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in Trigonometry by (41.3k points)

The number of integral values of k for which the equation sin-1x + tan-1x = sin-1sinx + 2k - 1 has a real solution is

(A) 1

(B) 3

(C) 2

(D) 4

1 Answer

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Best answer

Correct option  (D) 4

Explanation: 

sin-1x + tan-1x - sin-1 sin x = 2k - 1

The range of sin-1x + tan-1x - sin-1sinx is

[-3π/4 - 1, 3π/4 + 1].

Then

-3π/4 - 1 ≤ 2k - 1 ≤ 3π/4 + 1

 - 3π/8 ≤ k ≤ 3π/8 + 1

Hence, the integral values of k are -1, 0, 1, 2.

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