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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Limits

NDA 2020 · Mathematics · Limits

If _ x arrow 1 (x^4-1)/(x-1)= _ x arrow k (x^3-k^3)/(x^2-k^2) ,...

If lim⁡x→1x4−1x−1=lim⁡x→kx3−k3x2−k2\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}, where k≠0k \neq 0, then what is the value of kk ?

Options

  1. A23\frac{2}{3}
  2. B43\frac{4}{3}
  3. C83\frac{8}{3}
  4. D4

Correct answer

C. 83\frac{8}{3}

Step-by-step solution

lim⁡x→1x4−1x−1=lim⁡x→1(x−1)(x+1)(x2+1)(x−1)=lim⁡x→1(x+1)(x2+1)=(1+1)(1+1)=4\begin{aligned} \lim _{x \rightarrow 1} \frac{x^4-1}{x-1} & =\lim _{x \rightarrow 1} \frac{(x-1)(x+1)\left(x^2+1\right)}{(x-1)} \\ & =\lim _{x \rightarrow 1}(x+1)\left(x^2+1\right)=(1+1)(1+1)=4 \end{aligned} Now, lim⁡x→kx3−k3x2−k2=4lim⁡x→k(x−k)(x2+k2+xk)(x−k)(x+k)=4\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}=4\lim _{x \rightarrow k} \frac{(x-k)\left(x^2+k^2+x k\right)}{(x-k)(x+k)}=4lim⁡x→k(x2+k2+xk)(x+k)=4k2+k2+k⋅kk+k=4⇒32k=4⇒k=83\begin{aligned} & \lim _{x \rightarrow k} \frac{\left(x^2+k^2+x k\right)}{(x+k)}=4 \\ & \frac{k^2+k^2+k \cdot k}{k+k}=4 \Rightarrow \frac{3}{2} k=4 \Rightarrow k=\frac{8}{3}\end{aligned}$$
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