NIMCET 2012 Question Paper with solutions

By given condition, we get
Let f (x) = x + $\frac{1}{\mathit{x}}$ ... (i)
On differentiating w.r.t x, we get f . (x) = 1 - $\frac{1}{{\mathit{x}}^2}$
For max or min of f(x),
Put f . (x) = 0
⇒ 1 - $\frac{1}{{\mathit{x}}^2}$ = 0
⇒ $\frac{\left({\mathit{x}}^2-1\right)}{{\mathit{x}}^2}$ = 0 (Since x ≠ 0)
⇒ (x - 1) (x + 1) = 0
⇒ x = 1 or - 1
Now , f '' (x) = $\frac{1}{{\mathit{x}}^3}$
at x = - 1 , f '' (-1) = - 2 (max)
at x = 1 , f '' (1) = 2 (min)
So, f(x) is min at (x = 1) and its minimum value at (x = 1)is
f (1) = 1 + $\frac{1}{1}$ = 2
or Let f (x) = x + $\frac{1}{\mathit{x}}$
Since AM ≥ GM
⇒ $\frac{\mathit{x}+\frac{1}{\mathit{x}}}{2}$ ≥ ${\left(\mathit{x}.\frac{1}{\mathit{x}}\right)}^{\frac{1}{2}}$
⇒ $\left(\mathit{x}+\frac{1}{2}\right)$ ≥ 2
Min of f (x) is 2.