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# The number of solutions of the equation $\cos \left( {\pi \sqrt {x - 4} } \right) \cdot \cos \left( {\pi \sqrt x } \right) = 1$ is${\text{A}}{\text{. 1}} \\ {\text{B}}{\text{. 2}} \\ {\text{C}}{\text{. More than two}} \\ {\text{D}}{\text{. None of these}} \\$  Hint: In this question we have to find the number of solutions of the given equation. To solve this question the main point is that the value of $\cos x$ is less than or equal to 1. At $x=0$ the value of cos is always to be 1.

In this question we have been given the equation $\cos \left( {\pi \sqrt {x - 4} } \right) \cdot \cos \left( {\pi \sqrt x } \right) = 1$
The RHS is 1 and both terms in multiplication in LHS are in cosine.
Now both terms can only be less than or equal to 1. But if anything less than 1 is multiplied with something less than 1, then we can never get 1 in RHS.
So, $\cos \left( {\pi \sqrt {x - 4} } \right) = 1$ and $\cos \left( {\pi \sqrt x } \right) = 1$
$\Rightarrow \sqrt {x - 4} = 0$ and $\sqrt x = 0$
$\Rightarrow x = 4{\text{ and }}x = 0$
As x=4 is only given in the options
And hence, option A is correct.

Note: Whenever we face such types of problems the key point to remember is that we need to have a good grasp over trigonometric properties, some of which have been used above. We must remember that the maximum value of sine and cosine is 1. This helps in getting us the required condition and gets us on the right track to reach the answer.
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