 3 - \pi $
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
Suppose f(x) is a polynomial of degree four,
having critical points at –1, 0, 1. If
T = {x $ \in $ R | f(x) = f(0)}, then the sum of squares of all the elements of T is :
T = {x $ \in $ R | f(x) = f(0)}, then the sum of squares of all the elements of T is :
A.
6
B.
2
C.
8
D.
4
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
$\int\limits_{ - \pi }^\pi {\left| {\pi - \left| x \right|} \right|dx} $ is equal to :
A.
${\pi ^2}$
B.
2${\pi ^2}$
C.
$\sqrt 2 {\pi ^2}$
D.
${{{\pi ^2}} \over 2}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
Let a function ƒ : [0, 5] $ \to $ R be continuous,
ƒ(1) = 3 and F be defined as :
$F(x) = \int\limits_1^x {{t^2}g(t)dt} $ , where $g(t) = \int\limits_1^t {f(u)du} $
Then for the function F, the point x = 1 is :
$F(x) = \int\limits_1^x {{t^2}g(t)dt} $ , where $g(t) = \int\limits_1^t {f(u)du} $
Then for the function F, the point x = 1 is :
A.
a point of inflection.
B.
a point of local maxima.
C.
a point of local minima.
D.
not a critical point.
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
The value of
$\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$ is equal to :
$\int\limits_0^{2\pi } {{{x{{\sin }^8}x} \over {{{\sin }^8}x + {{\cos }^8}x}}} dx$ is equal to :
A.
4$\pi $
B.
2$\pi $
C.
$\pi $2
D.
2$\pi $2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
If for all real triplets (a, b, c), ƒ(x) = a + bx + cx2;
then $\int\limits_0^1 {f(x)dx} $ is equal to :
A.
${1 \over 6}\left\{ {f(0) + f(1) + 4f\left( {{1 \over 2}} \right)} \right\}$
B.
$2\left\{ 3{f(1) + 2f\left( {{1 \over 2}} \right)} \right\}$
C.
${1 \over 3}\left\{ {f(0) + f\left( {{1 \over 2}} \right)} \right\}$
D.
${1 \over 2}\left\{ {f(1) + 3f\left( {{1 \over 2}} \right)} \right\}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
If $I = \int\limits_1^2 {{{dx} \over {\sqrt {2{x^3} - 9{x^2} + 12x + 4} }}} $, then :
A.
${1 \over 16} < {I^2} < {1 \over 9}$
B.
${1 \over 8} < {I^2} < {1 \over 4}$
C.
${1 \over 9} < {I^2} < {1 \over 8}$
D.
${1 \over 6} < {I^2} < {1 \over 2}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$ is equal to
A.
$ - {1 \over 5}$
B.
$ - {1 \over 10}$
C.
0
D.
$ {1 \over 10}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
The value of $\alpha $ for which
$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$, is:
$4\alpha \int\limits_{ - 1}^2 {{e^{ - \alpha \left| x \right|}}dx} = 5$, is:
A.
${\log _e}2$
B.
${\log _e}\sqrt 2 $
C.
${\log _e}\left( {{4 \over 3}} \right)$
D.
${\log _e}\left( {{3 \over 2}} \right)$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
If $\theta $1
and $\theta $2
be respectively the smallest and the largest values of $\theta $ in (0, 2$\pi $) - {$\pi $} which satisfy
the equation,
2cot2$\theta $ - ${5 \over {\sin \theta }}$ + 4 = 0, then
$\int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta } $ is equal to :
2cot2$\theta $ - ${5 \over {\sin \theta }}$ + 4 = 0, then
$\int\limits_{{\theta _1}}^{{\theta _2}} {{{\cos }^2}3\theta d\theta } $ is equal to :
A.
${\pi \over 9}$
B.
${{2\pi } \over 3}$
C.
${{\pi } \over 3}$
D.
${\pi \over 3} + {1 \over 6}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
If ƒ(a + b + 1 - x) = ƒ(x), for all x, where a and b are fixed positive real numbers, then
${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$ is equal to:
${1 \over {a + b}}\int_a^b {x\left( {f(x) + f(x + 1)} \right)} dx$ is equal to:
A.
$\int_{a - 1}^{b - 1} {f(x+1)dx} $
B.
$\int_{a + 1}^{b + 1} {f(x + 1)dx} $
C.
$\int_{a - 1}^{b - 1} {f(x)dx} $
D.
$\int_{a + 1}^{b + 1} {f(x)dx} $
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
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
A value of $\alpha $ such that
$\int\limits_\alpha ^{\alpha + 1} {{{dx} \over {\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}}} = {\log _e}\left( {{9 \over 8}} \right)$ is :
$\int\limits_\alpha ^{\alpha + 1} {{{dx} \over {\left( {x + \alpha } \right)\left( {x + \alpha + 1} \right)}}} = {\log _e}\left( {{9 \over 8}} \right)$ is :
A.
2
B.
- 2
C.
${1 \over 2}$
D.
$-{1 \over 2}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
If $\int\limits_0^{{\pi \over 2}} {{{\cot x} \over {\cot x + \cos ecx}}} dx$ = m($\pi $ + n), then m.n is equal to
A.
- 1
B.
1
C.
$ - {1 \over 2}$
D.
${1 \over 2}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
Let f : R $ \to $ R be a continuously differentiable function such that f(2) = 6 and f'(2) = ${1 \over {48}}$. If $\int\limits_6^{f\left( x \right)} {4{t^3}} dt$ = (x - 2)g(x), then $\mathop {\lim }\limits_{x \to 2} g\left( x \right)$ is equal to :
A.
18
B.
36
C.
12
D.
24
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
The integral $\int\limits_{\pi /6}^{\pi /3} {{{\sec }^{2/3}}} x\cos e{c^{4/3}}xdx$ is equal to :
A.
${3^{{5 \over 3}}} - {3^{{1 \over 3}}}$
B.
${3^{{5 \over 6}}} - {3^{{2 \over 3}}}$
C.
${3^{{4 \over 3}}} - {3^{{1 \over 3}}}$
D.
${3^{{7 \over 6}}} - {3^{{5 \over 6}}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
The value of $\int\limits_0^{2\pi } {\left[ {\sin 2x\left( {1 + \cos 3x} \right)} \right]} dx$,
where [t] denotes the greatest integer function is :
where [t] denotes the greatest integer function is :
A.
2$\pi $
B.
$\pi $
C.
-2$\pi $
D.
-$\pi $