The minimum value of 3^{ x}+3^{ x} is

Correct Answer is : 2(3−12)2≤ft(3^{{-1}{{2}}})2(32−1)

Correct Answer is : 2(3−12)2≤ft(3^{{-1}{{2}}})2(32−1)

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Indian Institute of Foreign Trade 2010 Solved Paper

Question : 16 of 135

Marks: +1, -0

3^{\sin x}+3^{\cos x}

Solution:

Using AM ≥.GM, We can say that

\frac{3^{\sin x}+3^{\cos x}}{2} \ge \sqrt{3^{\sin x} \times 3^{\cos x}}

\Rightarrow 3^{\sin x}+3^{\cos x} \ge 2\times 3^{\frac{\sin x+\cos x}{2}}

.....(1)

We know that -

\sqrt{A^2+B^2} \le A\sin x+B\cos x \le \sqrt{A^2+B^2}

Therefore,

-\sqrt{1^2+1^2} \le \sin x+\cos x \le \sqrt{1^2+1^2}

\Rightarrow -\sqrt{2} \le \sin x+\cos x

Hence, we can say that the minimum value of

\sin x+\cos x = -\sqrt{2}

...(2)

From equation (1) and (2) we can say that,

\Rightarrow 3^{\sin x}+3^{\cos x} \ge .2 \times 3^{\frac{-\sqrt{2}}{2}}

\Rightarrow 3^{\sin x}+3^{\cos x} \ge .2 \times 3^{\frac{-\sqrt{1}}{2}}

Therefore, option B is the correct answer.

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