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Definite Integration question

2021 · 1 Sep · Shift 2 · Q24
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  5. /2021 · 1 Sep · Shift 2 · Q24

Definite Integration question

2021 · 1 Sep · Shift 2 · Q24

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f : R →\to→ R be a continuous function. Then lim⁡x→π4π4∫2sec⁡2xf(x) dxx2−π216\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}x→4π​lim​x2−16π2​4π​2∫sec2x​f(x)dx​ is equal to :
  1. A
    f (2)
  2. B
    2f (2)
  3. C
    2f (2)\left( {\sqrt 2 } \right)(2​)
  4. D
    4f (2)
View written solutionFree

Correct answer: B

  1. Interpret the integral carefully

The expression is

lim⁡x→π/4π4∫2sec⁡2xf(t) dtx2−π2/16.\lim_{x\to \pi/4} \frac{\frac{\pi}{4}\int_{2}^{\sec^2 x} f(t)\,dt}{x^2-\pi^2/16}.x→π/4lim​x2−π2/164π​∫2sec2x​f(t)dt​.

Since the integration variable is dummy, I wrote it as ttt instead of xxx to avoid confusion.

So we need to evaluate

L=lim⁡x→π/4π4∫2sec⁡2xf(t) dtx2−(π/4)2.L=\lim_{x\to \pi/4} \frac{\frac{\pi}{4}\int_{2}^{\sec^2 x} f(t)\,dt}{x^2-(\pi/4)^2}.L=x→π/4lim​x2−(π/4)24π​∫2sec2x​f(t)dt​.
  1. Check the form of the limit

As x→π/4x\to \pi/4x→π/4,

sec⁡2(π4)=2.\sec^2\left(\frac\pi4\right)=2.sec2(4π​)=2.

Hence

∫2sec⁡2xf(t) dt→∫22f(t) dt=0.\int_2^{\sec^2 x} f(t)\,dt \to \int_2^2 f(t)\,dt =0.∫2sec2x​f(t)dt→∫22​f(t)dt=0.

Also,

x2−π216→0.x^2-\frac{\pi^2}{16} \to 0.x2−16π2​→0.

So the limit is of the form 00\frac0000​.

  1. Apply L'Hospital's Rule

Differentiate numerator and denominator with respect to xxx.

Numerator

Let

N(x)=π4∫2sec⁡2xf(t) dt.N(x)=\frac\pi4\int_2^{\sec^2 x} f(t)\,dt.N(x)=4π​∫2sec2x​f(t)dt.

By the Fundamental Theorem of Calculus and chain rule,

N′(x)=π4 f(sec⁡2x)⋅ddx(sec⁡2x).N'(x)=\frac\pi4\,f(\sec^2 x)\cdot \frac{d}{dx}(\sec^2 x).N′(x)=4π​f(sec2x)⋅dxd​(sec2x).

Now,

ddx(sec⁡2x)=2sec⁡2xtan⁡x.\frac{d}{dx}(\sec^2 x)=2\sec^2 x\tan x.dxd​(sec2x)=2sec2xtanx.

Thus

N′(x)=π4 f(sec⁡2x)⋅2sec⁡2xtan⁡x.N'(x)=\frac\pi4\,f(\sec^2 x)\cdot 2\sec^2 x\tan x.N′(x)=4π​f(sec2x)⋅2sec2xtanx.

Denominator

Let

D(x)=x2−π216.D(x)=x^2-\frac{\pi^2}{16}.D(x)=x2−16π2​.

Then

D′(x)=2x.D'(x)=2x.D′(x)=2x.

Therefore,

L=lim⁡x→π/4π4 f(sec⁡2x)⋅2sec⁡2xtan⁡x2x.L=\lim_{x\to \pi/4} \frac{\frac\pi4\,f(\sec^2 x)\cdot 2\sec^2 x\tan x}{2x}.L=x→π/4lim​2x4π​f(sec2x)⋅2sec2xtanx​.
  1. Substitute x=π/4x=\pi/4x=π/4

Using

sec⁡2(π4)=2,tan⁡(π4)=1,2x∣x=π/4=π2,\sec^2\left(\frac\pi4\right)=2, \qquad \tan\left(\frac\pi4\right)=1, \qquad 2x\Big|_{x=\pi/4}=\frac\pi2,sec2(4π​)=2,tan(4π​)=1,2x​x=π/4​=2π​,

we get

L=π4⋅f(2)⋅2⋅2⋅1π/2.L=\frac{\frac\pi4\cdot f(2)\cdot 2\cdot 2\cdot 1}{\pi/2}.L=π/24π​⋅f(2)⋅2⋅2⋅1​.

Simplify the numerator:

π4⋅4f(2)=πf(2).\frac\pi4\cdot 4f(2)=\pi f(2).4π​⋅4f(2)=πf(2).

So

L=πf(2)π/2=2f(2).L=\frac{\pi f(2)}{\pi/2}=2f(2).L=π/2πf(2)​=2f(2).
  1. Match with the options

Hence the correct option is

B: 2f(2).\boxed{\text{B: }2f(2)}.B: 2f(2)​.
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