Question

Let
f(x) = ab.sin(3x) + b/1 - a² cos(3x)
where 0 < a < 1, then maximum f(x)​

Answer 1

Answer:

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Answer 2

Given:

A function f(x) = ab.sin(3x) + b/1 - a² cos(3x) .

To find:

Maximum value of f(x) i.e.  f max.

Solution:

Maximum value of Sin x =1

Also, maximum value of cos x =1

But, f(x) is of the form p.sin x + q.cos x, In such cases maximum value of function is $\sqrt{p^{2}+q^{2} }$

=> F max = $\sqrt{(ab)^{2}+(\frac{b}{1-a^{2} }) ^{2} }$

Hence, maximum  value of f(x) is $\sqrt{(ab)^{2}+(\frac{b}{1-a^{2} }) ^{2} }$.