JEE MATHEMATICS MCQ: FUNCTIONS

1 Medium

1 of 1

Application +4 | -1

f(x)=\cos (\log x)

, then the value of

f(x) \cdot f(4)-\frac{1}{2}\left[f\left(\frac{x}{4}\right)+f(4 x)\right]

View answer and explanation

Correct option: C

The correct answer is 0 because when substituting

f(x) = \cos(\log x)

into the expression and simplifying using trigonometric identities, the result is 0.

More Questions from FUNCTIONS

Practice All Chapter MCQs →

f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

g(f(x))=|\sin x|

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}