Quadratic EquationHard
Question
The set of values of ' a ' for which the equation $\cos^{4}x - \sin^{4}x + cos2x + a^{2} + a = 0$ will have atleast one real solution is
Options
A.$\lbrack - 2,1\rbrack$
B.$\lbrack - 1,2\rbrack$
C.$\lbrack - 1,1\rbrack$
D.$\lbrack 1,2\rbrack$
Solution
$- 2cos2x = a^{2} + a$
For equation to have solution
$$\begin{matrix} & - 2 \leq a^{2} + a \leq 2 \\ \Rightarrow & a^{2} + a - 2 \leq 0 \\ \Rightarrow & a \in \lbrack - 2,1\rbrack \end{matrix}$$
Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
Let a > 0, b > 0 and c > 0. Then, both the roots of the equation ax2 + bx + c = 0....If α, β be the roots of x2 + px - q = 0 and γ, δ be the roots of x2 + px + r = 0, q + r ≠ 0 th...If all the roots of the equation $x^{4} - 12x^{3} + ax^{2} + bx + 81 = 0$ where $a,b \in R$ are positive, then...The solution of the equation 2x2 + 3x − 9 is given by-...If the quadratic equation $ax^{2} - bx + 7 = 0$ does not have two distinct real roots, then the minimum value of $a + b$...