Mathematics

# If $\displaystyle \int_{0}^{a} \dfrac{dx}{\sqrt{x+a}+\sqrt{x}}=\displaystyle \int_{0}^{\pi/8} \dfrac{2 \tan \theta}{\sin 2 \theta} d\theta$, then value of $a$ is equal to $(a > 0)$

$\dfrac{9}{16}$

##### SOLUTION

$\\LHS=\int_{0}^{a}(\dfrac{1}{\sqrt{x+a}+\sqrt{x}})dx\\=\int_{0}^{a}(\dfrac{1}{\sqrt{x+a}+\sqrt{x}})\times(\dfrac{\sqrt{x+a}-\sqrt{x}}{\sqrt{x+a}-\sqrt{x}})dx\\=\int_{0}^{a}(\dfrac{\sqrt{x+a}-\sqrt{x}}{(x+a)-x})dx$

$\\=(\dfrac{1}{a})\left[\int_{0}^{a}\sqrt{x+a}dx-\int_{0}^{a}\sqrt{x}dx\right]\\=(\dfrac{1}{a})\left[(\dfrac{(x+a)^{(\frac{3}{2})}}{(\dfrac{3}{2})})\right]_{0}^{a}-(\dfrac{1}{a})\left[(\dfrac{x^{(\frac{3}{2})}}{(\dfrac{3}{2})})\right]_{0}^{a}$

$\\=(\dfrac{1}{a})\left[(2a)^{(\frac{3}{2})}-a^{(\frac{3}{2})}\right]-(\dfrac{2}{3a})\left[a^{(\frac{3}{2})}-0\right]\\=(\dfrac{2}{3a})\>a^{(\frac{3}{2})}[2\sqrt{2}-1-1]\\=(\dfrac{2}{3a})\>a^{(\frac{3}{2})}(2\sqrt{2}-2)\\=(\dfrac{4}{3a})(\sqrt{2}-1)\>a^{(\frac{3}{2})}$

$\\=(\dfrac{4}{3})(\sqrt{2}-1)\>a^{(\frac{1}{2})}$

$\\RHS=\int_{0}^{(\frac{\pi}{8})}(\dfrac{2tan\theta}{sin2\theta})d\theta\\=\int_{0}^{(\frac{\pi}{8})}(\dfrac{2sin\theta}{cos\theta\times2sin\theta\>cos\theta})d\theta\\=\int_{0}^{(\frac{\pi}{8})}sec^2\theta\>d\theta=[tan\theta]_{0}^{\pi/8}\\=tan(\dfrac{\pi}{8})=(\sqrt{2}-1)\\=as\>LHS=RHS\\\therefore\>(\dfrac{4}{3})(\sqrt{2}-1)a^{(\frac{1}{2})}=(\sqrt{2}-1)\\\therefore\>a=(\dfrac{9}{16})$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Hard
The value of $\displaystyle \int_{3}^{4}\sqrt {(4 - x)(x - 3)}dx$ is
• A. $\dfrac {\pi}{16}$
• B. $\dfrac {\pi}{4}$
• C. $\dfrac {\pi}{2}$
• D. $\dfrac {\pi}{8}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Passage Hard
If $n\rightarrow \infty$ then the limit of series in $n$ can be evaluated by following the rule : $\displaystyle \lim_{n\rightarrow \infty}\sum_{r=an+b}^{cn+d}\frac{1}{n}f\left ( \frac{r}{n} \right )=\int_{a}^{c}f(x)dx,$
where in $LHS$, $\dfrac{r}{n}$ is replaced by $x$,
$\dfrac{1}{n}$ by $dx$
and the lower and upper limits are $\lim_{n\rightarrow \infty }\dfrac{an+b}{n}\, and \, \lim_{n\rightarrow \infty }\dfrac{cn+d}{n}$ respectively.

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
$\int \dfrac {dx}{(\sqrt {1 + x^{2}} - x)^{n}}(n\neq \pm 1) = \dfrac {1}{2} \left (\dfrac {z^{n + 1}}{n + 1} + \dfrac {z^{n - 1}}{n - 1}\right ) + O$
where
• A. $z = x - \sqrt {1 + x^{2}}$
• B. $z = \sqrt {1 + x^{2}} - x$
• C. $z = x - \sqrt {1 - x^{2}}$
• D. $z = x + \sqrt {1 + x^{2}}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
$\lim _{ n\rightarrow \infty }{ \cfrac { 1 }{ \sqrt { n } \sqrt { n+1 } } +\cfrac { 1 }{ \sqrt { n } \sqrt { n+2 } } +....+\cfrac { 1 }{ \sqrt { n } \sqrt { 4n } } }$ is equal to
• A. $2$
• B. $4$
• C. $\sqrt { 2 } -1$
• D. $2\left( \sqrt { 5 } -1 \right)$

Solve $\int _ { 0 } ^ { \frac { \pi } { 2 } } \cos ^ { 2 } x d x$