Splitting cases

Calculus Level 3

f ( x ) = sin ⁡ 3 ( x ) + k sin ⁡ 2 ( x ) \large f(x) = \sin^3(x) + k \sin^2(x)

For variable k k independent of x x , consider the function f f on the interval − π 2 < x < π 2 -\frac\pi2 < x < \frac\pi2 as described above. Find the sum of all integral values of k k such that f ( x ) f(x) has exactly one minimum value and one maximum value.

No solution exist -1 0 1

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Ravi Dwivedi by Ravi Dwivedi , 24, India

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1 solution

One minimum and one maximum correspond to 2 points of extremum. This means that f ′ ( x ) = 0 f'(x) = 0 should have two solutions.

f ′ ( x ) = 3 sin ⁡ 2 x cos ⁡ x + 2 k sin ⁡ x cos ⁡ x = ( sin ⁡ x ) ( cos ⁡ x ) ( 3 sin ⁡ x + 2 k ) = 0 f'(x) = 3\sin^2 x\cos x + 2k\sin x\cos x = (\sin x)(\cos x)(3\sin x + 2k) = 0

cos ⁡ x ≠ 0 \cos x \neq 0 due to the domain.

Hence, sin ⁡ x = 0 \sin x = 0 and sin ⁡ x = − 2 k 3 \sin x = -\dfrac{2k}{3}

x = 0 x = 0 is one solution

For the other solution to exist:

− 1 ≤ − 2 k 3 ≤ 1 -1 \leq -\dfrac{2k}{3} \leq 1

The integral values of k k satisfying above inequation are − 1 , 1 -1,1 ( 0 is not included since it is already a solution ).

The sum of values of k k is 0 \large \color{#3D99F6}{\boxed{0}}

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