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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2026 · 21 January · Morning Shift · Mathematics · Inverse Trigonometric Functions

If the domain of the function f(x)= ^ -1 ((2 x-5)/(11-3 x))+ ^ -1...

If the domain of the function f(x)=cos⁡−1f(x)=\cos ^{-1}\left(\frac{2 x-5}{11-3 x}\right)+\sin ^{-1}(2x2−3x+1)\left(2 x^{2}-3 x+1\right) is the interval [α,β][\alpha, \beta], then α+2β\alpha+2 \beta is equal to :

Options

  1. A3
  2. B5
  3. C1
  4. D2

Correct answer

A. 3

Step-by-step solution

For cos⁡−1(2x−511−3x)\cos^{-1}\left(\frac{2x-5}{11-3x}\right): −1≤2x−511−3x≤1-1 \leq \frac{2x-5}{11-3x} \leq 1 gives x≤165x \leq \frac{16}{5} or x>6x > 6.
For sin⁡−1(2x2−3x+1)\sin^{-1}(2x^2 - 3x + 1): −1≤2x2−3x+1≤1-1 \leq 2x^2 - 3x + 1 \leq 1.
Lower bound always satisfied (discriminant <0< 0).
Upper bound gives x(2x−3)≤0x(2x-3) \leq 0, so 0≤x≤320 \leq x \leq \frac{3}{2}.
Intersection of both conditions: [0,32][0, \frac{3}{2}].
Thus α=0\alpha = 0, β=32\beta = \frac{3}{2}.
α+2β=0+3=3\alpha + 2\beta = 0 + 3 = 3.
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