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When   \lim _{x \rightarrow-1} \frac{x^2-m x+12 n}{x+1}=11 , then which of the following is true?

Option: 1

m^{2}-n^{2}=186


Option: 2

m^{2}-n^{2}=168


Option: 3

m^{2}-n^{2}=64


Option: 4

Cannot be determined


Answers (1)

best_answer

Apply the L' Hospital's Rule, when  \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}, \text { etc. }  [an intermediate form] in the following way

  • Differentiate \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}, provided that the limit \lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)} exists.

  • Otherwise, go on differentiating till the limit with the determinate form is achieved.

The provided limit is

\lim _{x \rightarrow-1} \frac{x^2-m x+12 n}{x+1}=11

Here use the L’ Hospital rule and also refer to the following.

  • When x\rightarrow -1, it implies (x+1)=0 .

  • The limit will take the form of   \frac{0}{0}   , when x\rightarrow -1 .

Now, derive the following:

\begin{aligned} & \left.\left(x^2-m x+12 n\right)\right|_{x--1}=0 \\ & (-1)^2-a(-1)+12 b=0 \\ & 1+a+12 b=0 \\ & -a=1+12 b \quad \ldots(i) \end{aligned}

Use the equation (i) to calculate the following:

\begin{aligned} & x^2-m x+12 n \\ & =x^2+(1+12 n) x+12 n \\ & =x^2+x+12 n x+12 n \\ & =x(x+1)+12 n(x+1) \\ & \therefore x^2-m x+12 n=(x+1)(x+12 n) \end{aligned}

Use the equation (ii) to rewrite the limit.

\begin{aligned} & \lim _{x \rightarrow-1} \frac{x^2-m x+12 n}{x+1}=11 \\ & \lim _{x \rightarrow-1} \frac{(x+1)(x+12 n)}{(x+1)}=11 \\ & \lim _{x \rightarrow-1}(x+12 n)=11 \quad[0 \quad x \neq-1] \\ & -1+12 n=11 \\ & 12 n=12 \\ & n=1 \quad \ldots(\text { iii }) \end{aligned}

From the equations (ii) and (iii) the following is evident.

m\\=1-12n\\=-1-12\\=-3

Now, find the values of the following expressions.

  • (m+n)=(-13+1)=-12\\
  • (m-n)=(-13-1)=-14\\
  • m^{2}-n^{2}=(m+n)(m-n)=(-12)\times(-14)=168
Posted by

Ritika Kankaria

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