The number of real values of a satisfying the | vedclass

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

Question

(C) Given equation is $a^2 - 2a\sin x + 1 = 0$. For $a$ to be a real value,the discriminant $D$ of this quadratic equation in $a$ must be greater than

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given equation is $a^2 - 2a\sin x + 1 = 0$. For $a$ to be a real value,the discriminant $D$ of this quadratic equation in $a$ must be greater than or equal to $0$. $D = (-2\sin x)^2 - 4(1)(1) = 4\sin^2 x - 4 = 4(\sin^2 x - 1) = -4\cos^2 x$. Since $D \ge 0$,we must have $-4\cos^2 x \ge 0$,which implies $\cos^2 x \le 0$. Since $\cos^2 x$ cannot be negative,the only possibility is $\cos^2 x = 0$,which means $\cos x = 0$. If $\cos x = 0$,then $\sin x = 1$ or $\sin x = -1$. Case $1$: If $\sin x = 1$,the equation becomes $a^2 - 2a + 1 = 0$,which is $(a - 1)^2 = 0$,giving $a = 1$. Case $2$: If $\sin x = -1$,the equation becomes $a^2 + 2a + 1 = 0$,which is $(a + 1)^2 = 0$,giving $a = -1$. Thus,the real values of $a$ are $1$ and $-1$. There are $2$ such real values.

The number of real values of $a$ satisfying the equation $a^2 - 2a\sin x + 1 = 0$ is

Similar Questions

What is the nature of the roots of the equation $a^2x^2 + (a + b)x - b^2 = 0$?

Let $E$ denote the set of all integers $a$ such that the point of intersection of the parabola $y = x^2 + 2ax + 2021$ with the $x$-axis has rational coordinates. The largest element of $E$ is

If $x = \sqrt{7} + \sqrt{3}$ and $xy = 4$,then $x^4 + y^4 =$

If $\frac{k}{kx+3}+\frac{3}{3x-k}=\frac{12x+5}{(kx+3)(3x-k)}$ for all $x \in R - \{-\frac{3}{k}, \frac{k}{3}\}$,then both the roots of the equation $kx^2-7x+3=0$ are

The quadratic equation $2x^{2} - (a^{3} + 8a - 1)x + a^{2} - 4a = 0$ has roots of opposite signs. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module