Question

$\begin{array}{cccccc}& ColumnI& & Column2& & Column3\\ \left(I\right)& f\left(x\right)=tan\left(2ta{n}^{-1}\sqrt{\frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}}+1}}\right)& \left(i\right)& \int f\left(x\right)dx=\frac{4}{3}{x}^{\frac{3}{4}}+C& \left(P\right)& {\int }_0^{1}f\left(x\right)dx=\frac{4}{3}\\ \left(II\right)& f\left(x\right)=cot\left(2ta{n}^{-1}\sqrt{\frac{\sqrt{1+\sqrt{x}}-\sqrt[4]{4x}}{\sqrt{1+\sqrt{x}}+\sqrt[4]{4x}}}\right)& \left(ii\right)& \int f\left(x\right)dx=\frac{4}{5}{x}^{\frac{5}{4}}+C& \left(Q\right)& {\int }_0^{1}f\left(x\right)dx=\frac{4}{5}\\ \left(III\right)& f\left(x\right)\left(\frac{1-tan\left(\frac{1}{2}si{n}^{-1}\right(\frac{1-\sqrt{x}}{1+\sqrt{x}}\left)\right)}{1+tan\left(\frac{1}{2}si{n}^{-1}\right(\frac{1-\sqrt{x}}{1+\sqrt{x}}\left)\right)}\right)& \left(iii\right)& \int f\left(x\right)dx=\frac{2}{3}{x}^{\frac{3}{4}}+C& \left(R\right)& {\int }_0^{1}f\left(x\right)dx=\frac{2}{3}\\ \left(IV\right)& f\left(x\right)=\sqrt{x}tan\left(2ta{n}^{-1}\left(\frac{\sqrt{\sqrt{1+\sqrt{x}+1}}-\sqrt{\sqrt{1+\sqrt{x}-1}}}{\sqrt{\sqrt{1+\sqrt{x}+1}}+\sqrt{\sqrt{1+\sqrt{x}-1}}}\right)\right)& \left(iv\right)& \int f\left(x\right)dx=\frac{2}{5}{x}^{\frac{5}{4}}+C& \left(S\right)& {\int }_0^{1}f\left(x\right)dx=\frac{2}{5}\end{array}$

Which of the following options is only correct combination?

• A. (II), (i), (P)
• B. (II), (iv), (S)
• C. (III), (i), (P)
• D. (IV), (ii), (Q)

Answer 1

The correct option is D

(IV), (ii), (Q)

Let $\sqrt{x}=ta{n}^2\theta$
$x=ta{n}^4\theta \Rightarrow tan\theta =x^{\frac{1}{4}}\Rightarrow dx=4ta{n}^3\theta se{c}^2\theta d\theta \therefore \sqrt{1+\sqrt{x}}=sec\theta$

$\left(A\right)\int tan\left(2ta{n}^{-1}\sqrt{\frac{\sqrt{1+\sqrt{x}}-1}{\sqrt{1+\sqrt{x}+1}}}\right)dx=\int tan\theta .4ta{n}^3\theta se{c}^2\theta d\theta =\frac{4}{5}ta{n}^5\theta +C=\frac{4}{5}\left(x^{\frac{5}{4}}\right)+C$

$\left(B\right)\int cot\left(2ta{n}^{-1}\sqrt{\frac{\sqrt{1+\sqrt{x}}-\sqrt[4]{4x}}{\sqrt{1+\sqrt{x}}+\sqrt[4]{4x}}}\right)dx=\int cot\left(2ta{n}^{-1}\right)\sqrt{(sec\theta -tan\theta {)}^2}4ta{n}^3\theta se{c}^2\theta d\theta =\int cot\left(2ta{n}^{-1}\left(\frac{1-sin\theta }{cos\theta }\right)\right)4ta{n}^3\theta se{c}^2\theta d\theta \int tan\theta 4ta{n}^3\theta se{c}^2\theta d\theta =\frac{4}{5}ta{n}^5\theta +C=\frac{4}{5}{x}^{\frac{5}{4}}+C$

$\left(C\right)\int \frac{1-tan\left(\frac{1}{2}si{n}^{-1}\right(\frac{1-\sqrt{x}}{1+\sqrt{x}}\left)\right)}{1+tanx\left(\frac{1}{2}si{n}^{-1}\right(\frac{1-\sqrt{x}}{1+\sqrt{x}}\left)\right)}dx=\int \frac{1-tan(\frac{\pi }{4}-\theta )}{1+tan(\frac{\pi }{4}-\theta )}4ta{n}^3\theta se{c}^2\theta d\theta =\int tan\theta 4ta{n}^3\theta se{c}^2\theta d\theta =\frac{4}{5}ta{n}^5\theta +C=\frac{4}{5}{x}^{\frac{5}{4}}+C$

$\left(D\right)\int \sqrt{x}tan\left(2ta{n}^{-1}\left(\frac{\sqrt{\sqrt{1+\sqrt{x}+1}}-\sqrt{\sqrt{1+\sqrt{x}-1}}}{\sqrt{1+\sqrt{x}+1}+\sqrt{\sqrt{1+\sqrt{x}-1}}}\right)\right)dx=intta{n}^2\theta tan\left(2ta{n}^{-1}\left(\frac{cos\frac{\theta }{2}-sin\frac{\theta }{2}}{cos\frac{\theta }{2}+sin\frac{\theta }{2}}\right)\right)4ta{n}^3\theta se{c}^2\theta d\theta =\int ta{n}^2\theta tan(\frac{\pi }{2}-\theta )4ta{n}^3\theta se{c}^2\theta d\theta =\frac{4}{5}ta{n}^5\theta +C=\frac{4}{5}{x}^{\frac{5}{4}}+C$