Differential Calculus jee mains questions | JEE Main Maths Quadratic Equations Previous Year Question

(39)

Let $f (x) = 1 + x (λ^{2} - x^{2})$

$f (x) = - x^{3} + (λ^{2} x + 1)$

$\Rightarrow f' (x) = - 3 x^{2} + λ^{2}$

Put $f' (x) = 0$

$\Rightarrow λ^{2} - 3 x^{2} = 0 \Rightarrow (λ - \sqrt{3} x) (λ + \sqrt{3} x) = 0$

$x = \pm \frac{λ}{\sqrt{3}}, f'' (x) = - 6 x$

So, $x = \frac{- λ}{\sqrt{3}}$ is point of minima.

Now, $- \frac{λ}{\sqrt{3}}$ should satisfy the given condition $\frac{x^{2} + x + 2}{x^{2} + 5 x + 6} < 0$

$i.e., \frac{1}{(x + 2) (x + 3)} < 0$ $[∵ x^{2} + x + 2 > 0]$

$\Rightarrow x \in (- 3, - 2)$

$\Rightarrow - 3 < - \frac{λ}{\sqrt{3}} < - 2 \Rightarrow - 3 \sqrt{3} < - λ < - 2 \sqrt{3}$

$2 \sqrt{3} < λ < 3 \sqrt{3} \Rightarrow λ \in (2 \sqrt{3}, 3 \sqrt{3}) \equiv (α, β)$ $(∵ Given)$

$∴$ ${(2 \sqrt{3})}^{2} + {(3 \sqrt{3})}^{2} = 12 + 27 = 39$

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