Verified Solution • 2025–26 Session moderate

QJEEM17Mae17e: JEE Main 2017 Mathematics - Class 12 JEEM17Mae17e Solution: equals

lim⁡x→π2cot⁡x−cos⁡x(π−2x)3\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}} equals

Multiple Choice Options

  • A 116{1 \over {16}}
  • B 18{1 \over 8}
  • C 14{1 \over {4}}
  • D 124{1 \over {24}}

Step-by-Step Verified Solution

lim⁡x→π2cot⁡x−cos⁡x(π−2x)3\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}

= lim⁡x→π218.cos⁡x(1−sin⁡x)sin⁡x(π2−x)3\mathop {\lim }\limits_{x \to {\pi \over 2}} {1 \over 8}.{{\cos x\left( {1 - \sin x} \right)} \over {\sin x{{\left( {{\pi \over 2} - x} \right)}^3}}}

Put π2−x{{\pi \over 2} - x} = t

⇒\Rightarrow x →\to π2{{\pi \over 2}}

t →\to 0

= lim⁡t→018.cos⁡(π2−t)(1−sin⁡(π2−t))sin⁡(π2−t)(π2−π2+t)3\mathop {\lim }\limits_{t \to 0} {1 \over 8}.{{\cos \left( {{\pi \over 2} - t} \right)\left( {1 - \sin \left( {{\pi \over 2} - t} \right)} \right)} \over {\sin \left( {{\pi \over 2} - t} \right){{\left( {{\pi \over 2} - {\pi \over 2} + t} \right)}^3}}}

= 18lim⁡t→0sin⁡t(1−cos⁡t)cos⁡t(t)3\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {1 - \cos t} \right)} \over {\cos t{{\left( t \right)}^3}}}

= 18lim⁡t→0sin⁡t(2sin⁡2t2)cos⁡t(t)3\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {2{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}

= 14lim⁡t→0sin⁡t(sin⁡2t2)cos⁡t(t)3\mathop {{1 \over 4}\lim }\limits_{t \to 0} {{\sin t\left( {{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}

= lim⁡t→0(sin⁡tt)(sin⁡t2t2)21cos⁡t14\mathop {\lim }\limits_{t \to 0} \left( {{{\sin t} \over t}} \right){\left( {{{\sin {t \over 2}} \over {{t \over 2}}}} \right)^2}{1 \over {\cos t}}{1 \over 4}

= 14×14{1 \over 4} \times {1 \over 4}

= 116{1 \over {16}}

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