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Limits Continuity and Differentiability question

2019 · 10 Apr · Shift 1 · Q42
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Limits Continuity and Differentiability question

2019 · 10 Apr · Shift 1 · Q42

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If lim⁡x→1x4−1x−1=lim⁡x→kx3−k3x2−k2\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to k} {{{x^3} - {k^3}} \over {{x^2} - {k^2}}}x→1lim​x−1x4−1​=x→klim​x2−k2x3−k3​, then k is :
  1. A
    32{3 \over 2}23​
  2. B
    83{8 \over 3}38​
  3. C
    43{4 \over 3}34​
  4. D
    38{3 \over 8}83​
View written solutionFree

Correct answer: B

  1. Evaluate the first limit

We need

lim⁡x→1x4−1x−1.\lim_{x\to 1} \frac{x^4-1}{x-1}.x→1lim​x−1x4−1​.

Factor x4−1x^4-1x4−1:

x4−1=(x−1)(x+1)(x2+1).x^4-1=(x-1)(x+1)(x^2+1).x4−1=(x−1)(x+1)(x2+1).

So,

x4−1x−1=(x+1)(x2+1).\frac{x^4-1}{x-1}=(x+1)(x^2+1).x−1x4−1​=(x+1)(x2+1).

Now substitute x=1x=1x=1:

lim⁡x→1x4−1x−1=(1+1)(12+1)=2⋅2=4.\lim_{x\to 1} \frac{x^4-1}{x-1}=(1+1)(1^2+1)=2\cdot 2=4.x→1lim​x−1x4−1​=(1+1)(12+1)=2⋅2=4.
  1. Evaluate the second limit

We need

lim⁡x→kx3−k3x2−k2.\lim_{x\to k} \frac{x^3-k^3}{x^2-k^2}.x→klim​x2−k2x3−k3​.

Factor numerator and denominator:

x3−k3=(x−k)(x2+xk+k2),x^3-k^3=(x-k)(x^2+xk+k^2),x3−k3=(x−k)(x2+xk+k2), x2−k2=(x−k)(x+k).x^2-k^2=(x-k)(x+k).x2−k2=(x−k)(x+k).

Thus,

x3−k3x2−k2=x2+xk+k2x+k.\frac{x^3-k^3}{x^2-k^2}=\frac{x^2+xk+k^2}{x+k}.x2−k2x3−k3​=x+kx2+xk+k2​.

Taking x→kx\to kx→k,

lim⁡x→kx3−k3x2−k2=k2+k2+k2k+k=3k22k.\lim_{x\to k} \frac{x^3-k^3}{x^2-k^2} =\frac{k^2+k^2+k^2}{k+k} =\frac{3k^2}{2k}.x→klim​x2−k2x3−k3​=k+kk2+k2+k2​=2k3k2​.

For k≠0k\neq 0k=0,

lim⁡x→kx3−k3x2−k2=3k2.\lim_{x\to k} \frac{x^3-k^3}{x^2-k^2}=\frac{3k}{2}.x→klim​x2−k2x3−k3​=23k​.
  1. Equate the two limits

Given,

lim⁡x→1x4−1x−1=lim⁡x→kx3−k3x2−k2.\lim_{x\to 1} \frac{x^4-1}{x-1} = \lim_{x\to k} \frac{x^3-k^3}{x^2-k^2}.x→1lim​x−1x4−1​=x→klim​x2−k2x3−k3​.

So,

4=3k2.4=\frac{3k}{2}.4=23k​.

Hence,

k=83.k=\frac{8}{3}.k=38​.
  1. Check options
  • A: 32\frac{3}{2}23​
  • B: 83\frac{8}{3}38​
  • C: 43\frac{4}{3}34​
  • D: 38\frac{3}{8}83​

Therefore, the correct option is

83.\boxed{\frac{8}{3}}.38​​.
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