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Maths-

General

Easy

Question

If the equation $cot to the power of 4 end exponent invisible function application x minus 2 blank c o s e c to the power of 2 end exponent blank x plus a to the power of 2 end exponent equals 0$ has at least one solution, then the sum of all possible integral values of $´ a to the power of ´ end exponent$ is equal to

4
3
2
0

The correct answer is: 0

$cot to the power of 4 end exponent invisible function application x minus 2 open parentheses 1 plus cot to the power of 2 end exponent invisible function application x close parentheses plus a to the power of 2 end exponent equals 0$
$rightwards double arrow cot to the power of 4 end exponent invisible function application x minus 2 cot to the power of 2 end exponent invisible function application x plus a to the power of 2 end exponent minus 2 equals 0$
$rightwards double arrow open parentheses cot to the power of 2 end exponent invisible function application x minus 1 close parentheses to the power of 2 end exponent equals 3 minus a to the power of 2 end exponent$
$T o blank h a v e blank a t blank l e a s t blank o n e blank s o l u t i o n comma blank 3 minus a to the power of 2 end exponent greater or equal than 0$
$rightwards double arrow a to the power of 2 end exponent minus 3 less or equal than 0$
$a element of open square brackets negative square root of 3 comma square root of 3 close square brackets$
Integral values are $negative 1 comma blank 0 comma blank 1 semicolon$ therefore, the sum is 0

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