The number of integer values of k, for which the equation 2sinx = 2k + 1 has a solution, is

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NDA (Held On: 18 Apr 2021) Maths Previous Year paper

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Answer 1

Concept:

The minimum and maximum value of a sin x + b cos x

-\(\rm \sqrt{a^{2} + b^{2}}\) ≤ a sin x + b cos x ≤ \(\rm \sqrt{a^{2} + b^{2}}\)

Calculation:

As we know, -\(\rm \sqrt{a^{2} + b^{2}}\) ≤ a sin x + b cos x ≤ \(\rm \sqrt{a^{2} + b^{2}}\)

⇒ -\(\rm \sqrt{2^{2} + 0^{2}}\) ≤ 2sin x + 0 cos x ≤ \(\rm \sqrt{2^{2} + 0^{2}}\)

⇒ -2 ≤ 2sin x ≤ 2

⇒ -2 ≤ 2k + 1 ≤ 2

⇒ -3 ≤ 2k ≤ 1

⇒ -1.5 ≤ k ≤ 0.5

k = 0, -1

Hence, Option 3 is correct.