JEE MATHEMATICS MCQ: FUNCTIONS

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Application +4 | -1

f: R \rightarrow S

defined by

f(x)=\sin x-\sqrt{3} \cos x+1

is onto, then the interval of

S

View answer and explanation

Correct option: A

[-1,3]

Option A ([-1,3]) is correct because the range of the function f(x) = sin(x) - √3cos(x) + 1 is [-1, 3] when it's onto.

More Questions from FUNCTIONS

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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\}