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Definite Integration

JEE Mathematics · 230 questions · Page 13 of 23 · Click an option or "Show Solution" to reveal answer

Q121
Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be a function defined as f(x)=asin⁡(π[x]2)+[2−x],a∈Rf(x)=a \sin \left(\dfrac{\pi[x]}{2}\right)+[2-x], a \in \mathbb{R} where [t][t] is the greatest integer less than or equal to tt. If lim⁡x→−1f(x)\mathop {\lim }\limits_{x \to -1 } f(x) exists, then the value of ∫04f(x)dx\int\limits_{0}^{4} f(x) d x is equal to
A −-1
B −-2
C 1
D 2
Correct Answer
Option B
Solution
f(x)=asin⁡(π[x]2)+[2−x] a∈Rf(x) = a\sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]\,a \in R

Now, ∵\because

lim⁡x→−1f(x)\mathop {\lim }\limits_{x \to - 1} f(x)

exist ∴\therefore

lim⁡x→−1−f(x)=lim⁡x→−1+f(x)\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)
⇒asin⁡(−2π2)+3=asin⁡(−π2)+2\Rightarrow a\sin \left( {{{ - 2\pi } \over 2}} \right) + 3 = a\sin \left( {{{ - \pi } \over 2}} \right) + 2
⇒−a=1⇒a=−1\Rightarrow - a = 1 \Rightarrow a = - 1

Now,

∫04f(x)dx=∫04(−sin⁡(π[x]2)+[2−x])dx\int_0^4 {f(x)dx = \int_0^4 {\left( { - \sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]} \right)dx} }
=∫011dx+∫12−1dx+∫23−1dx+∫34(1−2)dx= \int_0^1 {1dx + \int_1^2 { - 1dx + \int_2^3 { - 1dx + \int_3^4 {(1 - 2)dx} } } }
=1−1−1−1=−2= 1 - 1 - 1 - 1 = - 2
Q122
Let I=∫π/4π/3(8sin⁡x−sin⁡2xx)dx I=\int_{\pi / 4}^{\pi / 3}\left(\dfrac{8 \sin x-\sin 2 x}{x}\right) d x . Then
A π2<I<3π4{\pi \over 2} < I < {{3\pi } \over 4}
B π5<I<5π12{\pi \over 5} < I < {{5\pi } \over {12}}
C 5π12<I<23π{{5\pi } \over {12}} < I < {{\sqrt 2 } \over 3}\pi
D 3π4<I<π{{3\pi } \over 4} < I < \pi
Correct Answer
Option C
Solution

I comes out around 1.536 which is not satisfied by any given options.

∫π/4π/38x−2xxdx>I>∫π/4π/38sin⁡x−2xxdx\int\limits_{\pi /4}^{\pi /3} {{{8x - 2x} \over x}dx > I > \int\limits_{\pi /4}^{\pi /3} {{{8\sin x - 2x} \over x}dx} }
π2>I>∫π/4π/3(8sin⁡xx−2)dx{\pi \over 2} > I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x} \over x} - 2} \right)dx}
sin⁡xx{{\sin x} \over x}

is decreasing in

(π4,π3)\left( {{\pi \over 4},{\pi \over 3}} \right)

so it attains maximum at

x=π4x = {\pi \over 4}
I>∫π/4π/3(8sin⁡x/3x/3−2)dxI > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x/3} \over {x/3}} - 2} \right)dx}
I>3−π6I > \sqrt 3 - {\pi \over 6}
Q123
Let a function f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be defined as : f(x)={∫0x(5−∣t−3∣)dt,x>4x2+bx,x≤4f(x)= \begin{cases}\int\limits_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x & , x \leq 4\end{cases} where b∈R\mathrm{b} \in \mathbb{R}. If ff is continuous at x=4x=4, then which of the following statements is NOT true?
A ff is not differentiable at x=4x=4
B f′(3)+f′(5)=354f^{\prime}(3)+f^{\prime}(5)=\dfrac{35}{4}
C ff is increasing in (−∞,18)∪(8,∞)\left(-\infty, \dfrac{1}{8}\right) \cup(8, \infty)
D ff has a local minima at x=18x=\dfrac{1}{8}
Correct Answer
Option C
Solution

∵\because f(x) is continuous at x = 4

⇒f(4−)=f(4+)\Rightarrow f({4^ - }) = f({4^ + })
⇒16+4b=∫04(5−∣t−3∣)dt\Rightarrow 16 + 4b = \int\limits_0^4 {(5 - |t - 3|)dt}
=∫03(2+t)dt+∫34(8−t)dt= \int\limits_0^3 {(2 + t)dt + \int\limits_3^4 {(8 - t)dt} }
=2t+t22)03+8t−t23]34= \left. {2t + {{{t^2}} \over 2}} \right)_0^3 + \left. {8t - {{{t^2}} \over 3}} \right]_3^4
=6+92−0+(32−8)−(24−92)= 6 + {9 \over 2} - 0 + (32 - 8) - \left( {24 - {9 \over 2}} \right)
16+4b=1516 + 4b = 15
⇒b=−14\Rightarrow b = {{ - 1} \over 4}
⇒f(x)={∫0x5−∣t−3∣ dtx>4x2−x4x≤4\Rightarrow f(x) = \left\{ \begin{array}{ll}{\int\limits_0^x {5 - |t - 3|\,dt} } & {x > 4} \\ {{x^2} - {x \over 4}} & {x \le 4} \end{array} \right.
⇒f′(x)={5−∣x−3∣x>42x−14x≤4\Rightarrow f'(x) = \left\{ \begin{array}{ll}{5 - |x - 3|} & {x > 4} \\ {2x - {1 \over 4}} & {x \le 4} \end{array} \right.
⇒f′(x)={8−xx>42x−14x≤4\Rightarrow f'(x) = \left\{ \begin{array}{ll}{8 - x} & {x > 4} \\ {2x - {1 \over 4}} & {x \le 4} \end{array} \right.
f′(x)f'(x)

f'(3) + f'(5) = 6 - {1 \over 4} = {{35} \over 4}

f'(x) = 0 \Rightarrow x = {1 \over 8}

havelocalminimahave local minima

\therefore$$ (C) is only incorrect option.

Q124
Let f(x)=2+∣x∣−∣x−1∣+∣x+1∣,x∈Rf(x)=2+|x|-|x-1|+|x+1|, x \in \mathbf{R}. Consider (S1):f′(−32)+f′(−12)+f′(12)+f′(32)=2(\mathrm{S} 1): f^{\prime}\left(-\dfrac{3}{2}\right)+f^{\prime}\left(-\dfrac{1}{2}\right)+f^{\prime}\left(\dfrac{1}{2}\right)+f^{\prime}\left(\dfrac{3}{2}\right)=2 (S2):∫−22f(x)dx=12(\mathrm{S} 2): \int\limits_{-2}^{2} f(x) \mathrm{d} x=12 Then,
A both (S1) and (S2) are correct
B both (S1) and (S2) are wrong
C only (S1) is correct
D only (S2) is correct
Correct Answer
Option D
Solution
f(x)=2+∣x∣−∣x−1∣+∣x+1∣, x∈Rf(x) = 2 + |x| - |x - 1| + |x + 1|,\,x \in R

∴\therefore

f(x) = \left\{ {\matrix{ { - x} & , & {x

\therefore

f'\left( { - {3 \over 2}} \right) + f'\left( { - {1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) + f'\left( {{3 \over 2}} \right) = - 1 + 1 + 3 + 1 = 4

andand

\int\limits_{ - 2}^2 {f(x)dx = \int\limits_{ - 2}^{ - 1} {f(x)dx + \int\limits_{ - 1}^0 {f(x)dx + \int\limits_0^1 {f(x)dx + \int\limits_1^2 {f(x)dx} } } } }

= \left[ { - {{{x^2}} \over 2}} \right]_2^{ - 1} + \left[ {{{{{(x + 2)}^2}} \over 2}} \right]_{ - 1}^0 + \left[ {{{{{(3x + 2)}^2}} \over 6}} \right]_0^1 + \left[ {{{{{(x + 4)}^2}} \over 2}} \right]_1^2

= {3 \over 2} + {3 \over 2} + {7 \over 2} + {{11} \over 2} = {{24} \over 2} = 12

\therefore$$ Only (S2) is correct

Q125
∫02(∣2x2−3x∣+[x−12])dx\int\limits_{0}^{2}\left(\left|2 x^{2}-3 x\right|+\left[x-\dfrac{1}{2}\right]\right) \mathrm{d} x, where [t] is the greatest integer function, is equal to :
A 76\dfrac{7}{6}
B 1912\dfrac{19}{12}
C 3112\dfrac{31}{12}
D 32\dfrac{3}{2}
Correct Answer
Option B
Solution
∫02∣2x2−3x∣dx+∫02[x−12]dx\int\limits_0^2 {|2{x^2} - 3x|dx + \int\limits_0^2 {\left[ {x - {1 \over 2}} \right]dx} }
=∫03/2(3x−2x2)dx+∫3/22(2x2−3x)dx+∫01/2−1dx+∫1/23/20 dx+∫3/221dx= \int\limits_0^{3/2} {(3x - 2{x^2})dx + \int\limits_{3/2}^2 {(2{x^2} - 3x)dx + \int\limits_0^{1/2} { - 1dx + \int\limits_{1/2}^{3/2} {0\,dx + \int\limits_{3/2}^2 {1dx} } } } }
=(3x22−2x33)∣03/2+(2x33−3x22)∣3/22−12+12= \left. {\left( {{{3{x^2}} \over 2} - {{2{x^3}} \over 3}} \right)} \right|_0^{3/2} + \left. {\left( {{{2{x^3}} \over 3} - {{3{x^2}} \over 2}} \right)} \right|_{3/2}^2 - {1 \over 2} + {1 \over 2}
=(278−2712)+(163−6−2712+278)= \left( {{{27} \over 8} - {{27} \over {12}}} \right) + \left( {{{16} \over 3} - 6 - {{27} \over {12}} + {{27} \over 8}} \right)
=1912= {{19} \over {12}}
Q126
The minimum value of the twice differentiable function f(x)=∫0xex−tf′(t)dt−(x2−x+1)exf(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}, x∈Rx \in \mathbf{R}, is :
A −2e-\dfrac{2}{\sqrt{\mathrm{e}}}
B −2e-2 \sqrt{\mathrm{e}}
C −e-\sqrt{\mathrm{e}}
D 2e\dfrac{2}{\sqrt{\mathrm{e}}}
Correct Answer
Option A
Solution
f(x)=∫0xex−tf′(t)dt−(x2−x+1)exf(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}}
f(x)=ex∫0xe−tf′(t)dt−(x2−x+1)exf(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}}
e−xf(x)=∫0xe−tf′(t)dt−(x2−x+1){e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)}

Differentiate on both side

e−xf′(x)+(−f(x)e−x)=e−xf′(x)−2x+1{e^{ - x}}f'(x) + ( - f(x){e^{ - x}}) = {e^{ - x}}f'(x) - 2x + 1
f(x)=ex(2x−1)f(x) = {e^x}(2x - 1)
f′(x)=ex(2)+ex(2x−1)f'(x) = {e^x}(2) + {e^x}(2x - 1)
=ex(2x+1)= {e^x}(2x + 1)
x=−12x = - {1 \over 2}
f′′(x)=ex(2)+(2x+1)exf''(x) = {e^x}(2) + (2x + 1){e^x}
=ex(2x+3)= {e^x}(2x + 3)

For

x=−12  f′′(x)>0x = - {1 \over 2}\,\,f''(x) > 0

⇒\Rightarrow Maxima ∴\therefore Max.

=e−12(−1−1)= {e^{ - {1 \over 2}}}( - 1 - 1)

∴\therefore

−2e- {2 \over {\sqrt e }}
Q127
The minimum value of the function f(x)=∫02e∣x−t∣dtf(x) = \int\limits_0^2 {{e^{|x - t|}}dt} is :
A 2
B 2(e−1)2(e-1)
C e(e−1)e(e-1)
D 2e−12e-1
Correct Answer
Option B
Solution
f(x)=∫02e∣x−t∣dtf(x)=\int_0^2 e^{|x-t|} d t

For x>2x>2

f(x)=∫02ex−tdt=ex(1−e−2)f(x)=\int_0^2 e^{x-t} d t=e^x\left(1-e^{-2}\right)

For x2x2

f(x)∣min⁡=e2−1\left.f(x)\right|_{\min =e^2-1}

For x<0\mathrm{x}<0

f(x)∣min⁡=e2−1\left.f(x)\right|_{\min =e^2-1}

For x∈[0,2]x \in[0,2]

f(x)∣min⁡=2(e−1)\left.f(x)\right|_{\min }=2(e-1)
Q128
Let In(x)=∫0x1(t2+5)ndt,n=1,2,3,….I_{n}(x)=\int_{0}^{x} \dfrac{1}{\left(t^{2}+5\right)^{n}} d t, n=1,2,3, \ldots . Then :
A 50I6−9I5=xI5′50 I_{6}-9 I_{5}=x I_{5}^{\prime}
B 50I6−11I5=xI5′50 I_{6}-11 I_{5}=x I_{5}^{\prime}
C 50I6−9I5=I5′50 I_{6}-9 I_{5}=I_{5}^{\prime}
D 50I6−11I5=I5′50 I_{6}-11 I_{5}=I_{5}^{\prime}
Correct Answer
Option A
Solution
In(x)=∫0x1(t2+5)ndt{I_n}(x) = \int\limits_0^x {{1 \over {{{({t^2} + 5)}^n}}}dt}
=∫0x1(t2+5)n⏟I×III dt= \int\limits_0^x {{1 \over {\underbrace {{{({t^2} + 5)}^n}}_I}} \times \mathop I\limits_{II} \,dt}
=t(t2+5)n∣0x−∫0x−2nt(t2+5)n+1×t dt= \left. {{t \over {{{({t^2} + 5)}^n}}}} \right|_0^x - \int\limits_0^x {{{ - 2nt} \over {{{({t^2} + 5)}^{n + 1}}}} \times t\,dt}
=x(x2+5)n+∫0x2n(t2+5−5(t2+5)n+1)dt= {x \over {{{({x^2} + 5)}^n}}} + \int\limits_0^x {2n\left( {{{{t^2} + 5 - 5} \over {{{({t^2} + 5)}^{n + 1}}}}} \right)dt}
In(x)=x(x2+5)n+2n In(x)−10n In+1(x){I_n}(x) = {x \over {{{({x^2} + 5)}^n}}} + 2n\,{I_n}(x) - 10n\,{I_{n + 1}}(x)
10n In+1(x)−(2n−1) In(x)=xIn′(x)10n\,{I_{n + 1}}(x) - (2n - 1)\,{I_n}(x) = xI{'_n}(x)

For

n=5n = 5
50I6(x)−9I5(x)=xI5′(x)50{I_6}(x) - 9{I_5}(x) = xI{'_5}(x)
Q129
The integral ∫0π213+2sin⁡x+cos⁡x dx\int\limits_{0}^{\dfrac{\pi}{2}} \dfrac{1}{3+2 \sin x+\cos x} \mathrm{~d} x is equal to :
A tan⁡−1(2)\tan ^{-1}(2)
B tan⁡−1(2)−π4\tan ^{-1}(2)-\dfrac{\pi}{4}
C 12tan⁡−1(2)−π8\dfrac{1}{2} \tan ^{-1}(2)-\dfrac{\pi}{8}
D 12\dfrac{1}{2}
Correct Answer
Option B
Solution
I=∫0π/213+2sin⁡x+cos⁡xdxI = \int\limits_0^{\pi /2} {{1 \over {3 + 2\sin x + \cos x}}dx}
=∫0π/2(1+tan⁡2x/2)dx3(1+tan⁡2x/2)+2(2tan⁡x/2)+(1−tan⁡2x/2)= \int\limits_0^{\pi /2} {{{(1 + {{\tan }^2}x/2)dx} \over {3(1 + {{\tan }^2}x/2) + 2(2\tan x/2) + (1 - {{\tan }^2}x/2)}}}

Let

tan⁡x/2=t⇒sec⁡2x/2dx=2dt\tan x/2 = t \Rightarrow {\sec ^2}x/2dx = 2dt
I=∫012dt4+2t2+4tI = \int\limits_0^1 {{{2dt} \over {4 + 2{t^2} + 4t}}}
=∫01dtt2+2t+2=∫01dt(t+1)2+1= \int\limits_0^1 {{{dt} \over {{t^2} + 2t + 2}} = \int\limits_0^1 {{{dt} \over {{{(t + 1)}^2} + 1}}} }
=tan⁡−1(t+1)∣01=tan⁡−12−π4= \left. {{{\tan }^{ - 1}}(t + 1)} \right|_0^1 = {\tan ^{ - 1}}2 - {\pi \over 4}
Q130
If f(α)=∫1αlog⁡10t1+tdt,α>0f(\alpha)=\int\limits_{1}^{\alpha} \dfrac{\log _{10} \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}, \alpha>0, then f(e3)+f(e−3)f\left(\mathrm{e}^{3}\right)+f\left(\mathrm{e}^{-3}\right) is equal to :
A 9
B 92\dfrac{9}{2}
C 9log⁡e(10)\dfrac{9}{\log _{e}(10)}
D 92log⁡e(10)\dfrac{9}{2 \log _{e}(10)}
Correct Answer
Option D
Solution
f(α)=∫1αlog⁡10t1+tdtf(\alpha ) = \int_1^\alpha {{{{{\log }_{10}}t} \over {1 + t}}dt}

...... (i)

f(1α)=∫11αlog⁡10t1+tdtf\left( {{1 \over \alpha }} \right) = \int_1^{{1 \over \alpha }} {{{{{\log }_{10}}t} \over {1 + t}}dt}

Substituting

t→1pt \to {1 \over p}
f(1α)=∫1αlog⁡10(1p)1+1p(−1p2)dpf\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}\left( {{1 \over p}} \right)} \over {1 + {1 \over p}}}\left( {{{ - 1} \over {{p^2}}}} \right)dp}
=∫1αlog⁡10pp(p+1)dp=∫1α(log⁡10tt−log⁡10tt+1)dt= \int_1^\alpha {{{{{\log }_{10}}p} \over {p(p + 1)}}dp = \int_1^\alpha {\left( {{{{{\log }_{10}}t} \over t} - {{{{\log }_{10}}t} \over {t + 1}}} \right)dt} }

....... (ii) By (i) + (ii)

f(α)+f(1α)=∫1αlog⁡10ttdt=∫1αln⁡tt . log⁡10e dtf(\alpha ) + f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}t} \over t}dt = \int_1^\alpha {{{\ln t} \over t}\,.\,{{\log }_{10}}e\,dt} }
=(ln⁡α)22log⁡e10= {{{{(\ln \alpha )}^2}} \over {2{{\log }_e}10}}
α=e3⇒f(e3)+f(e−3)=92log⁡e10\alpha = {e^3} \Rightarrow f({e^3}) + f({e^{ - 3}}) = {9 \over {2{{\log }_e}10}}
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