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  • When  <img alt="\lim _{x \rightarrow-1} \frac{x^2-a x+12 b}{x+1}=11" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crightarrow-1%7D%20%5Cfrac%7Bx%5E2-a

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

Option: 1

a=+v e, b=-v e


Option: 2

a=-v e, b=+v e


Option: 3

a=+v e, b=+v e


Option: 4

a=-v e, b=-v e


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-a x+12 b}{x+1}=11

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

  • When x\rightarrow -1 , it implies (x+1)\rightarrow 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-a x+12 b\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-a x+12 b \\ & =x^2+(1+12 b) x+12 b \\ & =x^2+x+12 b x+12 b \\ & =x(x+1)+12 b(x+1) \\ & \therefore x^2-a x+12 b=(x+1)(x+12 b) \end{aligned}

Use the equation (ii) to rewrite the limit.

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

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

\begin{aligned} & a \\ & =-1-12 b \\ & =-1-12 \\ & =-13 \\ \end{aligned}

Posted by

Ritika Harsh

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