Question

If we have a function as $f\left( x \right)=\dfrac{{{x}^{2}}-bx+25}{{{x}^{2}}-7x+10}$ for $x\ne 5$ is continuous at x = 5, then the value of f(5) is
\[\begin{align}
  & A.0 \\
 & B.1 \\
 & C.2 \\
 & D.3 \\
\end{align}\]

Answer 1

Hint: In this question, we are given f(x) to be continuous at x = 5 and we have to find the value of f(5). For this, we will have to find the value of $\displaystyle \lim_{x \to 5}f\left( x \right)$. Since putting 5 will give us 0 in the denominator, so f(x) should be in indeterminate form and hence the numerator should be zero. From this, we will find the value of b. After that, we will factorize numerator and denominator separately and then cancel out common factors. Putting the value of 5 in simplified f(x) will give us the value of f(5). We will use splitting the middle term method for factorization.

Complete step by step solution:
Here, we are given $f\left( x \right)=\dfrac{{{x}^{2}}-bx+25}{{{x}^{2}}-7x+10}$ for $x\ne 5$.
We need to find the value of f(5). Hence, we need to evaluate $\displaystyle \lim_{x \to 5}f\left( x \right)=\displaystyle \lim_{x \to 5}\dfrac{{{x}^{2}}-bx+25}{{{x}^{2}}-7x+10}$.
Now, let us put value of x as 5 in f(x), we get,
$\dfrac{{{\left( 5 \right)}^{2}}-b\left( 5 \right)+25}{{{\left( 5 \right)}^{2}}-7\left( 5 \right)+10}\Rightarrow \dfrac{50-5b}{0}$.
Therefore, $\displaystyle \lim_{x \to 5}f\left( x \right)=\dfrac{50-5b}{0}$.
So, it should be in indeterminate form that is $\left( \dfrac{0}{0}\text{form} \right)$. Hence, 50-5b should be equal to zero.
Therefore,
\[\begin{align}
  & \Rightarrow 50-5b=0 \\
 & \Rightarrow 5b=50 \\
 & \Rightarrow b=10 \\
\end{align}\]
Therefore, we get $f\left( x \right)=\dfrac{{{x}^{2}}-10x+25}{{{x}^{2}}-7x+10}$.
We still need to evaluate f(5) which means we need to evaluate $\displaystyle \lim_{x \to 5}\dfrac{{{x}^{2}}-10x+25}{{{x}^{2}}-7x+10}$.
Putting the value of x as 5 gives us $\dfrac{0}{0}$ form, so let us first simplify the equation.
$\displaystyle \lim_{x \to 5}f\left( x \right)=\displaystyle \lim_{x \to 5}\dfrac{{{x}^{2}}-10x+25}{{{x}^{2}}-7x+10}$.
Splitting the middle term in numerator as -10 = -5-5 and splitting the middle term in denominator as -7 = -5-2 we get:
\[\begin{align}
  & \Rightarrow \displaystyle \lim_{x \to 5}f\left( x \right)=\displaystyle \lim_{x \to 5}\dfrac{{{x}^{2}}-5x-5x+25}{{{x}^{2}}-5x-2x+10} \\
 & \Rightarrow \displaystyle \lim_{x \to 5}f\left( x \right)=\displaystyle \lim_{x \to 5}\dfrac{x\left( x-5 \right)-5\left( x-5 \right)}{x\left( x-5 \right)-2\left( x-5 \right)} \\
 & \Rightarrow \displaystyle \lim_{x \to 5}f\left( x \right)=\displaystyle \lim_{x \to 5}\dfrac{\left( x-5 \right)\left( x-5 \right)}{\left( x-2 \right)\left( x-5 \right)} \\
\end{align}\]
Cancelling out (x-5) from numerator and denominator we get:
\[\Rightarrow \displaystyle \lim_{x \to 5}f\left( x \right)=\displaystyle \lim_{x \to 5}\dfrac{\left( x-5 \right)}{\left( x-2 \right)}\]
Since, function f(x) is continuous at x = 5, therefore putting x = 5 in above equation we get:
\[\begin{align}
  & \Rightarrow f\left( 5 \right)=\dfrac{\left( 5-5 \right)}{\left( 5-2 \right)} \\
 & \Rightarrow f\left( 5 \right)=\dfrac{\left( 0 \right)}{\left( 3 \right)} \\
 & \Rightarrow f\left( 5 \right)=0 \\
\end{align}\]
Hence the value of f(5) is equal to 0. Hence, option A is the correct answer.

Note: Students should note that, we have considered $\displaystyle \lim_{x \to 5}f\left( x \right)=f\left( 5 \right)$ because f(x) is continuous at f(5). Initially (before factorization) we could say function was discontinuous but it was removable discontinuously and hence, we removed it by simplifying the function. While splitting the middle term for ${{x}^{2}}+ax+c$, we had to find two values p, q such that p+q = a and pq = c. For ${{x}^{2}}-10x+25$ we had chosen -5,-5 because $-5-5=-10\text{ and }\left( -5 \right)\times \left( -5 \right)=25$. For, ${{x}^{2}}-7x+10$ we had chosen -5,-2 because $-5-2=-7\text{ and }\left( -5 \right)\times \left( -2 \right)=10$.