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JEE Main 202118 MarchMorning Shift
Exam
JEE Main
Year
2021
Date
18 March
Shift
Morning Shift
Subject
Mathematics
Chapter
Continuity and Differentiability

JEE Main 2021 · 18 March · Morning Shift · Mathematics · Continuity and Differentiability

If f(x) = \ cases (1)/(| x |) & ; & | x | ≥ 1 a x ^ 2 + b & ; & | x...

If f\left(x\right) = \left\{\begin{cases}\dfrac{1}{| x |} & ; & | x | \ge 1 \\ a{x}^{2} + b & ; & | x | < 1\end{cases}\right) is differentiable at every point of the domain, then the values ofa and bb are respectively:

Options

  1. A12,12\dfrac{1}{2}, \dfrac{1}{2}
  2. B12,−32\dfrac{1}{2}, - \dfrac{3}{2}
  3. C52,−32\dfrac{5}{2}, - \dfrac{3}{2}
  4. D−12,32- \dfrac{1}{2}, \dfrac{3}{2}

Correct answer

D. −12,32- \dfrac{1}{2}, \dfrac{3}{2}

Step-by-step solution

f\left(x\right) = \left\{{1∣x∣,∣x∣≥1ax2+b,∣x∣<1\begin{cases}\dfrac{1}{| x |}, & | x | \ge 1 \\ a{x}^{2} + b, & | x | < 1\end{cases}\right)

at x=1x = 1 function must be continuous

So, 1=a+b…(1)1 = a + b \ldots \left(1\right)

differentiability at x=1x = 1

(−1x2)x=1=(2⋅ax)x=1{\left(- \dfrac{1}{{x}^{2}}\right)}_{x = 1} = ( 2 \cdot ax{)}_{x = 1}

⇒−1=2a⇒a=−12⇒ - 1 = 2a ⇒ a = - \dfrac{1}{2}

Put in (1)⇒b=1+12=32( 1 ) ⇒ b = 1 + \dfrac{1}{2} = \dfrac{3}{2}

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