Practice Start Test

Definite Integration

JEE Mathematics · 230 questions · Page 12 of 23 · Click an option or "Show Solution" to reveal answer

Q111
The value of the integral ∫−22∣x3+x∣(ex∣x∣+1)dx\int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {({e^{x|x|}} + 1)}}dx} is equal to :
A 5e2
B 3e−-2
C 4
D 6
Correct Answer
Option D
Solution
I=∫−22∣x3+x∣ex∣x∣+1dxI = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{x|x|}} + 1}}dx}

..... (i)

I=∫−22∣x3+x∣e−x∣x∣+1dxI = \int\limits_{ - 2}^2 {{{|{x^3} + x|} \over {{e^{ - x|x|}} + 1}}dx}

..... (ii)

2I=∫−22∣x3+x∣dx2I = \int\limits_{ - 2}^2 {|{x^3} + x|dx}
2I=2∫02(x3+x)dx2I =2 \int\limits_0^2 {({x^3} + x)dx}
I=∫02(x3+x)dxI = \int\limits_0^2 {({x^3} + x)dx}
=x44+x22]02= \left. {{{{x^4}} \over 4} + {{{x^2}} \over 2}} \right]_0^2
=(164+42)−0= \left( {{{16} \over 4} + {4 \over 2}} \right) - 0
=4+2=6= 4 + 2 = 6
Q112
If bn=∫0π2cos⁡2nxsin⁡xdx, n∈N{b_n} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx} \over {\sin x}}dx,\,n \in N} , then
A b3−b2, b4−b3, b5−b4{b_3} - {b_2},\,{b_4} - {b_3},\,{b_5} - {b_4} are in A.P. with common difference −-2
B 1b3−b2,1b4−b3,1b5−b4{1 \over {{b_3} - {b_2}}},{1 \over {{b_4} - {b_3}}},{1 \over {{b_5} - {b_4}}} are in an A.P. with common difference 2
C b3−b2, b4−b3, b5−b4{b_3} - {b_2},\,{b_4} - {b_3},\,{b_5} - {b_4} are in a G.P.
D 1b3−b2,1b4−b3,1b5−b4{1 \over {{b_3} - {b_2}}},{1 \over {{b_4} - {b_3}}},{1 \over {{b_5} - {b_4}}} are in an A.P. with common difference −-2
Correct Answer
Option D
Solution
bn−bn−1=∫0π2cos⁡2nx−cos⁡2(n−1)xsin⁡xdx{b_n} - {b_{n - 1}} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx - {{\cos }^2}(n - 1)x} \over {\sin x}}dx}
=∫0π2−sin⁡(2n−1)x . sin⁡xsin⁡xdx= \int_0^{{\pi \over 2}} {{{ - \sin (2n - 1)x\,.\,\sin x} \over {\sin x}}dx}
=cos⁡(2n−1)x2n−1∣0π/2=−12n−1= \left. {{{\cos (2n - 1)x} \over {2n - 1}}} \right|_0^{\pi /2} = - {1 \over {2n - 1}}

So,

b3−b2{b_3} - {b_2}

,

b4−b3{b_4} - {b_3}

,

b5−b4{b_5} - {b_4}

are in H.P.

⇒1b3−b2, 1b4−b3, 1b5−b4\Rightarrow {1 \over {{b_3} - {b_2}}},\,{1 \over {{b_4} - {b_3}}},\,{1 \over {{b_5} - {b_4}}}

are in A.P. with common difference −-2.

Q113
The value of ∫0πecos⁡xsin⁡x(1+cos⁡2x)(ecos⁡x+e−cos⁡x)dx\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {(1 + {{\cos }^2}x)({e^{\cos x}} + {e^{ - \cos x}})}}dx} is equal to:
A π24{{{\pi ^2}} \over 4}
B π22{{{\pi ^2}} \over 2}
C π4{\pi \over 4}
D π2{\pi \over 2}
Correct Answer
Option C
Solution
∫0πecos⁡xsin⁡x(1+cos⁡2x)(ecos⁡x+e−cos⁡x)dx\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {\left( {1 + {{\cos }^2}x} \right)\left( {{e^{\cos x}} + {e^{ - \cos x}}} \right)}}dx}

Let

cos⁡x=t\cos x = t
sin⁡xdx=dt\sin xdx = dt
=∫1−1−etdt(1+t2)(et+e−t)= \int\limits_1^{ - 1} {{{ - {e^t}dt} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}}
I=∫−11et(1+t2)(et+e−t)dtI = \int\limits_{ - 1}^1 {{{{e^t}} \over {\left( {1 + {t^2}} \right)\left( {{e^t} + {e^{ - t}}} \right)}}dt}

...... (i)

I=∫−11e−t(1+t2)(e−t+et)dtI = \int\limits_{ - 1}^1 {{{{e^{ - t}}} \over {\left( {1 + {t^2}} \right)\left( {{e^{ - t}} + {e^t}} \right)}}dt}

.... (ii) Adding (i) and (ii)

2I=∫−11dt1+t22I = \int\limits_{ - 1}^1 {{{dt} \over {1 + {t^2}}}}
2I=tan⁡−t∣−112I = \left. {{{\tan }^{ - t}}} \right|_{ - 1}^1
2I=π4−(−π4)2I = {\pi \over 4} - \left( { - {\pi \over 4}} \right)
2I=π22I = {\pi \over 2}
I=π4I = {\pi \over 4}
Q114
The value of the integral ∫−π/2π/2dx(1+ex)(sin⁡6x+cos⁡6x)\int\limits_{ - \pi /2}^{\pi /2} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}} is equal to
A 2π\pi
B 0
C π\pi
D π2{\pi \over 2}
Correct Answer
Option C
Solution
I=∫−π2π2dx(1+ex)(sin⁡6x+cos⁡6x)I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^x})({{\sin }^6}x + {{\cos }^6}x)}}}

...... (i)

I=∫−π2π2dx(1+e−x)(sin6x+cos⁡6x)I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {(1 + {e^{ - x}})(si{n^6}x + {{\cos }^6}x)}}}

..... (ii) (i) and (ii) From equation (i) & (ii)

2I=∫−π2π2dxsin⁡6x+cos⁡6x2I = \int\limits_{{{ - \pi } \over 2}}^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}}}
⇒I=∫0π2dxsin⁡6x+cos⁡6x=∫0π2dx1−34sin⁡22x\Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{dx} \over {{{\sin }^6}x + {{\cos }^6}x}} = \int\limits_0^{{\pi \over 2}} {{{dx} \over {1 - {3 \over 4}{{\sin }^2}2x}}} }
⇒I=∫0π24sec⁡22xdx4+tan⁡22x=2∫0π44sec⁡22x4+tan⁡22xdx\Rightarrow I = \int\limits_0^{{\pi \over 2}} {{{4{{\sec }^2}2xdx} \over {4 + {{\tan }^2}2x}} = 2\int\limits_0^{{\pi \over 4}} {{{4{{\sec }^2}2x} \over {4 + {{\tan }^2}2x}}dx} }

when x = 0, t = 0 Now,

tan⁡2x=t\tan 2x = t

when

x=π4,  t→∞x = {\pi \over 4},\,\,t \to \infty
2sec⁡22x dx=dt2{\sec ^2}2x\,dx = dt

∴\therefore

I=2∫0∞2dt4+t2=2(tan⁡−1t2)0∞I = 2\int\limits_0^\infty {{{2dt} \over {4 + {t^2}}} = 2\left( {{{\tan }^{ - 1}}{t \over 2}} \right)_0^\infty }
=2π2=π= 2{\pi \over 2} = \pi
Q115
lim⁡n→∞(n2(n2+1)(n+1)+n2(n2+4)(n+2)+n2(n2+9)(n+3)+  ....  +  n2(n2+n2)(n+n))\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + {{{n^2}} \over {({n^2} + 9)(n + 3)}} + \,\,....\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right) is equal to :
A π8+14log⁡e2{\pi \over 8} + {1 \over 4}{\log _e}2
B π4+18log⁡e2{\pi \over 4} + {1 \over 8}{\log _e}2
C π4−18log⁡e2{\pi \over 4} - {1 \over 8}{\log _e}2
D π8+log⁡e2{\pi \over 8} + {\log _e}\sqrt 2
Correct Answer
Option A
Solution
lim⁡n→∞(n2(n2+1)(n+1)+n2(n2+4)(n+2)+  ...  +  n2(n2+n2)(n+n))\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + \,\,...\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)
=lim⁡n→∞∑r=1nn2(n2+r2)(n+r)= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{n^2}} \over {({n^2} + {r^2})(n + r)}}}
=lim⁡n→∞∑r=1n1n1[1+(rn)2][1+(rn)]= \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{1 \over {\left[ {1 + {{\left( {{r \over n}} \right)}^2}} \right]\left[ {1 + \left( {{r \over n}} \right)} \right]}}}
=∫011(1+x2)(1+x)dx= \int\limits_0^1 {{1 \over {(1 + {x^2})(1 + x)}}dx}
=12∫01[11+x−(x−1)(1+x2)]dx= {1 \over 2}\int_0^1 {\left[ {{1 \over {1 + x}} - {{(x - 1)} \over {(1 + {x^2})}}} \right]dx}
=12[ln⁡(1+x)−12ln⁡(1+x2)+tan⁡−1x]01= {1 \over 2}\left[ {\ln (1 + x) - {1 \over 2}\ln \left( {1 + {x^2}} \right) + {{\tan }^{ - 1}}x} \right]_0^1
=12[π4+12ln⁡2]=π8+14ln⁡2= {1 \over 2}\left[ {{\pi \over 4} + {1 \over 2}\ln 2} \right] = {\pi \over 8} + {1 \over 4}\ln 2
Q116
lim⁡n→∞∑r=1nr2r2−7rn+6n2\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} is equal to :
A log⁡e(32){\log _e}\left( {{{\sqrt 3 } \over 2}} \right)
B log⁡e(334){\log _e}\left( {{{3\sqrt 3 } \over 4}} \right)
C log⁡e(274){\log _e}\left( {{{27} \over 4}} \right)
D log⁡e(43){\log _e}\left( {{4 \over 3}} \right)
Correct Answer
Option B
Solution
lim⁡n→α∑r=1nr2r2−7rn+6n2\mathop {\lim }\limits_{n \to \alpha } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}}
=lim⁡n→α1n∑r=1n(rn)2(rn)2−7(rn)+6= \mathop {\lim }\limits_{n \to \alpha } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)} \over {2{{\left( {{r \over n}} \right)}^2} - 7\left( {{r \over n}} \right) + 6}}}
=∫abf(x)dx= \int_a^b {f(x)dx}
a=lim⁡n→α(1n)=0a = \mathop {\lim }\limits_{n \to \alpha } \left( {{1 \over n}} \right) = 0
b=lim⁡n→α(nn)=1b = \mathop {\lim }\limits_{n \to \alpha } \left( {{n \over n}} \right) = 1

and

rn→x{r \over n} \to x
=∫01x2x2−7x+6dx= \int_0^1 {{x \over {2{x^2} - 7x + 6}}dx}
=∫01x2x2−3x−4x+6dx= \int_0^1 {{x \over {2{x^2} - 3x - 4x + 6}}dx}
=∫01x(2x−3)(x−2)dx= \int_0^1 {{x \over {(2x - 3)(x - 2)}}dx}
=∫01[A(2x−3)+B(x−2)]dx= \int_0^1 {\left[ {{A \over {(2x - 3)}} + {B \over {(x - 2)}}} \right]dx}
=∫01(−32x−3+2x−2)dx= \int_0^1 {\left( {{{ - 3} \over {2x - 3}} + {2 \over {x - 2}}} \right)dx}
=[−3log⁡∣2x−3∣2+2log⁡∣x−2∣]01= \left[ { - {{3\log |2x - 3|} \over 2} + 2\log |x - 2|} \right]_0^1
=−32[log⁡(−1)−log⁡(−3)]+2[log⁡(−1)−log⁡(−2)]= - {3 \over 2}\left[ {\log ( - 1) - \log ( - 3)} \right] + 2\left[ {\log ( - 1) - \log ( - 2)} \right]
=−32log⁡(13)+2log⁡(12)= - {3 \over 2}\log \left( {{1 \over 3}} \right) + 2\log \left( {{1 \over 2}} \right)
=+32log⁡3−2log⁡2= + {3 \over 2}\log 3 - 2\log 2
=log⁡33−log⁡4= \log \sqrt {{3^3}} - \log 4
=log⁡32×34= \log {{\sqrt {{3^2} \times 3} } \over 4}
=log⁡334= \log {{3\sqrt 3 } \over 4}
Q117
For any real number xx, let [x][x] denote the largest integer less than equal to xx. Let ff be a real valued function defined on the interval [−10,10][-10,10] by f(x)={x−[x], if [x] is odd 1+[x]−x, if [x] is even .f(x)=\left\{\begin{array}{l}x-[x], \text{ if }[x] \text{ is odd } \\ 1+[x]-x, \text{ if }[x] \text{ is even } .\end{array}\right. Then the value of π210∫−1010f(x)cos⁡πx dx\dfrac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x \,d x is :
A 4
B 2
C 1
D 0
Correct Answer
Option A
Solution

Case 1 : Let

0≤x<10 \le x < 1

then

[x]=0\left[ x \right] = 0

, which is even ∴\therefore

f(x)=1+[x]−xf(x) = 1 + \left[ x \right] - x
=1+0−x= 1 + 0 - x
=1−x= 1 - x

Case 2 : Let

1≤x<21 \le x < 2

then

[x]=1\left[ x \right] = 1

, which is odd ∴\therefore

f(x)=x−[x]f(x) = x - \left[ x \right]
=x−1= x - 1

Case 3 : Let

2≤x<32 \le x < 3

then

[x]=2\left[ x \right] = 2

, which is even ∴\therefore

f(x)=1+[x]−xf(x) = 1 + \left[ x \right] - x
=1+2−x= 1 + 2 - x
=3−x= 3 - x

Case 4 : Let

3≤x<43 \le x < 4

then

[x]=3\left[ x \right] = 3

, which is odd ∴\therefore

f(x)=x−[x]f(x) = x - \left[ x \right]
=x−3= x - 3

∴\therefore

f(x)={1−x;0≤x<1x−1;1≤x<23−x;2≤x<3x−3;3≤x<4f(x) = \left\{ \begin{array}{lll}{1 - x} & ; & {0 \le x < 1} \\ {x - 1} & ; & {1 \le x < 2} \\ {3 - x} & ; & {2 \le x < 3} \\ {x - 3} & ; & {3 \le x < 4} \end{array} \right.

∴\therefore

f(x)f(x)

is periodic and period of

f(x)=2f(x) = 2

And period of

cos⁡πx=2ππ=2\cos \pi x = {{2\pi } \over \pi } = 2

∴\therefore Period of

f(x)cos⁡πx=2f(x)\cos \pi x = 2

Now,

I=π210∫−1010f(x)cos⁡πx dxI = {{{\pi ^2}} \over {10}}\int_{ - 10}^{10} {f(x)\cos \pi x\,dx}
=π210∫−10−10+10×2f(x)cos⁡πx dx= {{{\pi ^2}} \over {10}}\int_{ - 10}^{ - 10 + 10 \times 2} {f(x)\cos \pi x\,dx}
=π210∫010×2f(x)cos⁡πx dx= {{{\pi ^2}} \over {10}}\int_0^{10 \times 2} {f(x)\cos \pi x\,dx}
=π210×10∫02f(x)cos⁡πx dx= {{{\pi ^2}} \over {10}} \times 10\int_0^2 {f(x)\cos \pi x\,dx}
=π2∫02f(x)cos⁡πx dx= {\pi ^2}\int_0^2 {f(x)\cos \pi x\,dx}

∴\therefore

I=π2[∫01f(x)cos⁡πx dx+∫12f(x)cos⁡πx dx]I = {\pi ^2}\left[ {\int_0^1 {f(x)\cos \pi x\,dx + \int_1^2 {f(x)\cos \pi x\,dx} } } \right]
=π2[∫01(1−x)cos⁡πx dx+∫12(x−1)cos⁡πx dx]= {\pi ^2}\left[ {\int_0^1 {(1 - x)\cos \pi x\,dx + \int_1^2 {(x - 1)\cos \pi x\,dx} } } \right]
=π2[∫01cos⁡πx dx−∫01xcos⁡πx dx+∫12xcos⁡πx dx−∫12cos⁡πx dx]= {\pi ^2}\left[ {\int_0^1 {\cos \pi x\,dx - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - \int_1^2 {\cos \pi x\,dx} } } } } \right]
=π2[1π[sin⁡πx]01−∫01xcos⁡πx dx+∫12xcos⁡πx dx−1π[sin⁡πx]12]= {\pi ^2}\left[ {{1 \over \pi }\left[ {\sin \pi x} \right]_0^1 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - {1 \over \pi }\left[ {\sin \pi x} \right]_1^2} } } \right]
=π2[0−∫01xcos⁡πx dx+∫12xcos⁡πx dx−0]= {\pi ^2}\left[ {0 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - 0} } } \right]
=π2[−[xsin⁡πxπ+1π2cos⁡πx]01+[xsin⁡πxπ+1π2cos⁡πx]12]= {\pi ^2}\left[ { - \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_0^1 + \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_1^2} \right]
[As ∫xcos⁡πx dx=x . ∫cos⁡πx−∫(1 . sin⁡πxπ)dx=x . sin⁡πxπ+1π2cos⁡πx+c]\left[ {\mathrm{As}\,\int {x\cos \pi x\,dx = x\,.\,\int {\cos \pi x - \int {\left( {1\,.\,{{\sin \pi x} \over \pi }} \right)dx = x\,.\,{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x + c} } } } \right]
=π2[−[(1 . sin⁡ππ+1π2 . cos⁡π)−(0+1π2 . cos⁡0)]+[(2 . sin⁡2ππ+1π2cos⁡2π)−(1 . sin⁡ππ+1π2cos⁡π)]]= {\pi ^2}\left[ { - \left[ {\left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\,.\,\cos \pi } \right) - \left( {0 + {1 \over {{\pi ^2}}}\,.\,\cos 0} \right)} \right] + \left[ {\left( {2\,.\,{{\sin 2\pi } \over \pi } + {1 \over {{\pi ^2}}}\cos 2\pi } \right) - \left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\cos \pi } \right)} \right]} \right]
=π2[−{(−1π2)−(1π2)}+((+1π2)−(−1π2)]= {\pi ^2}\left[ { - \left\{ {\left( { - {1 \over {{\pi ^2}}}} \right) - \left( {{1 \over {{\pi ^2}}}} \right)} \right\} + \left( {\left( { + {1 \over {{\pi ^2}}}} \right) - \left( { - {1 \over {{\pi ^2}}}} \right)} \right.} \right]
=π2[−(−2π2)+2π2]= {\pi ^2}\left[ { - \left( { - {2 \over {{\pi ^2}}}} \right) + {2 \over {{\pi ^2}}}} \right]
=π2[2π2+2π2]= {\pi ^2}\left[ {{2 \over {{\pi ^2}}} + {2 \over {{\pi ^2}}}} \right]
=π2×4π2= {\pi ^2} \times {4 \over {{\pi ^2}}}
=4= 4
Q118
lim⁡n→∞12n(11−12n+11−22n+11−32n+  ...  +  11−2n−12n)\mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,...\,\, + \,\,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right) is equal to
A 12\dfrac{1}{2}
B 1
C 2
D −-2
Correct Answer
Option C
Solution
I=lim⁡n→∞12n(11−12n+11−22n+11−32n+  .....  + 11−2n−12n)I = \mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,.....\,\, + \,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)

Let

2n=t{2^n} = t

and if

n→∞n \to \infty

then

t→∞t \to \infty
I=lim⁡n→∞1t(∑r=1t−111−rt)I = \mathop {\lim }\limits_{n \to \infty } {1 \over t}\left( {\sum\limits_{r = 1}^{t - 1} {{1 \over {\sqrt {1 - {r \over t}} }}} } \right)
I=∫01dx1−x=∫01dxx         (∫0af(x)dx=∫0af(a−x)dxI = \int\limits_0^1 {{{dx} \over {\sqrt {1 - x} }} = \int\limits_0^1 {{{dx} \over {\sqrt x }}\,\,\,\,\,\,\,\,\,\left( {\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)dx} } } \right.} }
=[2x12]01=2= \left[ {2{x^{{1 \over 2}}}} \right]_0^1 = 2
Q119
Let [t][t] denote the greatest integer less than or equal to tt. Then the value of the integral ∫−3101([sin⁡(πx)]+e[cos⁡(2πx)])dx\int_{-3}^{101}\left([\sin (\pi x)]+e^{[\cos (2 \pi x)]}\right) d x is equal to
A 52(1−e)e\dfrac{52(1-e)}{e}
B 52e\dfrac{52}{e}
C 52(2+e)e\dfrac{52(2+e)}{e}
D 104e\dfrac{104}{e}
Correct Answer
Option B
Solution
I=∫−3101([sin⁡(πx)]+e[cos⁡(2πx)])dxI = \int_{ - 3}^{101} {\left( {\left[ {\sin (\pi x)} \right] + {e^{[\cos (2\pi x)]}}} \right)dx}
[sin⁡πx][\sin \pi x]

is periodic with period 2 and

e[cos⁡(2πx)]{{e^{[\cos (2\pi x)]}}}

is periodic with period 1. So,

I=52∫02([sin⁡πx]+e[cos⁡2πx])dxI = 52\int_0^2 {\left( {\left[ {\sin \pi x} \right] + {e^{[\cos 2\pi x]}}} \right)dx}
=52{∫12−1 dx+∫1434e−1 dx+∫5474e−1 dx+∫014e0 dx+∫3454e0 dx+∫742e0 dx}= 52\left\{ {\int_1^2 { - 1} \,dx + \int_{{1 \over 4}}^{{3 \over 4}} {{e^{ - 1}}\,dx + \int_{{5 \over 4}}^{{7 \over 4}} {{e^{ - 1}}\,dx + \int_0^{{1 \over 4}} {{e^0}\,dx + \int_{{3 \over 4}}^{{5 \over 4}} {{e^0}\,dx + \int_{{7 \over 4}}^2 {{e^0}\,dx} } } } } } \right\}
=52e= {{52} \over e}
Q120
If a=lim⁡n→∞∑k=1n2nn2+k2a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} and f(x)=1−cos⁡x1+cos⁡xf(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} , x∈(0,1)x \in (0,1), then :
A 22f(a2)=f′(a2)2\sqrt 2 f\left( {{a \over 2}} \right) = f'\left( {{a \over 2}} \right)
B f(a2)f′(a2)=2f\left( {{a \over 2}} \right)f'\left( {{a \over 2}} \right) = \sqrt 2
C 2f(a2)=f′(a2)\sqrt 2 f\left( {{a \over 2}} \right) = f'\left( {{a \over 2}} \right)
D f(a2)=2f′(a2)f\left( {{a \over 2}} \right) = \sqrt 2 f'\left( {{a \over 2}} \right)
Correct Answer
Option C
Solution
a=lim⁡n→∞∑k=1n2nn2+k2a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}}
=lim⁡n→∞1n∑k=1n21+(kn)2= \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {{2 \over {1 + {{\left( {{k \over n}} \right)}^2}}}}
a=∫0121+x2dx=2tan⁡−1x∫01=π2a = \int\limits_0^1 {{2 \over {1 + {x^2}}}dx = 2{{\tan }^{ - 1}}x\int_0^1 { = {\pi \over 2}} }
f(x)=1−cos⁡x1+cos⁡x, x∈(0,1)f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} ,\,x \in (0,1)
f(x)=1−cos⁡xsin⁡x=cos⁡ec x−cot⁡xf(x) = {{1 - \cos x} \over {\sin x}} = \cos ec\,x - \cot x
f′(x)=cos⁡ec2x−cos⁡ec xcot⁡xf'(x) = \cos e{c^2}x - \cos ec\,x\cot x
f(a2)=f(π4)=2−1f′(a2)=f′(π4)=2−2}f′(a2)=2 . f(a2)\left. \begin{array}{ll}{f\left( {{a \over 2}} \right) = f\left( {{\pi \over 4}} \right) = \sqrt 2 - 1} \\ {f'\left( {{a \over 2}} \right) = f'\left( {{\pi \over 4}} \right) = 2 - \sqrt 2 } \end{array} \right\}f'\left( {{a \over 2}} \right) = \sqrt 2 \,.\,f\left( {{a \over 2}} \right)
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