If the equation tan^4x - 2sec^2x + [a]^2 = 0 | vedclass

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

Question

(C) Given the equation $	an^4x - 2\sec^2x + [a]^2 = 0$. Using the identity $\sec^2x = 1 + 	an^2x$,we get: $	an^4x - 2(1 + 	an^2x) + [a]^2 = 0$ $

Vedclass · Accepted Answer

$[-1, 2)$

Vedclass · Answer

(C) Given the equation $	an^4x - 2\sec^2x + [a]^2 = 0$. Using the identity $\sec^2x = 1 + 	an^2x$,we get: $	an^4x - 2(1 + 	an^2x) + [a]^2 = 0$ $	an^4x - 2	an^2x - 2 + [a]^2 = 0$ $(	an^2x - 1)^2 - 1 - 2 + [a]^2 = 0$ $(	an^2x - 1)^2 = 3 - [a]^2$ Since $(	an^2x - 1)^2 \geq 0$,we must have $3 - [a]^2 \geq 0$,which implies $[a]^2 \leq 3$. Thus,$[a] \in [-\sqrt{3}, \sqrt{3}]$. Since $[a]$ is an integer,$[a] \in \{-1, 0, 1\}$. If $[a] = -1$,then $-1 \leq a If $[a] = 0$,then $0 \leq a If $[a] = 1$,then $1 \leq a Combining these,the range of $a$ is $[-1, 2)$.

If the equation $\tan^4x - 2\sec^2x + [a]^2 = 0$ has at least one solution,then the complete range of $a$ (where $a \in R$) is:
(Note: $[k]$ denotes the greatest integer less than or equal to $k$)

Similar Questions

${\left( \frac{\cos A + \cos B}{\sin A - \sin B} \right)^n} + {\left( \frac{\sin A + \sin B}{\cos A - \cos B} \right)^n}$ (where $n$ is an integer) $=$

$\sum_{k=1}^3 \cos^2 \left((2k-1) \frac{\pi}{12}\right)$ is equal to

If $a \sin^2 \theta + b \cos^2 \theta = c$,then $\tan^2 \theta$ is equal to

If $\tanh x = \operatorname{sech} y = \frac{3}{5}$ and $e^{x+y}$ is an integer,then $e^{x+y} =$

If $\cos x = \frac{2 \cos y - 1}{2 - \cos y}$ where $x, y \in (0, \pi)$,then $\tan(x/2) \cot(y/2) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If the equation $\tan^4x - 2\sec^2x + [a]^2 = 0$ has at least one solution,then the complete range of $a$ (where $a \in R$) is: (Note: $[k]$ denotes the greatest integer less than or equal to $k$)