JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

Let

f(x)=\frac{\alpha x^{2}}{x+1}, x \neq-1

. The value of

\alpha

for which

f(a)=a,(a \neq 0)

View answer and explanation

Correct option: C

1+\frac{1}{a}

Option C (

{ }^8 {C}_4 \cdot 3

) is correct because it selects 4 people out of 8 and then considers the 3 possible pairings within that group of 4.

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

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