JEE Main 2023MathematicsInverse Trigonometric FunctionsHardMCQ

JEE Main 2023 — Inverse Trigonometric Functions Question with Solution

JEE Main 2023 (31 Jan Shift 2)

Question

Let a,b⊂0,2π be the largest interval for which sin-1sinθ-cos-1sinθ>0,θ∈0,2π, holds . If αx2+βx+sin-1x2-6x+10+cos-1x2-6x+10=0 and α-β=b-a, then α is equal to;

Choose an option

Show full solutionCorrect option: D
Correct answer
Dπ12

Step-by-step explanation

Given, sin-1sinθ-cos-1sinθ>0,θ∈0,2π

∵sin-1θ+cos-1θ=π2

∴sin-1sinθ-π2-sin-1sinθ>0

sin-1sinθ>π4⇒θ∈π4,3π4

a,b=π4,3π4⇒b-a=π2

Given b-a=α-β

∴ α-β=π2  .....(1)

Now αx2+βx+sin-1x2-6x+10+cos-1x2-6x+10=0

Now defining x2-6x+10=1+x-32≥1

Hence x=3 is the only possible solution

9α+3β+π2=0       …2

On solving equations 1 and 2 we get,

α=π12

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About this question

This is a previous-year question from JEE Main 2023, covering the Inverse Trigonometric Functions chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.