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Limits Continuity and Differentiability question

2021 · 26 Feb · Shift 2 · Q26
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  5. /2021 · 26 Feb · Shift 2 · Q26

Limits Continuity and Differentiability question

2021 · 26 Feb · Shift 2 · Q26

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
Let f(x) be a differentiable function at x = a with f'(a) = 2 and f(a) = 4. Then lim⁡x→axf(a)−af(x)x−a\mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}x→alim​x−axf(a)−af(x)​ equals :
  1. A
    4 −-− 2a
  2. B
    2a + 4
  3. C
    a + 4
  4. D
    2a −-− 4
View written solutionFree

Correct answer: A

  1. We need to evaluate
L=lim⁡x→axf(a)−af(x)x−a.L=\lim_{x\to a}\frac{x f(a)-a f(x)}{x-a}.L=x→alim​x−axf(a)−af(x)​.
  1. Given: f(a)=4,f′(a)=2.f(a)=4, \qquad f'(a)=2.f(a)=4,f′(a)=2. So,
L=lim⁡x→a4x−af(x)x−a.L=\lim_{x\to a}\frac{4x-a f(x)}{x-a}.L=x→alim​x−a4x−af(x)​.
  1. Rewrite the numerator by adding and subtracting af(a)af(a)af(a):
xf(a)−af(x)=f(a)(x−a)−a(f(x)−f(a)).xf(a)-af(x)=f(a)(x-a)-a\big(f(x)-f(a)\big).xf(a)−af(x)=f(a)(x−a)−a(f(x)−f(a)).

Hence,

L=lim⁡x→a[f(a)−af(x)−f(a)x−a].L=\lim_{x\to a}\left[ f(a)-a\frac{f(x)-f(a)}{x-a}\right].L=x→alim​[f(a)−ax−af(x)−f(a)​].
  1. Now use the differentiability of fff at x=ax=ax=a:
lim⁡x→af(x)−f(a)x−a=f′(a)=2.\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)=2.x→alim​x−af(x)−f(a)​=f′(a)=2.

Also, f(a)=4f(a)=4f(a)=4. Therefore,

L=4−a⋅2=4−2a.L=4-a\cdot 2=4-2a.L=4−a⋅2=4−2a.
  1. Compare with the options:
  • A: 4−2a4-2a4−2a ✅
  • B: 2a+42a+42a+4
  • C: a+4a+4a+4
  • D: 2a−42a-42a−4

So the correct option is A.

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