What is the maximum value of 16 θ - 12 ² θ?

Correct Answer is : 16/316/316/3

Correct Answer is : 16/316/316/3

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UPSC NDA I 2018 Mathematics Solved Paper

Question : 24 of 120

Marks: +1, -0

What is the maximum value of 16 \sin \theta - 12 \sin^2 \theta?

Solution:

Concept:

The maximum value of a quadratic function

f(t) = at^2 + bt + c

with

a < 0

occurs at

t = -\frac{b}{2a}

. Here, treat

\sin\theta

as the variable

t

Explanation:

Let

f(\theta) = 16\sin\theta - 12\sin^2\theta

Let

t = \sin\theta

, where

t \in [-1, 1]

Then

f(t) = 16t - 12t^2 = -12t^2 + 16t + 0

This is a downward parabola (

-12 < 0

), so its maximum is at the vertex

t = -\frac{b}{2a}

. Here,

a = -12

b = 16

t = -\frac{16}{2(-12)} = \frac{16}{24} = \frac{2}{3}

Since

\frac{2}{3}

lies in

[-1, 1]

, the maximum value is

f\!\left(\frac{2}{3}\right) = 16\left(\frac{2}{3}\right) - 12\left(\frac{2}{3}\right)^2 = \frac{32}{3} - 12 \cdot \frac{4}{9} = \frac{32}{3} - \frac{48}{9}

Convert to common denominator:

\frac{32}{3} = \frac{96}{9}

, so

f_{\text{max}} = \frac{96}{9} - \frac{48}{9} = \frac{48}{9} = \frac{16}{3}

Answer:

\frac{16}{3}

(Option C)

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