Question

If the equation $2\cos x+\cos 2\lambda x=3$ has only one solution, then $\lambda$ is

• A. 1
• B. a rational number
• C. an irrational number
• D. None of these

Answer 1

Answer: (C)

Given equation:

$2\cos x+\cos 2\lambda x=3$

$\Rightarrow 2\cos x+2{\cos }^2\lambda x-1=3$ [$\because \cos 2\theta =2{\cos }^2\theta -1$]

$\Rightarrow 2\cos x+2{\cos }^2\lambda x=3+1$

$\Rightarrow 2\cos x+2{\cos }^2\lambda x=4$

$\Rightarrow 2(\cos x+{\cos }^2\lambda x)=2\times 2$

$\Rightarrow \cos x+{\cos }^2\lambda x=2$ ---(1)

Since, the range of $\cos \theta$ is $-1\le \cos \theta \le 1$.

i.e., Maximum value of $\cos \theta$ is $1$.

$\Rightarrow$ $\cos x+{\cos }^2\lambda x=2$ exist only when,

$\cos x={\cos }^2\lambda x=1$

$\cos x=1$

Since, the general solution of $\cos \theta =1$ is $2n\pi ,n\in Z$

$\Rightarrow x=2n\pi ,n\in Z$

$\Rightarrow x=0,2\pi ,4\pi ,....$---(2)

Now,

${\cos }^2\lambda x=1$

$\Rightarrow {\cos }^2\lambda x={\cos }^2\left(2n\pi \right)$

Since, the general solution of ${\cos }^2\alpha ={\cos }^2\beta$ is $\alpha =n\pi \pm \beta$

i.e., $\lambda x=n\pi \pm 2n\pi$

$\Rightarrow \lambda x=\pi (n\pm 2n)$

It is given that, $2\cos x+\cos 2\lambda x=3$ has only one solution.

So, $\lambda$ exist for only the irrational numbers.

Hence, Option C is correct.