Question

The equation cos3x = 3a-4 possesses a solution if.
(a) $1\le a\le \dfrac{5}{3}$
(b) $a\ge 2$
(c) $a\ge 3$
(d) $a\in R$

Answer 1

Hint: Think of the range and domain of cos3x. After you use the range, try to solve the inequality you get to reach the answer.

Complete step-by-step answer:
Before starting with the solution to the above question, let us first discuss the approaches that might click once we look at the problem.
The first approach that might click is using the formula: $\cos 3A=4{{\cos }^{3}}A-3\cos A$ , but this would be useful if we have all the terms in the expression were trigonometric, unlike our question.

The next approach would be to solve using the range of cos3x, which is what we will be doing here.
So, let us start with the solution to get the possible values of a.
We know the cosine of an angle has its value lying in [-1,1].
So, the possible values of cos3x should also be in [-1,1].

According to the equation cos3x = 3a-4 , the value of 3a-4 should also be lying in [-1,1] to have possible solutions to the equation.
Now representing the above statement in the form of inequality, we get
$-1\le \cos 3x\le 1$
$\therefore -1\le 3a-4\le 1$
On adding 4 to the inequality, we get
$-1+4\le 3a-4+4\le 1+4$
$3\le 3a\le 5$
Dividing the inequality by 3, we get
$\dfrac{3}{3}\le \dfrac{3a}{3}\le \dfrac{5}{3}$
$\therefore 1\le a\le \dfrac{5}{3}$

Therefore, the correct answer is option (a) $1\le a\le \dfrac{5}{3}$ , i.e. a must be lying in $\left[ 1,\dfrac{5}{3} \right]$ for the equation cos3x = 3a-4, to possess a solution.

Note: While handling an inequality, we should always be careful as whenever you multiply or divide an inequality by a negative number, the sign of inequality gets reversed.