JEE Advanced · Mathematics · 27. Definite Integration
Match the following.

- A
(i) (a), (ii) (b), (iii) (d), (iv) (c)
- B
(i) (c), (ii) (a), (iii) (b), (iv) (d)
- C
(i) (a), (ii) (d), (iii) (b), (iv) (c)
- D
(i) (b), (ii) (a), (iii) (c), (iv) (d)
Answer & Solution
Correct Answer
(C)
(i) (a), (ii) (d), (iii) (b), (iv) (c)
Step-by-step Solution
Detailed explanation
(i) \(I=\int_0^{\pi / 2}(\sin x)^{\cos x}\left(\cos x \cot x-\log (\sin x)^{\sin x}\right) d x=\int_0^{\pi / 2} \frac{d}{d x}(\sin x)^{\cos x} d x=1\)
(ii) The point of intersection of \(-4 y^2=x\) and \(x-1=-5 y^2\) is \((-4,-1)\) and \((-4,1)\).
Hence, required area \(=2\left\{\int_0^1\left(1-5 y^2\right) d y-\int_0^1-4 y^2 d y\right\}=\frac{4}{3}\).
(iii) The point intersection of \(y=3^{x-1} \log x\) and \(y=x^x-1\) is \((1,0)\).
Hence, \(\quad \frac{d y}{d x}=\frac{3^{x-1}}{x}+3^{x-1} \cdot \log 3 \cdot \log x\)
\(\therefore \quad\left(\frac{d y}{d x}\right)_{(1,0)}=1\)
For
\[
\therefore \quad\left(\frac{d y}{d x}\right)_{(1,0)}=1
\]
If \(\theta\) is angle between the curves, then \(\tan \theta=0\).
\[
\because \quad \theta=0^{\circ}
\]
(iv)
\[
\begin{aligned}
& \frac{d y}{d x}=\frac{2}{x+y} \Rightarrow \frac{d x}{d y}-\frac{x}{2}=\frac{y}{2} \\
& \Rightarrow \quad x e^{-y / 2}=\frac{1}{2} \cdot \int y \cdot e^{-y / 2} d y \\
& \Rightarrow \quad x+y+2=k e^{y / 2} \\
& \Rightarrow \quad k=3 \\
& \therefore \quad x+y+2=3 e^{y / 2} \\
&
\end{aligned}
\]
[passing through \((1,0)\) ]
(ii) The point of intersection of \(-4 y^2=x\) and \(x-1=-5 y^2\) is \((-4,-1)\) and \((-4,1)\).
Hence, required area \(=2\left\{\int_0^1\left(1-5 y^2\right) d y-\int_0^1-4 y^2 d y\right\}=\frac{4}{3}\).
(iii) The point intersection of \(y=3^{x-1} \log x\) and \(y=x^x-1\) is \((1,0)\).
Hence, \(\quad \frac{d y}{d x}=\frac{3^{x-1}}{x}+3^{x-1} \cdot \log 3 \cdot \log x\)
\(\therefore \quad\left(\frac{d y}{d x}\right)_{(1,0)}=1\)
For
\[
\therefore \quad\left(\frac{d y}{d x}\right)_{(1,0)}=1
\]
If \(\theta\) is angle between the curves, then \(\tan \theta=0\).
\[
\because \quad \theta=0^{\circ}
\]
(iv)
\[
\begin{aligned}
& \frac{d y}{d x}=\frac{2}{x+y} \Rightarrow \frac{d x}{d y}-\frac{x}{2}=\frac{y}{2} \\
& \Rightarrow \quad x e^{-y / 2}=\frac{1}{2} \cdot \int y \cdot e^{-y / 2} d y \\
& \Rightarrow \quad x+y+2=k e^{y / 2} \\
& \Rightarrow \quad k=3 \\
& \therefore \quad x+y+2=3 e^{y / 2} \\
&
\end{aligned}
\]
[passing through \((1,0)\) ]
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