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  • When   <img alt="\lim _{y \rightarrow-1} \frac{y^2-p y+3 q}{y+1}=14" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7By%20%5Crightarrow-1%7D%20%5Cfrac%7By%5E2-p

When   \lim _{y \rightarrow-1} \frac{y^2-p y+3 q}{y+1}=14 , then which of the following is true?

Option: 1

p^{2}+q^{2}=281


Option: 2

p^{2}+q^{2}=181


Option: 3

p^{2}+q^{2}=218


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 _{y \rightarrow-1} \frac{y^2-p y+3 q}{y+1}=14

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

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

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

Now, derive the following:

 

 \begin{aligned} & \left.\left(y^2-p y+3 q\right)\right|_{y=-1}=0 \\ & (-1)^2-p(-1)+3 q=0 \\ & 1+p+3 q=0 \\ & -p=1+3 q \quad \ldots(i) \end{aligned}

Use the equation (i) to calculate the following:

\begin{aligned} & y^2-p y+3 q \\ & =y^2+(1+3 q) y+3 q \\ & =y^2+y+3 qy +3 q \\ & =y(y+1)+3 q(y+1) \\ & \therefore y^2-a y+2 q=(y+1)(y+3 q) \, \, \, \, \, \, \, ...(ii) \end{aligned}

Use the equation (ii) to rewrite the limit.

\begin{aligned} & \lim _{y \rightarrow-1} \frac{y^2-p y+3 q}{y+1}=14 \\ & \lim _{y \rightarrow-1} \frac{(y+1)(y+3 q)}{(y+1)}=14 \\ & \lim _{y \rightarrow-1}(y+3 q)=14 \quad[\quad y \neq-1] \\ & -1+3 q=14 \\ & 3 q=15 \\ & q=5 \quad \ldots(\text { iii }) \end{aligned}

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

\begin{aligned} & p \\ & =-1-3 q \\ & =-1-3 \times 5 \\ & =-16 \\ \end{aligned}

Now, find the values of the following expressions.

\begin{aligned} \cdot & (p+q)=(-16+5)=-11 \\ \cdot & (p-q)=(-16-5)=-21 \\ \cdot & \, \, p^2+q^2=\frac{(p+q)^2+(p-q)^2}{2}=\frac{(-11)^2+(-21)^2}{2}=\frac{121+441}{2}=\frac{562}{2}=281 \end{aligned}

 

Posted by

Irshad Anwar

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