Let the set of all positive values of λ, for which the point of local minimum of the function (1 + x

Q: Wait, why is the numerator $x^2+x+2$ always positive? Can't it be negative for some values of $x$?

That is a fantastic question! To check if a quadratic $ax^2+bx+c$ is always positive, we look at its discriminant $D = b^2 - 4ac$. Here, $D = 1^2 - 4(1)(2) = 1 - 8 = -7$. Since the discriminant is negative and the leading coefficient ($a=1$) is positive, the parabola never touches the x-axis and opens upwards. It stays strictly above the x-axis for all real $x$. It is a 'floating' parabola, always positive!

Q: Why do we reverse the inequality signs when multiplying by $-\sqrt{3}$?

This is the most common trap in JEE algebra! Think of it this way: $2 < 5$ is true. If you multiply both sides by $-1$, you get $-2$ and $-5$. Clearly, $-2$ is greater than $-5$. Multiplying by a negative number flips the direction of the inequality because it reflects the values across the origin on the number line. Always remember to flip the sign!

Q: How do we know for sure that $x = -\frac{\lambda}{\sqrt{3}}$ is the local minimum and not the local maximum?

We use the second derivative test. We found $f''(x) = -6x$. For a local minimum, the curve must be concave up, meaning $f''(x) > 0$. If we plug in $x = -\frac{\lambda}{\sqrt{3}}$, we get $f''(-\frac{\lambda}{\sqrt{3}}) = -6(-\frac{\lambda}{\sqrt{3}}) = 2\sqrt{3}\lambda$. Since $\lambda$ is positive, this result is positive. A positive second derivative confirms a local minimum!

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Cubic Landscape We are given the function $f(x) = 1 + x(\lambda^2 - x^2)$, which simplifies to $f(x) = 1 + \lambda^2 x - x^3$. The leading term is $-x^3$, indicating that as $x$ grows large, the function plunges toward negative infinity. This is an inverted cubic, a shape that rises to a peak, dips into a valley, and then falls away. ### The Calculus of Extremes To find the valley, we must locate where the slope of the curve is zero. We invoke the first derivative: $$f'(x) = \frac{d}{dx}(1 + \lambda^2 x - x^3) = \lambda^2 - 3x^2$$ Setting this to zero, we find our critical points: $3x^2 = \lambda^2$, which yields $x = \pm \frac{\lambda}{\sqrt{3}}$. We turn to the second derivative test to identify the minimum: $f''(x) = -6x$. For a local minimum, we require $f''(x) > 0$. Substituting our candidates, we see that only $x = -\frac{\lambda}{\sqrt{3}}$ yields a positive second derivative, as $f''(-\frac{\lambda}{\sqrt{3}}) = 2\sqrt{3}\lambda > 0$. This point represents our local minimum. ### The Rational Gatekeeper The problem requires this point to satisfy the following condition: $$\frac{x^2+x+2}{x^2+5x+6} < 0$$ Consider the numerator $x^2+x+2$. Its discriminant is $D = 1^2 - 4(1)(2) = -7$. Because $D < 0$ and the leading coefficient is positive, this numerator is always greater than zero for any real $x$. Since the numerator is always positive, the fraction is negative only if the denominator is negative. Thus, we solve the inequality: $$x^2+5x+6 < 0$$ Factoring this, we get $(x+2)(x+3) < 0$. Using the wavy curve method, we determine that $x$ must reside in the open interval $(-3, -2)$. ### The Synthesis We bridge the two worlds by requiring our local minimum $x = -\frac{\lambda}{\sqrt{3}}$ to fall within the interval $(-3, -2)$. We write: $$-3 < -\frac{\lambda}{\sqrt{3}} < -2$$ To isolate $\lambda$, we multiply the entire inequality by $-\sqrt{3}$. Remembering to reverse the inequality signs, we obtain: $$3\sqrt{3} > \lambda > 2\sqrt{3}$$ Thus, the range for $\lambda$ is $(2\sqrt{3}, 3\sqrt{3})$. ### The Final Victory We are given this range as $(\alpha, \beta)$. By direct comparison, we identify $\alpha = 2\sqrt{3}$ and $\beta = 3\sqrt{3}$. The final calculation is: $$\alpha^2 + \beta^2 = (2\sqrt{3})^2 + (3\sqrt{3})^2 = 12 + 27 = 39$$ The final result is **39**.

Animated Solution for Mathematics - Differentiation: Let the set of all positive values of $λ$ , for which the point of local minimum of the function $(1 + x (λ^{2} - x^{2}))$ satisfies $\frac{x ^{2} + x + 2}{x ^{2} + 5 x + 6} < 0$ , be $(α, β)$ . Then $α^{2} + β^{2}$ is equal to _________

Enter Numerical Value:

Visualized Solution

Define the Function $f (x)$

Finding the First Derivative $f^{'} (x)$

Locating Critical Points

The Second Derivative Test

Analyzing the Rational Inequality

Checking the Numerator

Solving the Denominator Inequality

Finding the Valid Interval for $x$

Linking $x$ and $λ$

Solving for $λ$

Identifying $α$ and $β$

Final Calculation

The Sigma Insight: Maxima and Minima

Analyzing the Cubic Landscape

The Calculus of Extremes

The Rational Gatekeeper

The Synthesis

The Final Victory

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