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JEE Advanced 2014Paper 1
Exam
JEE Advanced
Year
2014
Paper
Paper 1
Subject
Mathematics
Chapter
Limits

JEE Advanced 2014 · Paper 1 · Mathematics · Limits

The largest value of the non - negative integer a for which _ x → 1...

The largest value of the non - negative integer a for which lim⁡x→1{−ax+sin⁡(x−1)+ax+sin⁡(x−1)−1}1−x1−x=14\lim_{x \to 1} {\left\{\dfrac{- ax + \sin \left(x - 1\right) + a}{x + \sin \left(x - 1\right) - 1}\right\}}^{\dfrac{1 - x}{1 - \sqrt{x}}} = \dfrac{1}{4} is _________

Correct answer

0

Step-by-step solution

lim⁡x→1(−ax+sin⁡(x−1)+ax+sin⁡(x−1)−1)1−x1−x=14{\lim_{x \to 1} \left(\dfrac{- ax + \sin \left(x - 1\right) + a}{x + \sin \left(x - 1\right) - 1}\right)}^{\dfrac{1 - x}{1 - \sqrt{x}}} = \dfrac{1}{4}
lim⁡x→1(sin⁡(x−1)(x−1)−asin⁡(x−1)(x−1)+1)\lim_{x \to 1} {\left(\dfrac{\dfrac{\sin \left(x - 1\right)}{\left(x - 1\right)} - a}{\dfrac{\sin \left(x - 1\right)}{\left(x - 1\right)} + 1}\right)}^{\left(1 + \sqrt{x}\right)} = \dfrac{1}{4} ⇒ {\left(\dfrac{1 - a}{2}\right)}^{2} = \dfrac{1}{4}<br/><br />⇒ a = 0, a = 2<br/><br />⇒ a = 0\left(sincefora = 2, basewillbecome - ve\right)
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