Verified Solution • 2025–26 Session moderate

QJEEM21Ma736a: JEE Main 2021 Mathematics - Class 12 JEEM21Ma736a Solution: The minimum value of for which the equation has at least one…

The minimum value of α\alpha for which the
equation 4sin⁡x+11−sin⁡x=α{4 \over {\sin x}} + {1 \over {1 - \sin x}} = \alpha has at least one solution in (0,π2)\left( {0,{\pi \over 2}} \right) is .......

Step-by-Step Verified Solution

f(x)=4sin⁡x+11−sin⁡xf(x) = {4 \over {\sin x}} + {1 \over {1 - \sin x}}

Let sinx = t ∵\because x \in \left( {0,{\pi \over 2}} \right) \Rightarrow 0 < t < 1

f(t)=4t+11−tf(t) = {4 \over t} + {1 \over {1 - t}}

f′(t)=−4t2+1(1−t)2f'(t) = {{ - 4} \over {{t^2}}} + {1 \over {{{(1 - t)}^2}}}

=t2−4(1−t)2t2(1−t)2= {{{t^2} - 4{{(1 - t)}^2}} \over {{t^2}{{(1 - t)}^2}}}

=(t−2(1−t))(t+2(1−t))t2(1−t)2= {{(t - 2(1 - t))(t + 2(1 - t))} \over {{t^2}{{(1 - t)}^2}}}

=(3t−2)(2−t)t2(1−t)2= {{(3t - 2)(2 - t)} \over {{t^2}{{(1 - t)}^2}}}

fmin⁡{f_{\min }} at t=23t = {2 \over 3}

αmin⁡=f(23)=423+11−23{\alpha _{\min }} = f\left( {{2 \over 3}} \right) = {4 \over {{2 \over 3}}} + {1 \over {1 - {2 \over 3}}}

=6+3= 6 + 3

=9= 9

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