(a) Find the least value of ' a ' so that f(x)=2 x^2-a x+3 is an increasing function on [2,4].

\[\begin{aligned} & f(x)=2 x^{2}-a x+3 \Rightarrow f^{\prime}(x)=4 x-a \\ & \text { Now } 2 \leq x \leq 4 \Rightarrow 8-a \leq 4 x-a \leq 16-a \\ & \text { For } f \text { to be an increasing function, } f^{\prime}(x) \geq 0 \\ & \Rightarrow 8-a \geq 0 \Rightarrow a \leq 8 \\ & \therefore \text { Least value of } a \text { does not exist. } \end{aligned} \] \[\begin{aligned} & f(x)=x+\frac{1}{x} \Rightarrow f^{\prime}(x)=1-\frac{1}{x^{2}}=\frac{x^{2}-1}{x^{2}} \\ & \text { Now } \frac{x^{2}-1}{x^{2}} \geq 0 \text { for all } x \geq 1 \\ & \Rightarrow f^{\prime}(x) \geq 0 \Rightarrow f \text { is an increasing function. } \end{aligned} \]