Question

The limiting position of the point of intersection of the straight lines 3x + 5y = 1 and $(2+c)x+5c^{2}y=1$ as c tends to one is:

• A. $(\frac{2}{5},-\frac{1}{25})$
• B. $(\frac{1}{2},-\frac{1}{10})$
• C. $(\frac{3}{8},-\frac{1}{40})$
• D. none

Answer 1

The correct option is A $(\frac{2}{5},-\frac{1}{25})$

The given Straight lines are $3x+5y=1$ ...$\left(1\right)$

and $(2+c)x+5c^{2}y=1$ ....$\left(2\right)$

The value of $y$ from equation $\left(1\right)$ is $\frac{1-3x}{5}$

By putting it in equation $\left(2\right)$, we get,

$\Rightarrow (2+c)x+5c^2\left(\frac{1-3x}{5}\right)-1=0$ ...$\left(2\right)$

$\Rightarrow (2+c)x+c^2(1-3x)-1=0$

$\Rightarrow x=\frac{1-c^2}{2+c-3c^2}$ ..$\left(3\right)$

Now if we put, $\underset{c\to 1}{lim}$ in the equation $\left(3\right)$, we get,

$\Rightarrow \underset{c\to 1}{lim}Eq\left(3\right)=\frac{1-(1{)}^2}{2+\left(1\right)-3(1{)}^2}$

as we can see $\underset{c\to 1}{lim}Eq\left(3\right)$ is a $\frac{0}{0}$ form, Hence using L'Hospital rule of finding limit.

$\Rightarrow \underset{c\to 1}{lim}Eq\left(3\right)=\frac{-2c}{1-6c}=\frac{-2\left(1\right)}{1-6\left(1\right)}=\frac{2}{5}$

After using L'hospital rule The $\underset{c\to 1}{lim}Eq\left(3\right)$ becomes $x=\frac{2}{5}$

From Equation $\left(1\right)$, we know that $y=\frac{1-3x}{5}$,

$\Rightarrow y=\frac{-1}{25}$

So Correct option is $A$