QJEEM21Ma481e: JEE Main 2021 Mathematics - Class 11 JEEM21Ma481e Solution: The number of solutions of the equation is :

Question

Verified Solution • 2025–26 Session moderate

The number of solutions of the equation

{32^{{{\tan }^2}x}} + {32^{{{\sec }^2}x}} = 81,\,0 \le x \le {\pi \over 4}

is :

Accepted Answer

{(32)^{{{\tan }^2}x}} + {(32)^{{{\sec }^2}x}} = 81

\Rightarrow {(32)^{{{\tan }^2}x}} + {(32)^{1 + {{\tan }^2}x}} = 81

\Rightarrow {(32)^{{{\tan }^2}x}} = {{81} \over {33}}

taking log of base 32 both side,

\Rightarrow

tan²x =

{\log _{32}}\left( {{{81} \over {33}}} \right)

\Rightarrow

tan x =

\sqrt {{{\log }_{32}}\left( {{{81} \over {33}}} \right)}

As value of

\sqrt {{{\log }_{32}}\left( {{{81} \over {33}}} \right)}

belongs to (0, 1).

In interval

\,0 \le x \le {\pi \over 4}

only one solution.

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