The number of all possible 5-tuples $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})$ such that $a_{1} + a_{2} s i n x + a_{3} c o s x + a_{4} s i n 2 x + a_{5} c o s 2 x = 0$ holds for all $x$ is-

Zero

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1

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2

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Infinite

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Solution

The correct option is B $1$
$a_{1} + a_{2} s i n x + a_{3} c o s x + a_{4} s i n 2 x + a_{5} c o s 2 x = 0$
For $x = 0$ , we get
$a_{1} + a_{3} + a_{5} = 0$ ...(1)
For $x = π$ , we get
$a_{1} - a_{3} + a_{5} = 0$ ...(2)
For $x = \frac{π}{2}$ , we get
$a_{1} + a_{2} - a_{5} = 0$ ...(3)
For $x = \frac{3 π}{2}$ , we get
$a_{1} - a_{2} - a_{5} = 0$ ...(4)
Solving (1),(2),(3) and (4), we get
$a_{1} = 0, a_{2} = 0, a_{3} = 0, a_{5} = 0$
For $x = \frac{π}{4}$ , we get
$a_{1} + \frac{a_{2}}{\sqrt{2}} + \frac{a_{3}}{\sqrt{2}} + a_{4} = 0 a_{4} = 0$
Hence only zero is the only possible value

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