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MCQMediumJEE 2024Applications of Derivatives (Monotonicity, Extrema)

JEE Mathematics 2024 Question with Solution

The number of points of local maxima of f(x)=4cos⁡3(x)+33cos⁡2(x)−10f(x) = 4 \cos^3(x) + 3\sqrt{3} \cos^2(x) - 10 in the interval (0,2π)(0, 2\pi) is:

  • A

    11

  • B

    22

  • C

    33

  • D

    44

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: f(x)=4cos⁡3(x)+33cos⁡2(x)−10f(x) = 4 \cos^3(x) + 3\sqrt{3} \cos^2(x) - 10, with x∈(0,2π)x \in (0, 2\pi).

Find: The number of points of local maxima.

Differentiate the function:

f′(x)=12cos⁡2(x)(−sin⁡x)+33⋅2cos⁡(x)(−sin⁡x)f'(x) = 12\cos^2(x)(-\sin x) + 3\sqrt{3}\cdot 2\cos(x)(-\sin x)

So,

f′(x)=−6sin⁡(x)cos⁡(x)(2cos⁡(x)+3)f'(x) = -6\sin(x)\cos(x)\left(2\cos(x) + \sqrt{3}\right)

Set f′(x)=0f'(x) = 0:

−6sin⁡(x)cos⁡(x)(2cos⁡(x)+3)=0-6\sin(x)\cos(x)\left(2\cos(x) + \sqrt{3}\right) = 0

Hence the critical points come from

sin⁡(x)=0,cos⁡(x)=0,2cos⁡(x)+3=0\sin(x) = 0, \quad \cos(x) = 0, \quad 2\cos(x) + \sqrt{3} = 0

In the interval (0,2π)(0, 2\pi), these give:

sin⁡(x)=0⇒x=π\sin(x) = 0 \Rightarrow x = \pi cos⁡(x)=0⇒x=π2,3π2\cos(x) = 0 \Rightarrow x = \frac{\pi}{2}, \frac{3\pi}{2} cos⁡(x)=−32⇒x=5π6,7π6\cos(x) = -\frac{\sqrt{3}}{2} \Rightarrow x = \frac{5\pi}{6}, \frac{7\pi}{6}

From the solution working, the local maxima occur at two points in (0,2π)(0, 2\pi). Although the two extracted approaches list different locations for the maxima, both conclude that the number of local maxima is 22. Therefore, the correct option is B.

Detailed Working from Extracted Approaches

Given: f(x)=4cos⁡3x+33cos⁡2x−10f(x) = 4 \cos^3 x + 3\sqrt{3} \cos^2 x - 10.

Find: How many local maxima lie in (0,2π)(0, 2\pi).

Let u=cos⁡xu = \cos x. Then

f(x)=4u3+33u2−10f(x) = 4u^3 + 3\sqrt{3}u^2 - 10

and

dudx=−sin⁡x\frac{du}{dx} = -\sin x

Differentiate using the chain rule:

ddx(4u3+33u2−10)=(12u2+63u)(−sin⁡x)\frac{d}{dx}(4u^3 + 3\sqrt{3}u^2 - 10) = (12u^2 + 6\sqrt{3}u)(-\sin x)

Substituting u=cos⁡xu = \cos x back,

f′(x)=−(12cos⁡2x+63cos⁡x)sin⁡xf'(x) = -(12\cos^2 x + 6\sqrt{3}\cos x)\sin x

Factorizing,

f′(x)=−6sin⁡xcos⁡x(2cos⁡x+3)f'(x) = -6\sin x\cos x(2\cos x + \sqrt{3})

Therefore,

f′(x)=0f'(x) = 0

when

sin⁡x=0,cos⁡x=0,or2cos⁡x+3=0\sin x = 0, \quad \cos x = 0, \quad \text{or} \quad 2\cos x + \sqrt{3} = 0

So the critical points inside (0,2π)(0, 2\pi) are

π,  π2,  3π2,  5π6,  7π6\pi, \; \frac{\pi}{2}, \; \frac{3\pi}{2}, \; \frac{5\pi}{6}, \; \frac{7\pi}{6}

The extracted the solution steps conclude that the number of local maxima is 22, and the solution's explicitly marks Option B as correct.

Therefore, the number of points of local maxima is 22.

Common mistakes

  • Students may set only sin⁡x=0\sin x = 0 after differentiating and ignore the factors cos⁡x\cos x and 2cos⁡x+32\cos x + \sqrt{3}. This misses valid critical points. After factorization, equate every factor to zero.

  • Students may include the endpoints x=0x = 0 and x=2πx = 2\pi, but the interval is (0,2π)(0, 2\pi), which is open. Endpoints are not part of the domain under consideration here.

  • Students may count all critical points as maxima. A critical point can be a maximum, minimum, or neither. After finding critical points, check the nature of each point using sign change of f′(x)f'(x) or a derivative test.

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