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Definite Integration question

2022 · 28 Jun · Shift 2 · Q28
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  5. /2022 · 28 Jun · Shift 2 · Q28

Definite Integration question

2022 · 28 Jun · Shift 2 · Q28

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
Let f : R →\to→ R be a differentiable function such that f(π4)=2, f(π2)=0f\left( {{\pi \over 4}} \right) = \sqrt 2 ,\,f\left( {{\pi \over 2}} \right) = 0f(4π​)=2​,f(2π​)=0 and f′(π2)=1f'\left( {{\pi \over 2}} \right) = 1f′(2π​)=1 and let g(x)=∫xπ/4(f′(t)sec⁡t+tan⁡tsec⁡t f(t)) dtg(x) = \int_x^{\pi /4} {(f'(t)\sec t + \tan t\sec t\,f(t))\,dt}g(x)=∫xπ/4​(f′(t)sect+tantsectf(t))dt for x∈[π4,π2)x \in \left[ {{\pi \over 4},{\pi \over 2}} \right)x∈[4π​,2π​). Then lim⁡x→(π2)−g(x)\mathop {\lim }\limits_{x \to {{\left( {{\pi \over 2}} \right)}^ - }} g(x)x→(2π​)−lim​g(x) is equal to :
  1. A
    2
  2. B
    3
  3. C
    4
  4. D
    −-− 3
View written solutionFree

Correct answer: B

  1. Recognize the integrand as a derivative

We are given

g(x)=∫xπ/4(f′(t)sec⁡t+tan⁡tsec⁡t f(t))dt.g(x)=\int_x^{\pi/4}\left(f'(t)\sec t+\tan t\sec t\,f(t)\right)dt.g(x)=∫xπ/4​(f′(t)sect+tantsectf(t))dt.

Notice that

ddt(f(t)sec⁡t)=f′(t)sec⁡t+f(t)⋅(sec⁡ttan⁡t).\frac{d}{dt}\big(f(t)\sec t\big)=f'(t)\sec t+f(t)\cdot(\sec t\tan t).dtd​(f(t)sect)=f′(t)sect+f(t)⋅(secttant).

So the integrand is exactly

ddt(f(t)sec⁡t).\frac{d}{dt}\big(f(t)\sec t\big).dtd​(f(t)sect).

Hence,

g(x)=∫xπ/4ddt(f(t)sec⁡t) dt.g(x)=\int_x^{\pi/4} \frac{d}{dt}\big(f(t)\sec t\big)\,dt.g(x)=∫xπ/4​dtd​(f(t)sect)dt.
  1. Evaluate the integral using the Fundamental Theorem of Calculus

Thus,

g(x)=[f(t)sec⁡t]t=xt=π/4=f(π4)sec⁡(π4)−f(x)sec⁡x.g(x)=\big[f(t)\sec t\big]_{t=x}^{t=\pi/4} = f\left(\frac{\pi}{4}\right)\sec\left(\frac{\pi}{4}\right)-f(x)\sec x.g(x)=[f(t)sect]t=xt=π/4​=f(4π​)sec(4π​)−f(x)secx.

Given

f(π4)=2,sec⁡(π4)=2,f\left(\frac{\pi}{4}\right)=\sqrt{2}, \qquad \sec\left(\frac{\pi}{4}\right)=\sqrt{2},f(4π​)=2​,sec(4π​)=2​,

so

f(π4)sec⁡(π4)=2⋅2=2.f\left(\frac{\pi}{4}\right)\sec\left(\frac{\pi}{4}\right)=\sqrt{2}\cdot \sqrt{2}=2.f(4π​)sec(4π​)=2​⋅2​=2.

Therefore,

g(x)=2−f(x)sec⁡x.g(x)=2-f(x)\sec x.g(x)=2−f(x)secx.

So we need

lim⁡x→(π/2)−g(x)=2−lim⁡x→(π/2)−f(x)sec⁡x.\lim_{x\to (\pi/2)^-} g(x)=2-\lim_{x\to (\pi/2)^-} f(x)\sec x.x→(π/2)−lim​g(x)=2−x→(π/2)−lim​f(x)secx.
  1. Find lim⁡x→(π/2)−f(x)sec⁡x\lim_{x\to (\pi/2)^-} f(x)\sec xlimx→(π/2)−​f(x)secx

As x→π2−x\to \frac{\pi}{2}^-x→2π​−,

f(π2)=0,cos⁡(π2)=0,f\left(\frac{\pi}{2}\right)=0, \qquad \cos\left(\frac{\pi}{2}\right)=0,f(2π​)=0,cos(2π​)=0,

so

f(x)sec⁡x=f(x)cos⁡xf(x)\sec x=\frac{f(x)}{\cos x}f(x)secx=cosxf(x)​

is of the indeterminate form 00\frac{0}{0}00​.

Apply L'Hospital's Rule:

lim⁡x→(π/2)−f(x)cos⁡x=lim⁡x→(π/2)−f′(x)−sin⁡x.\lim_{x\to (\pi/2)^-}\frac{f(x)}{\cos x} =\lim_{x\to (\pi/2)^-}\frac{f'(x)}{-\sin x}.x→(π/2)−lim​cosxf(x)​=x→(π/2)−lim​−sinxf′(x)​.

Given

f′(π2)=1,f'\left(\frac{\pi}{2}\right)=1,f′(2π​)=1,

and

sin⁡(π2)=1,\sin\left(\frac{\pi}{2}\right)=1,sin(2π​)=1,

therefore

lim⁡x→(π/2)−f(x)cos⁡x=1−1=−1.\lim_{x\to (\pi/2)^-} \frac{f(x)}{\cos x} =\frac{1}{-1}=-1.x→(π/2)−lim​cosxf(x)​=−11​=−1.

Hence,

lim⁡x→(π/2)−f(x)sec⁡x=−1.\lim_{x\to (\pi/2)^-} f(x)\sec x=-1.x→(π/2)−lim​f(x)secx=−1.
  1. Compute the required limit

So,

lim⁡x→(π/2)−g(x)=2−(−1)=3.\lim_{x\to (\pi/2)^-} g(x)=2-(-1)=3.x→(π/2)−lim​g(x)=2−(−1)=3.
  1. Check options
  • A: 222 ❌
  • B: 333 ✅
  • C: 444 ❌
  • D: −3-3−3 ❌

Therefore, the correct answer is

3.\boxed{3}.3​.
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