Let the set of all positive values of lambda, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) First,solve the inequality $\frac{x^2+x+2}{x^2+5x+6} < 0$.Since the numerator $x^2+x+2$ has a discriminant $D = 1^2 - 4(1)(2) = -7 < 0$,it is alwa

Vedclass · Accepted Answer

$39$

Vedclass · Answer

(C) First,solve the inequality $\frac{x^2+x+2}{x^2+5x+6} Since the numerator $x^2+x+2$ has a discriminant $D = 1^2 - 4(1)(2) = -7 Thus,the inequality reduces to $\frac{1}{(x+2)(x+3)} This gives $x \in (-3, -2)$.Next,consider the function $f(x) = 1 + x\lambda^2 - x^3$.Find the derivative: $f'(x) = \lambda^2 - 3x^2$.Set $f'(x) = 0$ to find critical points: $3x^2 = \lambda^2 \Rightarrow x = \pm \frac{\lambda}{\sqrt{3}}$.Using the second derivative test: $f''(x) = -6x$.For local minimum,$f''(x) > 0$,so $-6x > 0 \Rightarrow x Thus,the point of local minimum is $x = -\frac{\lambda}{\sqrt{3}}$.Given that this point must satisfy $x \in (-3, -2)$,we have:$-3 Multiplying by $-1$ reverses the inequality:$2 $2\sqrt{3} So,$\alpha = 2\sqrt{3}$ and $\beta = 3\sqrt{3}$.Then $\alpha^2 + \beta^2 = (2\sqrt{3})^2 + (3\sqrt{3})^2 = 12 + 27 = 39$.

Let the set of all positive values of $\lambda$,for which the point of local minimum of the function $f(x) = 1 + x(\lambda^2 - x^2)$ satisfies $\frac{x^2+x+2}{x^2+5x+6} < 0$,be $(\alpha, \beta)$. Then $\alpha^2 + \beta^2$ is equal to:

Similar Questions

The function $f(x) = \frac{x}{2} + \frac{2}{x}$ has a local minimum at $x = $ ........

If $f(x)=x^2+ax+b$ has a minima at $x=3$ whose value is $5$,then the values of $a$ and $b$ are respectively.

Let $f(x) = 1 - \sqrt{x^2}$, where the square root is to be taken as positive. Then:

The maximum value of $xy$ subject to $x+y=7$ is

$A$ right circular cone is inscribed in a sphere of radius $3$ units. If the volume of the cone is maximum, then the semi-vertical angle of the cone is

Vedclass Test Series

Exam Paper Generator

Online Exam Module