JEE Advanced 2009 Paper 2
Exam JEE Advanced
Year 2009
Paper Paper 2
Subject Mathematics
Chapter Definite Integration JEE Advanced 2009 · Paper 2 · Mathematics · Definite Integration
If I_n=∫_ -π ^π ( n x)/((1+π^x) x) d x ; n=0,1,2 , , then
If
I n = ∫ − π π sin n x ( 1 + π x ) sin x d x ; n = 0 , 1 , 2 I_n=\int_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x} d x ; n=0,1,2 I n = ∫ − π π ( 1 + π x ) s i n x s i n n x d x ; n = 0 , 1 , 2 ,
… \ldots … , then
Options A I n = I n + 2 I_n=I_{n+2} I n = I n + 2 B ∑ m = 1 10 I 2 m + 1 = 10 π \sum_{m=1}^{10} I_{2 m+1}=10 \pi ∑ m = 1 10 I 2 m + 1 = 10 π C ∑ m = 1 10 I 2 m = 0 \sum_{m=1}^{10} I_{2 m}=0 ∑ m = 1 10 I 2 m = 0 D I n = I n + 1 I_n=I_{n+1} I n = I n + 1 Correct answer A. I n = I n + 2 I_n=I_{n+2} I n = I n + 2
Step-by-step solution
(a)
I n = ∫ − π π sin n x ( 1 + π x ) sin x d x I n = ∫ − π π π x sin n x ( 1 + π x ) sin x d x [ ∵ ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x ] 2 I n = ∫ − π π sin n x sin x d x \begin{aligned}& I_n=\int_{-\pi}^\pi \frac{\sin n x}{\left(1+\pi^x\right) \sin x}d x \\
& I_n=\int_{-\pi}^\pi \frac{\pi^x \sin n x}{\left(1+\pi^x\right) \sin x}d x \\
& {\left[\because \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\right]} \\
& 2 I_n=\int_{-\pi}^\pi \frac{\sin n x}{\sin x}d x
\end{aligned} I n = ∫ − π π ( 1 + π x ) sin x sin n x d x I n = ∫ − π π ( 1 + π x ) sin x π x sin n x d x [ ∵ ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x ] 2 I n = ∫ − π π sin x sin n x d x ⇒ 2 I n = 2 ∫ 0 π sin n x sin x d x ⇒ I n = ∫ 0 π sin n x sin x d x Now, \quadI n + 2 − I n = ∫ 0 π sin ( n + 2 ) x − sin n x sin x d x = ∫ 0 π 2 cos ( n + 1 ) x sin x sin x d x = 2 [ sin ( n + 1 ) x ( n + 1 ) ] 0 π = 0 ⇒ I n + 2 = I n (b) As I 3 = I 5 = … = I 21 ∴ ∑ m = 1 10 I 2 m + 1 = 10 I 3 = 10 ∫ 0 π sin 3 x sin x d x = 10 ∫ 0 π ( 3 − 4 sin 2 x ) d x = 10 [ 3 x − 2 x + sin 2 x ] 0 π = 10 π (c) As \quadI 2 = I 4 = … = I 20 ∴ ∑ m = 1 10 I 2 m = 10 ∫ 0 π sin 2 x sin x d x = 20 [ sin x ] 0 π = 0 \begin{aligned}& \Rightarrow \quad 2 I_n=2 \int_0^\pi \frac{\sin n x}{\sin x}d x \\
& \Rightarrow \quad I_n=\int_0^\pi \frac{\sin n x}{\sin x}d x \\
& \text{ Now, } \quadI_{n+2}-I_n \\
& =\int_0^\pi \frac{\sin (n+2) x-\sin n x}{\sin x}d x \\
& =\int_0^\pi \frac{2 \cos (n+1) x \sin x}{\sin x}d x \\
& =2\left[\frac{\sin (n+1) x}{(n+1)}\right]_0^\pi=0 \\
& \Rightarrow \quad I_{n+2}=I_n \\
& \text{ (b) As }I_3=I_5=\ldots=I_{21} \\
& \therefore \sum_{m=1}^{10} I_{2 m+1}=10 I_3=10 \int_0^\pi \frac{\sin 3 x}{\sin x}d x \\
& =10 \int_0^\pi\left(3-4 \sin ^2 x\right) d x \\
& =10[3 x-2 x+\sin 2 x]_0^\pi=10 \pi\\
& \text{ (c) As } \quadI_2=I_4=\ldots=I_{20} \\
& \therefore \quad \sum_{m=1}^{10} I_{2 m}=10 \int_0^\pi \frac{\sin 2 x}{\sin x}d x \\
& =20[\sin x]_0^\pi=0 \\
&
\end{aligned} ⇒ 2 I n = 2 ∫ 0 π sin x sin n x d x ⇒ I n = ∫ 0 π sin x sin n x d x Now, \quadI n + 2 − I n = ∫ 0 π sin x sin ( n + 2 ) x − sin n x d x = ∫ 0 π sin x 2 cos ( n + 1 ) x sin x d x = 2 [ ( n + 1 ) sin ( n + 1 ) x ] 0 π = 0 ⇒ I n + 2 = I n (b) As I 3 = I 5 = … = I 21 ∴ m = 1 ∑ 10 I 2 m + 1 = 10 I 3 = 10 ∫ 0 π sin x sin 3 x d x = 10 ∫ 0 π ( 3 − 4 sin 2 x ) d x = 10 [ 3 x − 2 x + sin 2 x ] 0 π = 10 π (c) As \quadI 2 = I 4 = … = I 20 ∴ m = 1 ∑ 10 I 2 m = 10 ∫ 0 π sin x sin 2 x d x = 20 [ sin x ] 0 π = 0
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