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Definite Integration question

2022 · 25 Jun · Shift 2 · Q31
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  5. /2022 · 25 Jun · Shift 2 · Q31

Definite Integration question

2022 · 25 Jun · Shift 2 · Q31

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
If bn=∫0π2cos⁡2nxsin⁡xdx, n∈N{b_n} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx} \over {\sin x}}dx,\,n \in N}bn​=∫02π​​sinxcos2nx​dx,n∈N, then
  1. A
    b3−b2, b4−b3, b5−b4{b_3} - {b_2},\,{b_4} - {b_3},\,{b_5} - {b_4}b3​−b2​,b4​−b3​,b5​−b4​ are in A.P. with common difference −-− 2
  2. B
    1b3−b2,1b4−b3,1b5−b4{1 \over {{b_3} - {b_2}}},{1 \over {{b_4} - {b_3}}},{1 \over {{b_5} - {b_4}}}b3​−b2​1​,b4​−b3​1​,b5​−b4​1​ are in an A.P. with common difference 2
  3. C
    b3−b2, b4−b3, b5−b4{b_3} - {b_2},\,{b_4} - {b_3},\,{b_5} - {b_4}b3​−b2​,b4​−b3​,b5​−b4​ are in a G.P.
  4. D
    1b3−b2,1b4−b3,1b5−b4{1 \over {{b_3} - {b_2}}},{1 \over {{b_4} - {b_3}}},{1 \over {{b_5} - {b_4}}}b3​−b2​1​,b4​−b3​1​,b5​−b4​1​ are in an A.P. with common difference −-− 2
View written solutionFree

Correct answer: D

  1. Interpret the integral carefully

We are given

bn=∫0π/2cos⁡2(nx)sin⁡x dx.b_n=\int_0^{\pi/2}\frac{\cos^2(nx)}{\sin x}\,dx.bn​=∫0π/2​sinxcos2(nx)​dx.

Using

cos⁡2(nx)=1+cos⁡(2nx)2,\cos^2(nx)=\frac{1+\cos(2nx)}{2},cos2(nx)=21+cos(2nx)​,

we get

bn=12∫0π/21+cos⁡(2nx)sin⁡x dx.b_n=\frac12\int_0^{\pi/2}\frac{1+\cos(2nx)}{\sin x}\,dx.bn​=21​∫0π/2​sinx1+cos(2nx)​dx.

Now use the identity

1+cos⁡(2nx)=2cos⁡2(nx),1+\cos(2nx)=2\cos^2(nx),1+cos(2nx)=2cos2(nx),

and more importantly,

1+cos⁡(2nx)=2cos⁡(nx+0)cos⁡(nx−0),1+\cos(2nx)=2\cos(nx+0)\cos(nx-0),1+cos(2nx)=2cos(nx+0)cos(nx−0),

but a better route is to simplify the difference directly.

  1. Find a formula for bn−bn−1b_n-b_{n-1}bn​−bn−1​

Consider

bn−bn−1=∫0π/2cos⁡2(nx)−cos⁡2((n−1)x)sin⁡x dx.b_n-b_{n-1}=\int_0^{\pi/2}\frac{\cos^2(nx)-\cos^2((n-1)x)}{\sin x}\,dx.bn​−bn−1​=∫0π/2​sinxcos2(nx)−cos2((n−1)x)​dx.

Use

cos⁡2A−cos⁡2B=cos⁡2A−cos⁡2B2.\cos^2 A-\cos^2 B=\frac{\cos 2A-\cos 2B}{2}.cos2A−cos2B=2cos2A−cos2B​.

So

bn−bn−1=12∫0π/2cos⁡(2nx)−cos⁡(2n−2)xsin⁡x dx.b_n-b_{n-1}=\frac12\int_0^{\pi/2}\frac{\cos(2nx)-\cos(2n-2)x}{\sin x}\,dx.bn​−bn−1​=21​∫0π/2​sinxcos(2nx)−cos(2n−2)x​dx.

Now apply

cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2.\cos C-\cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}.cosC−cosD=−2sin2C+D​sin2C−D​.

With

C=2nx,D=2(n−1)x,C=2nx, \qquad D=2(n-1)x,C=2nx,D=2(n−1)x,

we get

cos⁡(2nx)−cos⁡(2n−2)x=−2sin⁡((2n−1)x)sin⁡x.\cos(2nx)-\cos(2n-2)x=-2\sin((2n-1)x)\sin x.cos(2nx)−cos(2n−2)x=−2sin((2n−1)x)sinx.

Therefore,

bn−bn−1=12∫0π/2−2sin⁡((2n−1)x)sin⁡xsin⁡x dx=−∫0π/2sin⁡((2n−1)x) dx.b_n-b_{n-1}=\frac12\int_0^{\pi/2}\frac{-2\sin((2n-1)x)\sin x}{\sin x}\,dx =-\int_0^{\pi/2}\sin((2n-1)x)\,dx.bn​−bn−1​=21​∫0π/2​sinx−2sin((2n−1)x)sinx​dx=−∫0π/2​sin((2n−1)x)dx.

Hence

bn−bn−1=−[−cos⁡((2n−1)x)2n−1]0π/2.b_n-b_{n-1}=-\left[\frac{-\cos((2n-1)x)}{2n-1}\right]_0^{\pi/2}.bn​−bn−1​=−[2n−1−cos((2n−1)x)​]0π/2​.

So

bn−bn−1=cos⁡((2n−1)π2)−12n−1.b_n-b_{n-1}=\frac{\cos\left((2n-1)\frac{\pi}{2}\right)-1}{2n-1}.bn​−bn−1​=2n−1cos((2n−1)2π​)−1​.

Since 2n−12n-12n−1 is odd,

cos⁡((2n−1)π2)=0.\cos\left((2n-1)\frac{\pi}{2}\right)=0.cos((2n−1)2π​)=0.

Thus

bn−bn−1=−12n−1.b_n-b_{n-1}=-\frac{1}{2n-1}.bn​−bn−1​=−2n−11​.
  1. Compute the required consecutive differences

For n=3n=3n=3,

b3−b2=−15.b_3-b_2=-\frac{1}{5}.b3​−b2​=−51​.

For n=4n=4n=4,

b4−b3=−17.b_4-b_3=-\frac{1}{7}.b4​−b3​=−71​.

For n=5n=5n=5,

b5−b4=−19.b_5-b_4=-\frac{1}{9}.b5​−b4​=−91​.

So the three numbers are

−15, −17, −19.-\frac15, \,-\frac17, \,-\frac19.−51​,−71​,−91​.
  1. Check each option

Option A

Claims these are in A.P. with common difference −2-2−2.

But

−17−(−15)=235,-\frac17-\left(-\frac15\right)=\frac{2}{35},−71​−(−51​)=352​,

not −2-2−2. So A is false.

Option B

Take reciprocals:

1b3−b2=−5,1b4−b3=−7,1b5−b4=−9.\frac{1}{b_3-b_2}=-5, \quad \frac{1}{b_4-b_3}=-7, \quad \frac{1}{b_5-b_4}=-9.b3​−b2​1​=−5,b4​−b3​1​=−7,b5​−b4​1​=−9.

These are in A.P. with common difference

−7−(−5)=−2,-7-(-5)=-2,−7−(−5)=−2,

not 222. So B is false.

Option C

Check G.P. for

−15,−17,−19.-\frac15, -\frac17, -\frac19.−51​,−71​,−91​.

Ratios are

−1/7−1/5=57,−1/9−1/7=79,\frac{-1/7}{-1/5}=\frac57, \qquad \frac{-1/9}{-1/7}=\frac79,−1/5−1/7​=75​,−1/7−1/9​=97​,

not equal. So C is false.

Option D

Reciprocals are

−5,−7,−9,-5,-7,-9,−5,−7,−9,

which are in A.P. with common difference −2-2−2. So D is true.

  1. Final answer

The correct option is D.\boxed{D}.D​.

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