If sin alpha and cos alpha are the roots of t | vedclass

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

Question

(A) Given the quadratic equation $x^{2}-bx+c=0$ with roots $\sin \alpha$ and $\cos \alpha$.From the relation between roots and coefficients:$\sin \alp

Vedclass · Accepted Answer

$c \leq \frac{1}{2}$

Vedclass · Answer

(A) Given the quadratic equation $x^{2}-bx+c=0$ with roots $\sin \alpha$ and $\cos \alpha$.From the relation between roots and coefficients:$\sin \alpha + \cos \alpha = b$ $(i)$$\sin \alpha \cdot \cos \alpha = c$ (ii)We know that $(\sin \alpha + \cos \alpha)^{2} = \sin^{2} \alpha + \cos^{2} \alpha + 2 \sin \alpha \cos \alpha$.Substituting the values: $b^{2} = 1 + 2c$, which implies $c = \frac{b^{2}-1}{2}$.Since $-1 \leq \sin \alpha \leq 1$ and $-1 \leq \cos \alpha \leq 1$, we have $-\sqrt{2} \leq \sin \alpha + \cos \alpha \leq \sqrt{2}$, so $-\sqrt{2} \leq b \leq \sqrt{2}$.Also, $\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2c$.Since $-1 \leq \sin 2\alpha \leq 1$, we have $-1 \leq 2c \leq 1$, which implies $c \leq \frac{1}{2}$.

If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $x^{2}-bx+c=0$, then which of the following statements is/are correct?

Similar Questions

If the polynomial $P(x) = x^2 + ax + b$ has factors $(x - a)(x - b)$,where $a, b \in R$,then the value of $P(2)$ is:

If $\alpha, \beta$ are the roots of $a x^2+b x+c=0$,then the quadratic equation whose roots are $\sqrt{5} \alpha, \sqrt{5} \beta$ is

If the harmonic mean between the roots of $(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0$ is $4$,then the value of $b$ is

If $\alpha, \beta, \gamma$ are the roots of $x^3-2x^2+3x-4=0$,then the value of $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$ is

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-9x^2+23x-15=0$,then $\alpha^3+\beta^3+\gamma^3=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module