Published by:
CGP EDU Academic Team
Published on: August 13, 2026
Find the set of values of k for which x 2 – kx + sin –1 (sin 4) > 0 for all real x.
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
We know,
sin –1 (sin 4) = sin –1 (sin ( π – 4)) = π – 4 
∴ We have x 2 – kx + π – 4 > 0 for all x ∈ R
∴ D < 0, i.e. k 2 – 4( π –4) < 0
or k 2 + 4 (4 – π ) < 0
which is not true for any real k, {as k 2 + 4 (4– π ) > 0}
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