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JEE Main 20199 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2019 · 9 April · Evening Shift · Mathematics · Indefinite Integration

If ∫ e ^ x ( x xf(x) + ( x x + se c ^ 2 x))dx = e ^ x f(x) + C,...

If ∫esec⁡x(sec⁡xtan⁡xf(x)+(sec⁡xtan⁡x+sec2x))dx=esec⁡xf(x)+C,∫ {e}^{\sec x}\left(\sec x \tan xf\left(x\right) + \left(\sec x \tan x + se{c}^{2}x\right)\right)dx = {e}^{\sec x}f\left(x\right) + C, then a possible choice of f(x)f\left(x\right) is:

Options

  1. Asec⁡x−tan⁡x−12\sec x - \tan x - \dfrac{1}{2}
  2. Bsec⁡x+tan⁡x+12\sec x + \tan x + \dfrac{1}{2}
  3. Cxsecx+tan⁡x+12xsecx + \tan x + \dfrac{1}{2}
  4. Dsec⁡x+xtanx−12\sec x + xtanx - \dfrac{1}{2}

Correct answer

B. sec⁡x+tan⁡x+12\sec x + \tan x + \dfrac{1}{2}

Step-by-step solution

∫esecx(secxtanxf(x)+(secxtanx+sec⁡2x)dx=esecxf(x)+C∫ {e}^{sec x} ( sec xtan xf\left(x\right) + \left(sec xtan x + \sec^{2} x\right)dx = {e}^{sec x}f\left(x\right) + C

Differentiating both sides w.r.t ‘x’‘x’ we get

esec⁡x(sec⁡xtan⁡xf(x)+(sec⁡xtan⁡x+sec⁡2x))=(esec⁡x⋅sec⁡xtan⁡x)f(x)+esec⁡xf′(x){e}^{\sec x}\left(\sec x \tan xf\left(x\right) + \left(\sec x \tan x + \sec^{2}x\right)\right) = \left({e}^{\sec x} ⋅ \sec x \tan x\right)f ( x ) + {e}^{\sec x}{f}^{'}\left(x\right)

⇒f′(x)=sec⁡2x+tan⁡xsec⁡x⇒ {f}^{'}\left(x\right) = \sec^{2}x + \tan x \sec x

⇒f(x)=∫(sec⁡2x+tan⁡xsec⁡x)dx⇒ f\left(x\right) = ∫ \left(\sec^{2}x + \tan x \sec x\right)dx

⇒f(x)=tan⁡x+sec⁡x+c,c∈R⇒ f\left(x\right) = \tan x + \sec x + c, c \in R
Hence, possible choice is f(x)=sec⁡x+tan⁡x+12.f\left(x\right) = \sec x + \tan x + \dfrac{1}{2} .

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