QJEEM21Ma069d: JEE Main 2021 Mathematics - Class 12 JEEM21Ma069d Solution

Question

Verified Solution • 2025–26 Session moderate

If

\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}

and

\beta = \mathop {\lim }\limits_{x \to 0 } {(\cos x)^{\cot x}}

are the roots of the equation, ax² + bx

-

4 = 0, then the ordered pair (a, b) is :

Accepted Answer

\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{{{\tan }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}};{0 \over 0}

form

Using L Hospital rule

\alpha = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\tan }^2}x{{\sec }^2}x - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}

\alpha

=

-

4

\beta = \mathop {\lim }\limits_{x \to 0} {(\cos x)^{\cot x}} = {e^{\mathop {\lim }\limits_{x \to 0} {{(\cos x - 1)} \over {\tan x}}}}

\beta = {e^{\mathop {\lim }\limits_{x \to 0} {{ - (1 - \cos x)} \over {{x^2}}}.{{{x^2}} \over {{{\left( {{{\tan x} \over x}} \right)}^x}}}}}

\beta = {e^{\mathop {\lim }\limits_{x \to 0} \left( {{{ - 1} \over 2}} \right).{x \over 1}}} = {e^0} \Rightarrow \beta = 1

\alpha

=

-

4;

\beta

= 1

If ax² + bx

-

4 = 0 are the roots then

16a

-

4b

-

4 = 0 & a + b

-

4 = 0

\Rightarrow

a = 1 & b = 3

Step-by-Step Verified Solution