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 \fracx^2+x+2x^2+5 x+6\lt 0 , be (\alpha, \beta) . Then \alpha^2+\beta^2 is equal to \underline\hspace2cm .

Application of Derivatives PYQ — JEE Main 2024 · 9 Apr · Shift 1 · Q58 with Solution

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Correct answer: 39

Given function

We interpret the function as $f(x)=1+x(\lambda^2-x^2)=1+\lambda^2x-x^3,$ where $\lambda>0$ .

We need the point of local minimum of $f(x)$ to satisfy $\frac{x^2+x+2}{x^2+5x+6}<0.$

Find critical points of $f(x)$

Differentiate: $f'(x)=\lambda^2-3x^2.$

Set $f'(x)=0$ :

\implies x^2=\frac{\lambda^2}{3} \implies x=\pm \frac{\lambda}{\sqrt{3}}.$$ Now, $$f''(x)=-6x.$$ - At $x=\dfrac{\lambda}{\sqrt{3}}>0$, $$f''\left(\frac{\lambda}{\sqrt{3}}\right)=-6\cdot \frac{\lambda}{\sqrt{3}}<0,$$ so this is a **local maximum**. - At $x=-\dfrac{\lambda}{\sqrt{3}}<0$, $$f''\left(-\frac{\lambda}{\sqrt{3}}\right)= -6\left(-\frac{\lambda}{\sqrt{3}}\right)>0,$$ so this is a **local minimum**. Hence the point of local minimum occurs at $$x=-\frac{\lambda}{\sqrt{3}}.$$ --- 3. **Apply the inequality condition** We need $$\frac{x^2+x+2}{x^2+5x+6}<0$$ at the local minimum point. First factor the denominator: $$x^2+5x+6=(x+2)(x+3).$$ For the numerator, $$x^2+x+2.$$ Its discriminant is $$\Delta=1-8=-7<0,$$ and since the coefficient of $x^2$ is positive, we have $$x^2+x+2>0 \quad \text{for all real }x.$$ Therefore the fraction is negative iff the denominator is negative: $$(x+2)(x+3)<0.$$ This gives $$-3<x<-2.$$ Now substitute the local minimum point $x=-\dfrac{\lambda}{\sqrt{3}}$: $$-3< -\frac{\lambda}{\sqrt{3}} < -2.$$ Multiply by $-1$ throughout (reverse inequalities): $$3>\frac{\lambda}{\sqrt{3}}>2.$$ So, $$2<\frac{\lambda}{\sqrt{3}}<3.$$ Multiply by $\sqrt{3}$: $$2\sqrt{3}<\lambda<3\sqrt{3}.$$ Thus the set of positive values of $\lambda$ is $$(\alpha,\beta)=(2\sqrt{3},3\sqrt{3}).$$ --- 4. **Compute** $\alpha^2+\beta^2$ $$\alpha^2=(2\sqrt{3})^2=12,$$ $$\beta^2=(3\sqrt{3})^2=27.$$ Hence, $$\alpha^2+\beta^2=12+27=39.$$ --- 5. **Comparison with stored answer** Derived answer = $39$. Stored correct answer = $39$. They match.

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View written solutionFree

Correct answer: 39

Given function

We interpret the function as $f(x)=1+x(\lambda^2-x^2)=1+\lambda^2x-x^3,$ where $\lambda>0$ .

We need the point of local minimum of $f(x)$ to satisfy $\frac{x^2+x+2}{x^2+5x+6}<0.$

Find critical points of $f(x)$

Differentiate: $f'(x)=\lambda^2-3x^2.$

Set $f'(x)=0$ :

\implies x^2=\frac{\lambda^2}{3} \implies x=\pm \frac{\lambda}{\sqrt{3}}.$$ Now, $$f''(x)=-6x.$$ - At $x=\dfrac{\lambda}{\sqrt{3}}>0$, $$f''\left(\frac{\lambda}{\sqrt{3}}\right)=-6\cdot \frac{\lambda}{\sqrt{3}}<0,$$ so this is a **local maximum**. - At $x=-\dfrac{\lambda}{\sqrt{3}}<0$, $$f''\left(-\frac{\lambda}{\sqrt{3}}\right)= -6\left(-\frac{\lambda}{\sqrt{3}}\right)>0,$$ so this is a **local minimum**. Hence the point of local minimum occurs at $$x=-\frac{\lambda}{\sqrt{3}}.$$ --- 3. **Apply the inequality condition** We need $$\frac{x^2+x+2}{x^2+5x+6}<0$$ at the local minimum point. First factor the denominator: $$x^2+5x+6=(x+2)(x+3).$$ For the numerator, $$x^2+x+2.$$ Its discriminant is $$\Delta=1-8=-7<0,$$ and since the coefficient of $x^2$ is positive, we have $$x^2+x+2>0 \quad \text{for all real }x.$$ Therefore the fraction is negative iff the denominator is negative: $$(x+2)(x+3)<0.$$ This gives $$-3<x<-2.$$ Now substitute the local minimum point $x=-\dfrac{\lambda}{\sqrt{3}}$: $$-3< -\frac{\lambda}{\sqrt{3}} < -2.$$ Multiply by $-1$ throughout (reverse inequalities): $$3>\frac{\lambda}{\sqrt{3}}>2.$$ So, $$2<\frac{\lambda}{\sqrt{3}}<3.$$ Multiply by $\sqrt{3}$: $$2\sqrt{3}<\lambda<3\sqrt{3}.$$ Thus the set of positive values of $\lambda$ is $$(\alpha,\beta)=(2\sqrt{3},3\sqrt{3}).$$ --- 4. **Compute** $\alpha^2+\beta^2$ $$\alpha^2=(2\sqrt{3})^2=12,$$ $$\beta^2=(3\sqrt{3})^2=27.$$ Hence, $$\alpha^2+\beta^2=12+27=39.$$ --- 5. **Comparison with stored answer** Derived answer = $39$. Stored correct answer = $39$. They match.

Application of Derivatives question

Application of Derivatives question

More from Application of Derivatives

Application of Derivatives question

Application of Derivatives question

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