Practice Start Test

Definite Integration

JEE Mathematics · 230 questions · Page 14 of 23 · Click an option or "Show Solution" to reveal answer

Q131
If [t][t] denotes the greatest integer ≤t\leq t, then the value of ∫01[2x−∣3x2−5x+2∣+1]dx\int_{0}^{1}\left[2 x-\left|3 x^{2}-5 x+2\right|+1\right] \mathrm{d} x is :
A 37+13−46\dfrac{\sqrt{37}+\sqrt{13}-4}{6}
B 37−13−46\dfrac{\sqrt{37}-\sqrt{13}-4}{6}
C −37−13+46\dfrac{-\sqrt{37}-\sqrt{13}+4}{6}
D −37+13+46\dfrac{-\sqrt{37}+\sqrt{13}+4}{6}
Correct Answer
Option A
Solution
∫01[2x−∣3x2−5x+2∣+1]dx\int_0^1 {\left[ {2x - |3{x^2} - 5x + 2| + 1} \right]dx}
3x2−5x+2=03{x^2} - 5x + 2 = 0
⇒3x2−3x−2x+2=0\Rightarrow 3{x^2} - 3x - 2x + 2 = 0
⇒3x(x−1)−2(x−1)=0\Rightarrow 3x(x - 1) - 2(x - 1) = 0
⇒(x−1)(3x−2)=0\Rightarrow (x - 1)(3x - 2) = 0

∴\therefore In 0 to

23, ∣3x2−5x+2∣=3x2−5x+2{2 \over 3},\,|3{x^2} - 5x + 2| = 3{x^2} - 5x + 2

And in

23{2 \over 3}

to 1,

∣3x2−5x+2∣=−(3x2−5x+2)|3{x^2} - 5x + 2| = - (3{x^2} - 5x + 2)

∴\therefore

∫023[2x−3x2+5x−2+1]dx+∫231[2x+3x2−5x+2+1]dx\int_0^{{2 \over 3}} {\left[ {2x - 3{x^2} + 5x - 2 + 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {2x + 3{x^2} - 5x + 2 + 1} \right]dx} }
=∫023[−3x2+7x−1]dx+∫231[3x2−3x+3]dx= \int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} }

Now, let

f(x)=−3x2+7x−1f(x) = - 3{x^2} + 7x - 1
⇒f′(x)=−6x+7\Rightarrow f'(x) = - 6x + 7
=−6(x−76)= - 6\left( {x - {7 \over 6}} \right)
f(0)=−1f(0) = - 1
f(23)=−3×49+143−1f\left( {{2 \over 3}} \right) = - 3 \times {4 \over 9} + {{14} \over 3} - 1
=73=2.33= {7 \over 3} = 2.33

∴\therefore Range of

f(x)=[−1,73]f(x) = \left[ { - 1,{7 \over 3}} \right]

Integer value between

−1- 1

to

73=−1,0,1,2,73{7 \over 3} = - 1,0,1,2,{7 \over 3}

When

f(x)=−1f(x) = - 1
⇒−3x2+7x−1=−1\Rightarrow - 3{x^2} + 7x - 1 = - 1
⇒3x2−7x=0\Rightarrow 3{x^2} - 7x = 0
⇒x(3x−7)=0\Rightarrow x(3x - 7) = 0
⇒x=0,73\Rightarrow x = 0,{7 \over 3}

(not possible as

73∉(0,2){7 \over 3} \notin (0,2)

) When

f(x)=0f(x) = 0
⇒−3x2+7x−1=0\Rightarrow - 3{x^2} + 7x - 1 = 0
⇒3x2−7x+1=0\Rightarrow 3{x^2} - 7x + 1 = 0
⇒x=7 ± 49−126\Rightarrow x = {{7\, \pm \,\sqrt {49 - 12} } \over 6}
=7 ±376= {{7\, \pm \sqrt {37} } \over 6}

∴\therefore

x=7−376x = {{7 - \sqrt {37} } \over 6}

When

f(x)=1f(x) = 1
⇒−3x2+7x−1=1\Rightarrow - 3{x^2} + 7x - 1 = 1
⇒3x2−7x+2=0\Rightarrow 3{x^2} - 7x + 2 = 0
⇒x=7 ± 49−246\Rightarrow x = {{7\, \pm \,\sqrt {49 - 24} } \over 6}
=7 ± 256= {{7\, \pm \,\sqrt {25} } \over 6}
=7 ± 56= {{7\, \pm \,5} \over 6}
⇒x=2, 13\Rightarrow x = 2,\,{1 \over 3}

When

f(x)=2f(x) = 2
⇒−3x2+7x−1=2\Rightarrow - 3{x^2} + 7x - 1 = 2
⇒3x2−7x+3=0\Rightarrow 3{x^2} - 7x + 3 = 0
⇒x=7 ± 49−366\Rightarrow x = {{7\, \pm \,\sqrt {49 - 36} } \over 6}
=7 ± 136= {{7\, \pm \,\sqrt {13} } \over 6}

∴\therefore

x=7 −136x = {{7\, - \sqrt {13} } \over 6}

∴\therefore Possible values of

x=0, 7 − 376,13,7 − 136,23x = 0,\,{{7\, - \,\sqrt {37} } \over 6},{1 \over 3},{{7\, - \,\sqrt {13} } \over 6},{2 \over 3}

∴\therefore

∫023[−3x2+7x−1]dx\int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx}
=∫07 − 376(−1)dx+∫7 − 376130 dx+∫137 − 136(1) dx+∫7 − 136232 dx= \int_0^{{{7\, - \,\sqrt {37} } \over 6}} {( - 1)dx + \int_{{{7\, - \,\sqrt {37} } \over 6}}^{{1 \over 3}} {0\,dx + \int_{{1 \over 3}}^{{{7\, - \,\sqrt {13} } \over 6}} {(1)\,dx + \int_{{{7\, - \,\sqrt {13} } \over 6}}^{{2 \over 3}} {2\,dx} } } }
=−7 + 376+0+7 − 136−13+2[23−7 − 136]= {{ - 7\, + \,\sqrt {37} } \over 6} + 0 + {{7\, - \,\sqrt {13} } \over 6} - {1 \over 3} + 2\left[ {{2 \over 3} - {{7\, - \,\sqrt {13} } \over 6}} \right]
=−7+37+7−13−2+8−14+2136= {{ - 7 + \sqrt {37} + 7 - \sqrt {13} - 2 + 8 - 14 + 2\sqrt {13} } \over 6}
=37+13−86= {{\sqrt {37} + \sqrt {13} - 8} \over 6}

Now,

∫231[3x2−3x+3]dx\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx}

Let

f(x)=3x2−3x+3f(x) = 3{x^2} - 3x + 3
f′(x)=6x−3=3(2x−1)f'(x) = 6x - 3 = 3(2x - 1)
f′(x)=0⇒2x−1=0⇒x=12f'(x) = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = {1 \over 2}
f(23)=3(49−23+1)f\left( {{2 \over 3}} \right) = 3\left( {{4 \over 9} - {2 \over 3} + 1} \right)
=73=2.3= {7 \over 3} = 2.3
f(1)=3(1−1+1)f(1) = 3(1 - 1 + 1)
=3= 3

No integer values present in between

73{7 \over 3}

and 3. ∴\therefore Possible value of

x=23, 1x = {2 \over 3},\,1

So, integration don't break anywhere. ∴\therefore

∫231[3x2−3x+3]dx\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx}
=∫231[2.33] dx= \int_{{2 \over 3}}^1 {\left[ {2.33} \right]\,dx}
=∫2312 dx=2(1−23)=23= \int_{{2 \over 3}}^1 {2\,dx = 2\left( {1 - {2 \over 3}} \right) = {2 \over 3}}

∴\therefore Value of Integration

=37+13−16+23= {{\sqrt {37} + \sqrt {13} - 1} \over 6} + {2 \over 3}
=37+13−8+46= {{\sqrt {37} + \sqrt {13} - 8 + 4} \over 6}
=37+13−46= {{\sqrt {37} + \sqrt {13} - 4} \over 6}
Q132
The value of the integral ∫−π4π4x+π42−cos⁡2xdx\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{x + {\pi \over 4}} \over {2 - \cos 2x}}dx} is :
A π263{{{\pi ^2}} \over {6\sqrt 3 }}
B π26{{{\pi ^2}} \over 6}
C π233{{{\pi ^2}} \over {3\sqrt 3 }}
D π2123{{{\pi ^2}} \over {12\sqrt 3 }}
Correct Answer
Option A
Solution
I=∫−π4π4x+π4(2−cos⁡2x)dxI=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{(2-\cos 2 x)} d x

Using ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b f(x) d x=\int_a^b f(a+b-x) d x

I=∫−π4π4(−x+π42−cos⁡2x)dxI=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{-x+\frac{\pi}{4}}{2-\cos 2 x}\right) d x

∴2I=∫−π4π4πdx2(2−cos⁡2x)⇒I=2π4∫0π4(dx2−1−tan⁡2x1+tan⁡2x)⇒I=π2∫0π4(1+tan⁡2x1+3tan⁡2x)dx\begin{aligned} & \therefore 2 I=\int_{-\dfrac{\pi}{4}}^{\dfrac{\pi}{4}} \dfrac{\pi d x}{2(2-\cos 2 x)} \\\\ & \Rightarrow I=\dfrac{2 \pi}{4} \int_0^{\dfrac{\pi}{4}}\left(\dfrac{d x}{\left.\dfrac{2-1-\tan ^2 x}{1+\tan ^2 x}\right)}\right. \\\\ & \Rightarrow I=\dfrac{\pi}{2} \int_0^{\dfrac{\pi}{4}}\left(\dfrac{1+\tan ^2 x}{1+3 \tan ^2 x}\right) d x\end{aligned} Now,

tan⁡x=t=π2∫01dt1+3t2=π2[tan⁡−1(3t)3]01=π23(π3)=π263\begin{aligned} & \tan x=t \\\\ & =\frac{\pi}{2} \int_0^1 \frac{d t}{1+3 t^2} \\\\ & =\frac{\pi}{2}\left[\frac{\tan ^{-1}(\sqrt{3} t)}{\sqrt{3}}\right]_0^1 \\\\ & =\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}\right)=\frac{\pi^2}{6 \sqrt{3}} \end{aligned}
Q133
Let α>0\alpha>0. If ∫0αxx+α−x dx=16+20215\int\limits_0^\alpha \dfrac{x}{\sqrt{x+\alpha}-\sqrt{x}} \mathrm{~d} x=\dfrac{16+20 \sqrt{2}}{15}, then α\alpha is equal to :
A 4
B 2
C 222 \sqrt{2}
D 2\sqrt{2}
Correct Answer
Option B
Solution

I=∫0αxx+α−xdx,α>0I=\int\limits_{0}^{\alpha} \dfrac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x, \alpha>0

=1α∫0αx(x+α+x)dx\begin{aligned} & =\frac{1}{\alpha} \int\limits_{0}^{\alpha} x(\sqrt{x+\alpha}+\sqrt{x}) d x \end{aligned}
=1α∫0α(xx+α+xx)dx=1α∫0α[(x+α−α)x+α+x32]dx\begin{aligned}& = {1 \over \alpha }\int\limits_0^\alpha {\left( {x\sqrt {x + \alpha } + x\sqrt x } \right)} dx \\ & = {1 \over \alpha }\int\limits_0^\alpha {\left[ {\left( {x + \alpha - \alpha } \right)\sqrt {x + \alpha } + {x^{{3 \over 2}}}} \right]} dx\end{aligned}
=1α∫0α[(x+α)3/2−α(x+α)1/2+x3/2]dx=1α[25(x+α)5/2−α23(x+α)3/2+25x5/2]0α=1α(25(2α)5/2−2α3(2α)3/2+25α5/2−25α5/2+23α5/2)=1α(27/2α5/25−25/2α5/23+23α5/2)=α3/2(27/25−25/23+23)=α3/215(242−202+10)=α3/215(42+10)\begin{aligned} & =\frac{1}{\alpha}\int\limits_0^\alpha \left[(x+\alpha)^{3 / 2}-\alpha(x+\alpha)^{1 / 2}+x^{3 / 2}\right] d x \\\\ & =\frac{1}{\alpha}\left[\frac{2}{5}(x+\alpha)^{5 / 2}-\alpha \frac{2}{3}(x+\alpha)^{3 / 2}+\frac{2}{5} x^{5 / 2}\right]_0^\alpha \\\\ & =\frac{1}{\alpha}\left(\frac{2}{5}(2 \alpha)^{5 / 2}-\frac{2 \alpha}{3}(2 \alpha)^{3 / 2}+\frac{2}{5} \alpha^{5 / 2}-\frac{2}{5} \alpha^{5 / 2}+\frac{2}{3} \alpha^{5 / 2}\right) \\\\ & =\frac{1}{\alpha}\left(\frac{2^{7 / 2} \alpha^{5 / 2}}{5}-\frac{2^{5 / 2} \alpha^{5 / 2}}{3}+\frac{2}{3} \alpha^{5 / 2}\right)=\alpha^{3 / 2}\left(\frac{2^{7 / 2}}{5}-\frac{2^{5 / 2}}{3}+\frac{2}{3}\right) \\\\ & =\frac{\alpha^{3 / 2}}{15}(24 \sqrt{2}-20 \sqrt{2}+10)=\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10) \end{aligned}

Now, α3/215(42+10)=16+20215=22(42+10)15\dfrac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)=\dfrac{16+20 \sqrt{2}}{15}=\dfrac{2 \sqrt{2}(4 \sqrt{2}+10)}{15}

⇒α3/2=22=23/2\Rightarrow \alpha^{3 / 2}=2 \sqrt{2}=2^{3 / 2}
⇒α=2\Rightarrow \alpha=2
Q134
If ϕ(x)=1x∫π4x(42sin⁡t−3ϕ′(t))dt,x>0\phi(x)=\dfrac{1}{\sqrt{x}} \int\limits_{\dfrac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t, x>0, then ∅′(π4)\emptyset^{\prime}\left(\dfrac{\pi}{4}\right) is equal to :
A 46+π\dfrac{4}{6+\sqrt{\pi}}
B 46−π\dfrac{4}{6-\sqrt{\pi}}
C 8π\dfrac{8}{\sqrt{\pi}}
D 86+π\dfrac{8}{6+\sqrt{\pi}}
Correct Answer
Option D
Solution

ϕ(x)=1x∫π/4x(42sin⁡t−3ϕ′(t))dt\phi(x)=\dfrac{1}{\sqrt{x}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t ⇒ϕ′(x)=−12x3/2∫π/4x(42sin⁡t−3ϕ′(t))dt\Rightarrow \phi^{\prime}(x)=\dfrac{-1}{2 x^{3 / 2}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t +1x(42sin⁡(x)−3ϕ′(x))+\dfrac{1}{\sqrt{x}}\left(4 \sqrt{2} \sin (x)-3 \phi^{\prime}(x)\right) At, x=π4x=\dfrac{\pi}{4} ϕ′(π4)=−12(π4)3/2×0+4π(42×12−3ϕ′(π4))\phi^{\prime}\left(\dfrac{\pi}{4}\right)=\dfrac{-1}{2\left(\dfrac{\pi}{4}\right)^{3 / 2}} \times 0+\sqrt{\dfrac{4}{\pi}}\left(4 \sqrt{2} \times \dfrac{1}{\sqrt{2}}-3 \phi^{\prime}\left(\dfrac{\pi}{4}\right)\right) ⇒ϕ′(π4)[1+6π]=2π×4\Rightarrow \phi^{\prime}\left(\dfrac{\pi}{4}\right)\left[1+\dfrac{6}{\sqrt{\pi}}\right]=\dfrac{2}{\sqrt{\pi}} \times 4 ⇒ϕ′(π4)=86+x\Rightarrow \phi^{\prime}\left(\dfrac{\pi}{4}\right)=\dfrac{8}{6+\sqrt{x}}

Q135
Let α∈(0,1)\alpha \in (0,1) and β=log⁡e(1−α)\beta = {\log _e}(1 - \alpha ). Let Pn(x)=x+x22+x33 + ... + xnn,x∈(0,1){P_n}(x) = x + {{{x^2}} \over 2} + {{{x^3}} \over 3}\, + \,...\, + \,{{{x^n}} \over n},x \in (0,1). Then the integral ∫0αt501−tdt\int\limits_0^\alpha {{{{t^{50}}} \over {1 - t}}dt} is equal to
A −(β+P50(α)) - \left( {\beta + {P_{50}}\left( \alpha \right)} \right)
B β−P50(α)\beta - {P_{50}}(\alpha )
C P50(α)−β{P_{50}}(\alpha ) - \beta
D β+P50−(α)\beta + {P_{50}} - (\alpha )
Correct Answer
Option A
Solution

∫0αt501−tdt\int_{0}^{\alpha} \dfrac{t^{50}}{1-t} d t

=∫0αt50−1+11−t=−∫0α(1−t501−t−11−t)dt=−∫0α(1+t+…..+t49)+∫0α11−tdt\begin{aligned} & = \int_{0}^{\alpha} \frac{t^{50}-1+1}{1-t}\\\\ & =-\int_{0}^{\alpha}\left(\frac{1-t^{50}}{1-t}-\frac{1}{1-t}\right) d t\\\\ & =-\int_{0}^{\alpha}\left(1+t+\ldots . .+t^{49}\right)+\int_{0}^{\alpha} \frac{1}{1-t} d t \end{aligned}

=−(α5050+α4949+…..+α11)+(ln⁡(1−f)−1)0α=-\left(\dfrac{\alpha^{50}}{50}+\dfrac{\alpha^{49}}{49}+\ldots . .+\dfrac{\alpha^{1}}{1}\right)+\left(\dfrac{\ln (1-\mathrm{f})}{-1}\right)_{0}^{\alpha} =−P50(α)−ln⁡(1−α)=-\mathrm{P}_{50}(\alpha)-\ln (1-\alpha) =−P50(α)−β=-\mathrm{P}_{50}(\alpha)-\beta

Q136
The value of ∫π3π2(2+3sin⁡x)sin⁡x(1+cos⁡x)dx\int\limits_{\dfrac{\pi}{3}}^{\dfrac{\pi}{2}} \dfrac{(2+3 \sin x)}{\sin x(1+\cos x)} d x is equal to :
A 103−3+log⁡e3\dfrac{10}{3}-\sqrt{3}+\log _{e} \sqrt{3}
B 72−3−log⁡e3\dfrac{7}{2}-\sqrt{3}-\log _{e} \sqrt{3}
C 103−3−log⁡e3\dfrac{10}{3}-\sqrt{3}-\log _{e} \sqrt{3}
D −2+33+log⁡e3-2+3\sqrt{3}+\log _{e} \sqrt{3}
Correct Answer
Option A
Solution

Let I =

∫π3π2(2+3sin⁡x)sin⁡x(1+cos⁡x)dx\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x
=∫π/3π/22sin⁡x(1+cos⁡x)dx+∫π/3π/231+cos⁡xdx=\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin x(1+\cos x)} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x
=∫π/3π/22(1−cos⁡x)sin⁡x(1−cos⁡2x)dx+∫π/3π/231+cos⁡xdx\begin{aligned} =\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin x\left(1-\cos ^2 x\right)} & d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x \end{aligned}
=∫π/3π/22(1−cos⁡x)sin⁡3xdx+∫π/3π/232cos⁡2x2dx=\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin ^3 x} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2 \cos ^2 \frac{x}{2}} d x
=∫π/3π/22sin⁡3xdx−∫π/3π/22cot⁡xcosec⁡2xdx+∫π/3π/232sec⁡2x2dx\begin{array}{r} =\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin ^3 x} d x-\int\limits_{\pi / 3}^{\pi / 2} 2 \cot x \operatorname{cosec}^2 x d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2} \sec ^2 \frac{x}{2} d x \end{array}
=2I1−2I2+32I3=2 I_1-2 I_2+\frac{3}{2} I_3

Now,

I1=∫π/3π/2cosec⁡3xdx=∫π/3π/21+cot⁡2xcosec⁡2xdx\begin{aligned} I_1 & =\int\limits_{\pi / 3}^{\pi / 2} \operatorname{cosec}^3 x d x \\\\ & =\int\limits_{\pi / 3}^{\pi / 2} \sqrt{1+\cot ^2 x} \operatorname{cosec}^2 x d x \end{aligned}

Put cot⁡x=t⇒−cosec⁡2xdx=dt\cot x=t \Rightarrow-\operatorname{cosec}^2 x d x=d t When, x=π3⇒t=13 and x=π2⇒t=0x=\dfrac{\pi}{3} \Rightarrow t=\dfrac{1}{\sqrt{3}} \text{ and } x=\dfrac{\pi}{2} \Rightarrow t=0

∴I1=−∫1301+t2dt=∫0131+t2dt=[t21+t2+12log⁡[t+1+t2]]013=13+12log⁡3\begin{aligned} \therefore & I_1=-\int\limits_{\frac{1}{\sqrt{3}}}^0 \sqrt{1+t^2} d t \\\\ = & \int\limits_0^{\frac{1}{\sqrt{3}}} \sqrt{1+t^2} d t \\\\ = & {\left[\frac{t}{2} \sqrt{1+t^2}+\frac{1}{2} \log \left[t+\sqrt{1+t^2}\right]\right]_0^{\frac{1}{\sqrt{3}}} } \\\\ = & \frac{1}{3}+\frac{1}{2} \log \sqrt{3} \end{aligned}
I2=∫π/3π/2cot⁡xcosec⁡2xdxI_2=\int\limits_{\pi / 3}^{\pi / 2} \cot x \operatorname{cosec}^2 x d x

Put cot⁡x=t⇒cosec⁡2xdx=−dt\cot x=t \Rightarrow \operatorname{cosec}^2 x d x=-d t When, x=π3x=\dfrac{\pi}{3} ⇒t=13\Rightarrow t=\dfrac{1}{\sqrt{3}} and x=π2⇒t=0x=\dfrac{\pi}{2} \Rightarrow t=0

∴I2=−∫130tdt=∫013ttdt=[t22]013=16\therefore \begin{aligned} I_2 & =-\int\limits_{\frac{1}{\sqrt{3}}}^0 t d t \\\\ & =\int\limits_0^{\frac{1}{\sqrt{3}} t} t d t=\left[\frac{t^2}{2}\right]_0^{\frac{1}{\sqrt{3}}}=\frac{1}{6} \end{aligned}
I3=∫π3π2sec⁡2x2dx=2(tan⁡π4−tan⁡π6)=2(1−13)\begin{aligned} & I_3=\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sec ^2 \frac{x}{2} d x \\\\ = & 2\left(\tan \frac{\pi}{4}-\tan \frac{\pi}{6}\right)=2\left(1-\frac{1}{\sqrt{3}}\right) \end{aligned}
∴I=2I1−2I2+32I3=2(13+12log⁡3)−2(16)+32(2(1−13))\begin{aligned} & \therefore \quad I=2 I_1-2 I_2+\frac{3}{2} I_3 \\\\ & =2\left(\frac{1}{3}+\frac{1}{2} \log \sqrt{3}\right)-2\left(\frac{1}{6}\right) +\frac{3}{2}\left(2\left(1-\frac{1}{\sqrt{3}}\right)\right) \end{aligned}
=23+log⁡3−13+3−3=103−3+log⁡e3\begin{aligned} & =\frac{2}{3}+\log \sqrt{3}-\frac{1}{3}+3-\sqrt{3} \\\\ & =\frac{10}{3}-\sqrt{3}+\log _e \sqrt{3} \end{aligned}
Q137
lim⁡n→∞3n{4+(2+1n)2+(2+2n)2+…+(3−1n)2}\lim\limits_{n \rightarrow \infty} \dfrac{3}{n}\left\{4+\left(2+\dfrac{1}{n}\right)^2+\left(2+\dfrac{2}{n}\right)^2+\ldots+\left(3-\dfrac{1}{n}\right)^2\right\} is equal to :
A 0
B 193\dfrac{19}{3}
C 19
D 12
Correct Answer
Option C
Solution
lim⁡n→∞∑r=0n−13n(2+rn)2\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{3 \over n}{{\left( {2 + {r \over n}} \right)}^2}}
=∫013(2+x)2 dx= \int_0^1 {3{{(2 + x)}^2}\,dx}
=3 . (2+x)33∣01= \left. {3\,.\,{{{{(2 + x)}^3}} \over 3}} \right|_0^1
=33−23=19= {3^3} - {2^3} = 19
Q138
If [t] denotes the greatest integer ≤t\le \mathrm{t}, then the value of 3(e−1)e∫12x2e[x]+[x3]dx{{3(e - 1)} \over e}\int\limits_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} is :
A e8−e\mathrm{e^8-e}
B e7−1\mathrm{e^7-1}
C e9−e\mathrm{e^9-e}
D e8−1\mathrm{e^8-1}
Correct Answer
Option A
Solution
I=3(e−1)e∫12x2e[x]+[x3]dxI = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx}
=3(e−1)e∫12x2e1+[x3]dx= {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{1 + [{x^3}]}}dx}

(∵\because

[x]=1[x] = 1

when

x∈(12)x \in (12)

)

=3(e−1)∫12x2e[x3]dx= 3(e - 1)\int_1^2 {{x^2}{e^{[{x^3}]}}dx}

Let

x3=t{x^3} = t
I=(e−1)∫18e[t]dtI = (e - 1)\int_1^8 {{e^{[t]}}dt}
=(e−1)(e1+e2+e3 + ... + e7)= ({e^{ - 1}})({e^1} + {e^2} + {e^3}\, + \,...\, + \,{e^7})
=(e−1)e(e7−1)e−1= (e - 1)e{{({e^7} - 1)} \over {e - 1}}
=e8−e= {e^8} - e
Q139
The value of the integral ∫12(t4+1t6+1)dt\int_1^2 {\left( {{{{t^4} + 1} \over {{t^6} + 1}}} \right)dt} is
A tan⁡−112−13tan⁡−18+π3{\tan ^{ - 1}}{1 \over 2} - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}
B tan⁡−12−13tan⁡−18+π3{\tan ^{ - 1}}2 - {1 \over 3}{\tan ^{ - 1}}8 + {\pi \over 3}
C tan⁡−12+13tan⁡−18−π3{\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}
D tan⁡−112+13tan⁡−18−π3{\tan ^{ - 1}}{1 \over 2} + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}
Correct Answer
Option C
Solution
∫12t4+1t6+1dt\int_1^2 {{{{t^4} + 1} \over {{t^6} + 1}}dt}
=∫12(t2+1)2t6+1dt−2∫12t2t6+1dt= \int_1^2 {{{{{({t^2} + 1)}^2}} \over {{t^6} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{t^6} + 1}}dt} }
=∫12t2+1t4−t2+1dt−2∫12t2(t3)2+1dt= \int_1^2 {{{{t^2} + 1} \over {{t^4} - {t^2} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{{({t^3})}^2} + 1}}dt} }
=tan⁡−1(2t+3)+tan⁡−1(2t−3)∣12−23tan⁡−1(t3)∣12= \left. {{{\tan }^{ - 1}}(2t + \sqrt 3 ) + {{\tan }^{ - 1}}(2t - \sqrt 3 )} \right|_1^2 - \left. {{2 \over 3}{{\tan }^{ - 1}}({t^3})} \right|_1^2
=tan⁡−1(4+3)+tan⁡−1(4−3)−tan⁡−1(2+3)−tan⁡−1(2+3)−tan⁡−1(23)−23(tan⁡−18−tan⁡−11)= {\tan ^{ - 1}}(4 + \sqrt 3 ) + {\tan ^{ - 1}}(4 - \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2\sqrt 3 ) - {2 \over 3}({\tan ^{ - 1}}8 - {\tan ^{ - 1}}1)
=tan⁡−12+13tan⁡−18−π3= {\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}
Q140
The value of the integral ∫1/22tan⁡−1xxdx\int\limits_{1/2}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} is equal to :
A π2log⁡e2{\pi \over 2}{\log _e}2
B π4log⁡e2{\pi \over 4}{\log _e}2
C 12log⁡e2{1 \over 2}{\log _e}2
D πlog⁡e2\pi {\log _e}2
Correct Answer
Option A
Solution
I=∫122tan⁡−1xxdxI = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx}

..... (i)

x→1xx \to {1 \over x}
I=∫1221xtan⁡−11xdxI = \int\limits_{{1 \over 2}}^2 {{1 \over x}{{\tan }^{ - 1}}{1 \over x}dx}

..... (ii)

2I=∫1221x . π2dx2I = \int\limits_{{1 \over 2}}^2 {{1 \over x}\,.\,{\pi \over 2}dx}
=π2ln⁡x∣122=πln⁡2= \left. {{\pi \over 2}\ln x} \right|_{{1 \over 2}}^2 = \pi \ln 2
⇒I=π2ln⁡2\Rightarrow I = {\pi \over 2}\ln 2
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