Question

If $f:[0,5] \rightarrow[7,13]$ be a surjective linear map such that $f(1)>f(2)$ and $f \left(2 \alpha^2-\alpha-1\right)+13= f \left(\alpha^2+1\right)+ f (0)$ for some $\alpha \in[0,5]$, then value of $f ^{-1}(\alpha+5)$ equals

• A. 1
• B. 3
• C. 0
• D. 5

Answer 1

Answer: (D)

Clearly $f$ is one-one & decreasing $\Rightarrow f (5)=7, f (0)=13$
$\therefore f \left(2 \alpha^2-\alpha-1\right)= f \left(\alpha^2+1\right) \Rightarrow \alpha=-1,2$
$\text { since } \alpha \in[0,5] \Rightarrow \alpha=2 $
$\Rightarrow f ^{-1}(\alpha+5)= f ^{-1}(7)=5$