JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

Let

\mathrm{f}

be an odd function defined on the set of real numbers such that for

x \geq 0, f(x)=3 \sin x+4 \cos x

.
Then

\mathrm{f}(\mathrm{x})

\mathrm{x}=-\frac{11 \pi}{6}

is equal to:

View answer and explanation

Correct option: C

\frac{3}{2}-2 \sqrt{3}

Option C (

\frac{{3}}{{2}} - 2 \sqrt{{3}}

) is correct because it is the value of the odd function f(x) at x = -11π/6, calculated using the properties of sine and cosine.

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\}