Let f( x ) = x^2 + 1 \over x^2 and g( x ) = x - 1 \over x , x \in R - \ - 1,0,1 \ . If h( x ) = f( x ) \over g( x ) , then the local minimum value of h(x) is

Application of Derivatives PYQ — JEE Main 2018 · Shift 0 · Q43 with Solution

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Correct answer: A

Given $f(x)=x^2+\frac{1}{x^2},\qquad g(x)=x-\frac{1}{x},$ with $x\in \mathbb R\setminus\{-1,0,1\}.$

Then $h(x)=\frac{f(x)}{g(x)}=\frac{x^2+\frac{1}{x^2}}{x-\frac{1}{x}}.$

Multiply numerator and denominator by $x^2$ : $h(x)=\frac{x^4+1}{x^3-x}.$

A more useful simplification is obtained by dividing by $x$ :

Let $t=x-\frac{1}{x}.$ Then $(x-\frac1x)^2=x^2+\frac1{x^2}-2,$ so $x^2+\frac1{x^2}=t^2+2.$ Hence

where $t\neq 0$ because $x\neq \pm 1$ .

So we only need to minimize $\phi(t)=t+\frac{2}{t},\qquad t\in \mathbb R\setminus\{0\}.$

Differentiate: $\phi'(t)=1-\frac{2}{t^2}.$

Set $\phi'(t)=0$ :

\implies t^2=2 \implies t=\pm \sqrt2.$$ Now, $$\phi''(t)=\frac{4}{t^3}.$$ - At $t=\sqrt2$, $$\phi''(\sqrt2)>0,$$ so this gives a **local minimum**. - At $t=-\sqrt2$, $$\phi''(-\sqrt2)<0,$$ so this gives a local maximum. --- 3. **Compute the local minimum value** At $t=\sqrt2$, $$\phi(\sqrt2)=\sqrt2+\frac{2}{\sqrt2}=\sqrt2+\sqrt2=2\sqrt2.$$ Thus the local minimum value of $h(x)$ is $$\boxed{2\sqrt2}.$$ --- 4. **Check attainability from the original variable** We should ensure that $t=\sqrt2$ is possible for some allowed real $x$. Solve $$x-\frac1x=\sqrt2 \implies x^2-\sqrt2 x-1=0.$$ This quadratic has real roots, and they are not $0,\pm1$. So the value is indeed attained. --- 5. **Option check** - A: $2\sqrt2$ ✅ - B: $3$ ❌ - C: $-3$ ❌ - D: $-2\sqrt2$ ❌ Therefore the correct option is **A**.