JEE MATHEMATICS MCQ: FUNCTIONS

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Analyzing +4 | -1

The second degree polynomial

f\left( x \right)

, satisfying

f\left( 0 \right) = 0,\;f\left( 1 \right) = 1,\;f'\left( x \right) > 0\;for\;all\;x \in \left( {0,\;1} \right)

View answer and explanation

Correct option: C

f\left( x \right) = ax + \left( {1 - a} \right){x^2},\;a \in \left( {0,\;2} \right)

Option C (

f(x) = ax + (1 - a)x^2, a \in (0, 2)

) is correct because it satisfies the given conditions: f(0) = 0, f(1) = 1, and f'(x) > 0 for x in (0, 1).

More Questions from FUNCTIONS

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

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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\}