Let \alpha_\theta and \beta_\theta be the distinct roots of 2 x^2+(\cos \theta) x-1=0, \theta \in(0,2 \pi) . If m and M are the minimum and the maximum values of \alpha_\theta^4+\beta_\theta^4 , then 16(M+m) equals :

Quadratic Equation and Inequalities PYQ — JEE Main 2025 · 22 Jan · Shift 2 · Q26 with Solution

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Correct answer: C

Given quadratic and relations between roots

The equation is

\cos\theta)x-1=0, \qquad \theta\in(0,2\pi).$$ Let its roots be $\alpha_\theta,\beta_\theta$. Using Vieta's formulas: $$\alpha+\beta=-\frac{\cos\theta}{2}, \qquad \alpha\beta=-\frac{1}{2}.$$ We need the minimum and maximum of $$\alpha^4+\beta^4.$$ --- 2. **Express $\alpha^4+\beta^4$ in terms of $\alpha+\beta$ and $\alpha\beta$** Use $$\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2\alpha^2\beta^2.$$ Now, $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta.$$ Substitute the Vieta values: $$\alpha+\beta=-\frac{\cos\theta}{2}, \qquad \alpha\beta=-\frac{1}{2}.$$ So, $$\alpha^2+\beta^2=\left(-\frac{\cos\theta}{2}\right)^2-2\left(-\frac{1}{2}\right) =\frac{\cos^2\theta}{4}+1.$$ Also, $$\alpha^2\beta^2=(\alpha\beta)^2=\frac{1}{4}.$$ Hence, $$\alpha^4+\beta^4=\left(1+\frac{\cos^2\theta}{4}\right)^2-2\cdot \frac14.$$ That is, $$\alpha^4+\beta^4=\left(1+\frac{\cos^2\theta}{4}\right)^2-\frac12.$$ Expand: $$=1+\frac{\cos^2\theta}{2}+\frac{\cos^4\theta}{16}-\frac12$$ $$=\frac12+\frac{\cos^2\theta}{2}+\frac{\cos^4\theta}{16}.$$ Let $$t=\cos^2\theta.$$ Since $\theta\in(0,2\pi)$, we have $$t\in[0,1].$$ Therefore $$f(t)=\alpha^4+\beta^4=\frac12+\frac{t}{2}+\frac{t^2}{16}, \qquad t\in[0,1].$$ --- 3. **Find minimum and maximum** Now $$f'(t)=\frac12+\frac{t}{8}>0 \quad \text{for } t\in[0,1].$$ So $f(t)$ is strictly increasing on $[0,1]$. Thus, - minimum occurs at $t=0$, - maximum occurs at $t=1$. So, $$m=f(0)=\frac12,$$ $$M=f(1)=\frac12+\frac12+\frac{1}{16}=1+\frac{1}{16}=\frac{17}{16}.$$ --- 4. **Compute $16(M+m)$** $$M+m=\frac{17}{16}+\frac12=\frac{17}{16}+\frac{8}{16}=\frac{25}{16}.$$ Hence, $$16(M+m)=25.$$ --- 5. **Option check** The correct option is: $$\boxed{\text{C: }25}.$$ This matches the stored correct answer.

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View written solutionFree

Correct answer: C

Given quadratic and relations between roots

The equation is

\cos\theta)x-1=0, \qquad \theta\in(0,2\pi).$$ Let its roots be $\alpha_\theta,\beta_\theta$. Using Vieta's formulas: $$\alpha+\beta=-\frac{\cos\theta}{2}, \qquad \alpha\beta=-\frac{1}{2}.$$ We need the minimum and maximum of $$\alpha^4+\beta^4.$$ --- 2. **Express $\alpha^4+\beta^4$ in terms of $\alpha+\beta$ and $\alpha\beta$** Use $$\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2\alpha^2\beta^2.$$ Now, $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta.$$ Substitute the Vieta values: $$\alpha+\beta=-\frac{\cos\theta}{2}, \qquad \alpha\beta=-\frac{1}{2}.$$ So, $$\alpha^2+\beta^2=\left(-\frac{\cos\theta}{2}\right)^2-2\left(-\frac{1}{2}\right) =\frac{\cos^2\theta}{4}+1.$$ Also, $$\alpha^2\beta^2=(\alpha\beta)^2=\frac{1}{4}.$$ Hence, $$\alpha^4+\beta^4=\left(1+\frac{\cos^2\theta}{4}\right)^2-2\cdot \frac14.$$ That is, $$\alpha^4+\beta^4=\left(1+\frac{\cos^2\theta}{4}\right)^2-\frac12.$$ Expand: $$=1+\frac{\cos^2\theta}{2}+\frac{\cos^4\theta}{16}-\frac12$$ $$=\frac12+\frac{\cos^2\theta}{2}+\frac{\cos^4\theta}{16}.$$ Let $$t=\cos^2\theta.$$ Since $\theta\in(0,2\pi)$, we have $$t\in[0,1].$$ Therefore $$f(t)=\alpha^4+\beta^4=\frac12+\frac{t}{2}+\frac{t^2}{16}, \qquad t\in[0,1].$$ --- 3. **Find minimum and maximum** Now $$f'(t)=\frac12+\frac{t}{8}>0 \quad \text{for } t\in[0,1].$$ So $f(t)$ is strictly increasing on $[0,1]$. Thus, - minimum occurs at $t=0$, - maximum occurs at $t=1$. So, $$m=f(0)=\frac12,$$ $$M=f(1)=\frac12+\frac12+\frac{1}{16}=1+\frac{1}{16}=\frac{17}{16}.$$ --- 4. **Compute $16(M+m)$** $$M+m=\frac{17}{16}+\frac12=\frac{17}{16}+\frac{8}{16}=\frac{25}{16}.$$ Hence, $$16(M+m)=25.$$ --- 5. **Option check** The correct option is: $$\boxed{\text{C: }25}.$$ This matches the stored correct answer.

Quadratic Equation and Inequalities question

Quadratic Equation and Inequalities question

More from Quadratic Equation and Inequalities

Quadratic Equation and Inequalities question

Quadratic Equation and Inequalities question

More from Quadratic Equation and Inequalities