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JEE Main 201815 AprilEvening ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Evening Shift
Mode
Online
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main 2018 · 15 April · Evening Shift · Online · Mathematics · Indefinite Integration

∫ 2 x+5 7-6 x-x^2 d x=A 7-6 x-x^2 + B ^ -1 ((x+3)/(4))+ C (where C...

∫2x+57−6x−x2\int \frac{2 x+5}{\sqrt{7-6 x-x^2}}d x=A 7−6x−x2+\sqrt{7-6 x-x^2}+B \sin ^{-1}(x+34)+\left(\frac{x+3}{4}\right)+C (where CC is a constant of integration), then the ordered pair (A,B)(A, B) is equal to

Options

  1. A(−2,−1)(-2,-1)
  2. B(2,−1)(2,-1)
  3. C(−2,1)(-2,1)
  4. D(2,1)(2,1)

Correct answer

A. (−2,−1)(-2,-1)

Step-by-step solution

∵7−6x−x2=16−(x+3)2\because 7-6 x-x^2=16-(x+3)^2 and ddx(7−6x−x2)=−2x−6\frac{d}{d x}\left(7-6 x-x^2\right)=-2 x-6 So, ∫2x+57−6x−x2dx=∫2x+67−6x−x2dx\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=\int \frac{2 x+6}{\sqrt{7-6 x-x^2}} d x -∫116−(x+3)2\int \frac{1}{\sqrt{16-(x+3)^2}}d x =-2 7−6x−x2−sin⁡−1(x+34)+\sqrt{7-6 x-x^2}-\sin ^{-1}\left(\frac{x+3}{4}\right)+C Therefore, A=−2,&B=−1A=-2, \& B=-1.
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