NIMCET 2017 Question Paper with solutions

let f (x) = 4 $\mathit{s}\mathit{i}{\mathit{n}}^2$ x + 3 $\mathit{c}\mathit{o}{\mathit{s}}^2$ x + sin (x/2)+cos or (x/2)
= 3 + $\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{x}+\mathit{s}\mathit{i}\mathit{n}\frac{\mathit{x}}{2}+\mathit{c}\mathit{o}\mathit{s}\frac{\mathit{x}}{2}$
= 3 + $\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{x}+\sqrt{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}$
∴ f'(x) = 2sinxcosx + $\frac{1}{2\sqrt{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}}$ . cosx
= cosx $\left[2\mathit{s}\mathit{i}\mathit{n}\mathit{x}+\frac{1}{2\sqrt{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}}\right]$
Now, f'(x) = 0
⇒ cosx $\left[2\mathit{s}\mathit{i}\mathit{n}\mathit{x}+\frac{1}{2\sqrt{1+\mathit{s}\mathit{i}\mathit{n}\mathit{x}}}\right]$ = 0
⇒ cos x = 0
⇒ x = $\frac{\mathit{\pi }}{2}$
Again, f''$\left(\frac{\mathit{\pi }}{2}\right)$ < 0
∴ f(x) attains maximum value at x = $\frac{\mathit{\pi }}{2}$
∴ ${\mathit{f}}_{\max }$ = f $\left(\frac{\mathit{\pi }}{2}\right)$
= 3 + $\mathit{s}\mathit{i}{\mathit{n}}^2\frac{\mathit{\pi }}{2}+\sqrt{1+\mathit{s}\mathit{i}\mathit{n}\frac{\mathit{\pi }}{2}}$
= 3 + 1 + $\sqrt{2}$ = 4 + $\sqrt{2}$