Question

If \[f\left( x \right)=\dfrac{{{x}^{2}}-9}{{{x}^{2}}-2x-3},x\ne 3\] is continuous at x=3, then which one of the following is correct?
a) f(3)=0
b) f(3)=1.5
c) f(3)=3
d) f(3)=-1.5

Answer 1

Hint:Simplify the numerator and denominator of $f(x)$ and cancel out the like terms. By using limits and derivatives, apply the limit $x\rightarrow3$ and find the value of $f(3)$.

Complete step-by-step answer:
Given, \[f\left( x \right)=\dfrac{{{x}^{2}}-9}{{{x}^{2}}-2x-3}\]
From the question it’s told that function is continuous at x=3 i.e. there will be no breakage of graph.
\[\begin{align}
  & \therefore \underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=f\left( 3 \right) \\
 & \Rightarrow \underset{x\to 3}{\mathop{\lim }}\,\dfrac{{{x}^{2}}-9}{{{x}^{2}}-2x-3}......\left( 1 \right) \\
\end{align}\]
We know \[\left( {{x}^{2}}-9 \right)\], can be written as \[\left( {{x}^{2}}-{{3}^{2}} \right)\]
where, \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]
 \[\therefore {{x}^{2}}-{{3}^{2}}=\left( x-3 \right)\left( x+3 \right)......\left( 2 \right)\]
Similarly, \[{{x}^{2}}+2x-3\] can be simplified by finding the roots we can find the roots by using the quadratic equation, which is of the form \[a{{x}^{2}}+bx+c=0\].
\[\therefore \]Comparing \[\left( a{{x}^{2}}+bx+c \right)\]and \[\left( {{x}^{2}}+2x-3 \right)\]we can find that a=1, b=2 and c=-3.
Substituting the values in the quadratic equation,
\[\begin{align}
  & \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\Rightarrow \dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 1\times \left( -3 \right)}}{2\times 1} \\
 & =\dfrac{-2\pm \sqrt{4+12}}{2}=\dfrac{-2\pm \sqrt{16}}{2}=\dfrac{-2\pm 4}{2} \\
\end{align}\]
\[\therefore \]We get two roots as \[\left( \dfrac{-2+4}{2} \right)\]and \[\left( \dfrac{-2-4}{2} \right)\]
\[\therefore \]Roots are (1, -3).
\[\therefore \]\[\left( {{x}^{2}}+2x-3 \right)\]becomes \[\left( x-1 \right)\left( x+3 \right).......(3)\]
Substitute (2) and (3) in equation (1)
\[\Rightarrow \underset{x\to 3}{\mathop{\lim }}\,\dfrac{{{x}^{2}}-9}{{{x}^{2}}+2x-3}=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\left( x+3 \right)\left( x-3 \right)}{\left( x-1 \right)\left( x+3 \right)}\]
Cancel out $(x+3)$ in the numerator and denominator
= \[\underset{x\to 3}{\mathop{\lim }}\,\dfrac{x-3}{x-1}\]
Put x=3
\[=\dfrac{3-3}{3-1}=\dfrac{0}{2}=0\]
Any fraction with zero as numerator and any number as denominator is always equal to zero.
\[\therefore f\left( 3 \right)=0\]
Hence, the correct option is (a) \[f\left( 3 \right)=0\]

Note: Find the roots of both numerator and denominator and then apply f(3) where x=3, to get the desired output.For these types of problems always try to factorize the given equation and simplify to cancel any common terms of both in numerator and denominator