Q27 / 83 · NTA Abhyas (JEE Main) 2020 · Mathematics · Trigonometric Equations

The equation $t a n^{4} x - 2 s e c^{2} x + a^{2} = 0$ will have at least one solution, if

A
$|a| \leq 4$
B
$|a| \leq 2$
Correct answer
C
$|a| \leq \sqrt{3}$
D
$|a| > 2$

Solution

t a n^{4} x - 2 s e c^{2} x + a^{2} = 0

t a n^{4} x - 2 (1 + t a n^{2} x) + a^{2} = 0

t a n^{4} x - 2 t a n^{2} x + 1 = 3 - a^{2}

{(t a n^{2} x - 1)}^{2} \geq 0 \Rightarrow 3 - a^{2} \geq 0

\Rightarrow a^{2} - 3 \leq 0 \Rightarrow |a| \leq \sqrt{3}

Hence, the correct answer is option: 'C'.

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