If the sum of the squares of the roots of the | vedclass

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

Question

(D) Given the quadratic equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$.Let the roots be $x_1$ and $x_2$.From the properties of roots,$x_1

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) Given the quadratic equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$.Let the roots be $x_1$ and $x_2$.From the properties of roots,$x_1 + x_2 = \sin \alpha - 2$ and $x_1 x_2 = -(1 + \sin \alpha)$.The sum of the squares of the roots is given by $S = x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2$.Substituting the values: $S = (\sin \alpha - 2)^2 - 2(-(1 + \sin \alpha)) = \sin^2 \alpha - 4 \sin \alpha + 4 + 2 + 2 \sin \alpha = \sin^2 \alpha - 2 \sin \alpha + 6$.To find the minimum value,we can rewrite this as $S = (\sin \alpha - 1)^2 + 5$.Since the range of $\sin \alpha$ is $[-1, 1]$,the expression $( \sin \alpha - 1)^2$ is minimum when $\sin \alpha = 1$.Thus,$\sin \alpha = 1$ implies $\alpha = \frac{\pi}{2}$.

If the sum of the squares of the roots of the equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$ is least,then $\alpha$ is

Similar Questions

If $\alpha$ and $\beta$ are the roots of $x^2 - ax + b = 0$ and if $\alpha^n + \beta^n = V_n$,then -

If $(3x - 2y) : (2x + 3y) = 5 : 6$,then one of the values of $\left(\frac{\sqrt[3]{x} + \sqrt[3]{y}}{\sqrt[3]{x} - \sqrt[3]{y}}\right)^{2}$ is

Solve the given two equations and select the correct option.
$I.$ $12x^2 - 47x + 40 = 0$
$II.$ $4y^2 + 3y - 10 = 0$

If $\alpha, \beta, \gamma$ are the roots of $x^3 - x - 2 = 0$,then the value of $\alpha^5 + \beta^5 + \gamma^5$ is-

Roots of $2 k x^{2}+5 k x+2=0$ are equal if $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module