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UPSC NDA I and II 2020 Math Solved Paper

Question : 111 of 120

Marks: +1, -0

3cos(A+\frac{\pi }{3})

where A ∈ R ?

Solution:

Concept:

The cosine function

\cos \theta

has a range from

-1

1

Multiplying by a positive constant scales this range linearly.

Explanation:

Let

f(A) = 3\cos\left(A + \frac{\pi}{3}\right)

, where

A \in \mathbb{R}

For any real angle

\theta

, we know

-1 \le \cos \theta \le 1

Here

\theta = A + \frac{\pi}{3}

takes all real values as

A

varies.

Thus

-1 \le \cos\left(A + \frac{\pi}{3}\right) \le 1

Multiply the entire inequality by

3

(positive, so direction unchanged):

-3 \le 3\cos\left(A + \frac{\pi}{3}\right) \le 3

Hence the minimum value of

f(A)

is the lower bound

-3

Answer:

-3

(Option A)

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