The equation ${cos}^{8} x + b {cos}^{4} x + 1 = 0$ will have a solution of $b$ belongs to

(- \infty, 2]

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[2, \infty)

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(- \infty, - 2]

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None of these

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Solution

The correct option is D $(- \infty, - 2]$
$c o s^{8} x + b c o s^{4} x + 1 = 0$
$\Rightarrow b = - (c o s^{4} x + \frac{1}{c o s^{4} x})$
we know for any positive number $a$ , $(a + \frac{1}{a}) \geq 2$
thus for having solution Range of $b$ is,
$b ϵ (- \infty, - 2]$
Hence, option 'C' is correct.

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The equation cos8x+bcos4x+1=0 will have a solution of b belongs to

The correct option is D (−∞,−2]cos8x+bcos4x+1=0⇒b=−(cos4x+1cos4x)we know for any positive number a, (a+1a)≥2thus for having solution Range of b is,bϵ(−∞,−2]Hence, option 'C' is correct.

The equation ${cos}^{8} x + b {cos}^{4} x + 1 = 0$ will have a solution of $b$ belongs to

The correct option is D $(- \infty, - 2]$
$c o s^{8} x + b c o s^{4} x + 1 = 0$
$\Rightarrow b = - (c o s^{4} x + \frac{1}{c o s^{4} x})$
we know for any positive number $a$ , $(a + \frac{1}{a}) \geq 2$
thus for having solution Range of $b$ is,
$b ϵ (- \infty, - 2]$
Hence, option 'C' is correct.