NIMCET 2017 Question Paper with solutions

∴ ${\mathit{x}}^2$ + f (x) = $\mathit{a}{\mathit{x}}^4+\mathit{b}{\mathit{x}}^3+3{\mathit{x}}^2$
⇒ f (x) = $\mathit{a}{\mathit{x}}^4+\mathit{b}{\mathit{x}}^3+2{\mathit{x}}^2$
⇒ f ' (x) = $4\mathit{a}{\mathit{x}}^3+3\mathit{b}{\mathit{x}}^2$ + 4x
also f ' (x) = 0 at x = 1 and 2
∴ f ' (x) = 0 ⇒ 4a + 3b + 4 = 0 ... (i)
and f ' (2) = 0
⇒ 32a + 12b + 8 = 0 ... (ii)
On solving Eqs. (i) and (ii), we get
a = $\frac{1}{2}$ and b = - 2
∴ f (x) = $\frac{1}{2}{\mathit{x}}^4-2{\mathit{x}}^3+3{\mathit{x}}^2$
∴ f (2) = $\frac{1}{2}{\left(2\right)}^4-2{\left(2\right)}^3+3{\left(2\right)}^2$ = 8 - 16 + 8 = 0