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JEE Main 202126 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
26 February
Shift
Evening Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2021 · 26 February · Evening Shift · Mathematics · Limits

Let f(x) be a differentiable function at x = a with f ^ ' (a) = 2...

Let f(x)beadifferentiablefunctionatf\left(x\right) be a differentiable function atx = awithwith{f}^{'}\left(a\right) = 2 and f(a)=4f\left(a\right) = 4. Then lim⁡x→axf(a)−af(x)x−a\lim_{x \to a}\dfrac{xf\left(a\right) - af\left(x\right)}{x - a} equals:

Options

  1. Aa+4a + 4
  2. B2a−42a - 4
  3. C4−2a4 - 2a
  4. D2a+42a + 4

Correct answer

C. 4−2a4 - 2a

Step-by-step solution

f′(a)=2,f(a)=4{f}^{'}\left(a\right) = 2, f\left(a\right) = 4

lim⁡x→axf(a)−af(x)x−a\lim_{x \to a}\dfrac{xf\left(a\right) - af\left(x\right)}{x - a}

⇒lim⁡x→af(a)−af′(x)1⇒ \lim_{x \to a}\dfrac{f\left(a\right) - a{f}^{'}\left(x\right)}{1} (L'Hospitals rule)

=f(a)−af′(a)= f\left(a\right) - a{f}^{'}\left(a\right)

=4−2a= 4 - 2a

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