Question

If f(x) = Sin(x)Cos(x) is continuous and differentiable in interval (0, x) then

(a) 1<\(\frac{Cos(x)Sin(x)}{x}\) <Sin(2x)

(b) 1<\(\frac{Cos(x)Sin(x)}{x}\) <Cos(2x)

(c) 1<\(\frac{Cos(x)Sin(x)}{x}\) <xCos(2x)

(d) 1<\(\frac{Cos(x)Sin(x)}{x}\) <1+Cos(2x)

I have been asked this question in a job interview.

My query is from Lagrange’s Mean Value Theorem in portion Differential Calculus of Engineering Mathematics

Answer 1

Right choice is (b) 1<\(\frac{Cos(x)Sin(x)}{x}\) <Cos(2x)

For explanation I would say: f(x) = sin(x)cos(x)

Given f(x) is continuous and differentiable in interval (0, x),

Applying mean value theorem in interval (0, x)

f’(c) = Cos(2c) = [f(x)-f(0)]/[x-0] = \(\frac{Cos(x)Sin(x)}{x}\) ……………………. (1)

Now, Given

0 < c < x

Multiplying by 2 and taking Cos, We get

1 < Cos(2c) < Cos(2x)

1 < \(\frac{Cos(x)Sin(x)}{x}\) < Cos(2x).