Question

Find the maximum integral value of ‘a’ for which the equation \[asinx + cos2x = 2a - 7\] has a solution?

Answer 1

Hint: We will first convert the equation in terms of sine function. We will get a quadratic function in sine, then by using the quadratic formula we will find the value of sine function. We will use the range of sine functions to find the solution of a.

Complete step-by-step answer:
We have given the equation \[asinx + cos2x = 2a - 7\] , to find the solution
We will try to convert the whole equation in terms of one function.
$ \Rightarrow asinx + cos2x = 2a - 7$
We will convert cos term into sin form by using the formula $cos2x = 1 - 2{\sin ^2}x$
$ \Rightarrow asinx + 1 - 2{\sin ^2}x = 2a - 7$
We get a quadratic function in terms of sin function
We know that the solution of as quadratic function $a{x^2} + bx + c$ is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
We will find the solution of the equation $2{\sin ^2}x - a\sin x + 2a - 8 = 0$
$ \Rightarrow sinx = \dfrac{{a \pm (a - 8)}}{4}$
We will first take negative sign and then positive sign
$ \Rightarrow sinx = 2$ and $sinx = \dfrac{{a - 4}}{2}$
We know that the range of sin function is -1 to 1
So, the value of $sinx = 2$ is not possible, so we will take in consideration the solution $sinx = \dfrac{{a - 4}}{2}$
We have $ - 1 \leqslant sinx \leqslant 1$
$ \Rightarrow \; - 1 \leqslant \dfrac{{a - 4}}{2} \leqslant 1$
We will multiply both side by 2
$ \Rightarrow \; - 2 \leqslant a - 4 \leqslant 2$
We will add 4 in whole equation
$ \Rightarrow \; - 2 + 4 \leqslant a \leqslant 2 + 4$
$ \Rightarrow \;2 \leqslant a \leqslant 6$
Hence, the maximum integral value of a is 6.

Note: The most important step in this question is to convert cos in terms of sin because then we get a quadratic function in terms of sin. We should be familiar with quadratic functions and their solutions. And then finally it also includes the concept of inequality.