If a real valued function $f:[a, \infty) \rightarrow[b, \infty)$ defined by $f(x)$ $=2 x^2-3 x+5$ is a bijection, then $3 a+2 b=$

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Aicharya · Accepted Answer

$\because f:[a, \infty) ightarrow[b, \infty) 	ext { and } f(x)=2 x^2-3 x+5$ Now, $f^{\prime}(x)=4 x-3$; If $f^{\prime}(x)=0 \Rightarrow x=\frac{3}{4}$ Since, $f(x)$ is bijection function. So, $a=\frac{3}{4}$ and $f\left(\frac{3}{4}ight)=b \Rightarrow b=\frac{31}{8}$ Now, $3 a+2 b=\frac{9}{4}+\frac{31}{4}=10$.

If a real valued function $f:[a, \infty) \rightarrow[b, \infty)$ defined by $f(x)$ $=2 x^2-3 x+5$ is a…

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