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Limits Continuity and Differentiability question

2022 · 29 Jul · Shift 2 · Q26
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  5. /2022 · 29 Jul · Shift 2 · Q26

Limits Continuity and Differentiability question

2022 · 29 Jul · Shift 2 · Q26

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
 Let the function f(x)={log⁡e(1+5x)−log⁡e(1+αx)x; if xeq010; if x=0 be continuous at x=0.\text { Let the function } f(x)=\left\{\begin{array}{cl} \frac{\log _{e}(1+5 x)-\log _{e}(1+\alpha x)}{x} & ;\text { if } x eq 0 \\ 10 & ; \text { if } x=0 \end{array} \text { be continuous at } x=0 .\right. Let the function f(x)={xloge​(1+5x)−loge​(1+αx)​10​; if xeq0; if x=0​ be continuous at x=0. Then α\alphaα is equal to
  1. A
    10
  2. B
    −-− 10
  3. C
    5
  4. D
    −-− 5
View written solutionFree

Correct answer: D

  1. For continuity at x=0x=0x=0, we need
lim⁡x→0f(x)=f(0)=10.\lim_{x\to 0} f(x)=f(0)=10.x→0lim​f(x)=f(0)=10.
  1. For x≠0x\ne 0x=0,
f(x)=ln⁡(1+5x)−ln⁡(1+αx)x.f(x)=\frac{\ln(1+5x)-\ln(1+\alpha x)}{x}.f(x)=xln(1+5x)−ln(1+αx)​.

So we evaluate

lim⁡x→0ln⁡(1+5x)−ln⁡(1+αx)x.\lim_{x\to 0}\frac{\ln(1+5x)-\ln(1+\alpha x)}{x}.x→0lim​xln(1+5x)−ln(1+αx)​.
  1. Use the standard expansion near x=0x=0x=0:
ln⁡(1+kx)=kx+o(x).\ln(1+kx)=kx+o(x).ln(1+kx)=kx+o(x).

Hence,

ln⁡(1+5x)−ln⁡(1+αx)=5x−αx+o(x)=(5−α)x+o(x).\ln(1+5x)-\ln(1+\alpha x)=5x-\alpha x+o(x)=(5-\alpha)x+o(x).ln(1+5x)−ln(1+αx)=5x−αx+o(x)=(5−α)x+o(x).

Dividing by xxx,

lim⁡x→0ln⁡(1+5x)−ln⁡(1+αx)x=5−α.\lim_{x\to 0}\frac{\ln(1+5x)-\ln(1+\alpha x)}{x}=5-\alpha.x→0lim​xln(1+5x)−ln(1+αx)​=5−α.
  1. Since continuity requires this limit to equal 101010,
5−α=10.5-\alpha=10.5−α=10.

Therefore,

α=−5.\alpha=-5.α=−5.
  1. Checking options:
  • A: 101010 ❌
  • B: −10-10−10 ❌
  • C: 555 ❌
  • D: −5-5−5 ✅

Therefore, the correct answer is D.

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