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Question: If \(\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - ax + b}}{{x - 1}} = 5\) , then \(a{\text{ }} +...

If lim⁡x→1x2−ax+bx−1=5\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - ax + b}}{{x - 1}} = 5 , then a + ba{\text{ }} + {\text{ }}b is equal to
(A) −7 - 7
(B) −4 - 4
(C) 55
(D) 11

Explanation

Solution

Since the limit is finite quantity x−1x - 1 should be a factor of both numerator and denominator. So factorize the numerator. Then calculate values of a and b. After that we calculate a + ba{\text{ }} + {\text{ }}b. First we use the factorization method of the numerator. Then we put factors of the numerator in fraction given in question. Hence, we can find values of a and b.

Complete step-by-step answer:
From the value of the given limit we know that the limit is a finite quantity. So x−1x - 1 must be a factor of both numerator and denominator. Thus by factor theorem we know that for x=1x = 1 numerator should be 0.

12−a(1)+b=0 1−a+b=0......(1)  {1^2} - a(1) + b = 0 \\\ 1 - a + b = 0......(1) \\\

Also from equation (1) , b = a − 1b{\text{ }} = {\text{ }}a{\text{ }}-{\text{ }}1 substituting the value of ‘b’ in the numerator , we get :

x2−ax+b =x2−ax+(a−1) =x2−ax+a−1 =(x2−1)−a(x−1) =(x−1)(x+1)−a(x−1) =(x−1)(x+1−a)  {x^2} - ax + b \\\ = {x^2} - ax + (a - 1) \\\ = {x^2} - ax + a - 1 \\\ = ({x^2} - 1) - a(x - 1) \\\ = (x - 1)(x + 1) - a(x - 1) \\\ = (x - 1)(x + 1 - a) \\\ ⇒lim⁡x→1x2−ax+bx−1=5 ⇒lim⁡x→1(x−1)(x+1−a)x−1=5 ⇒lim⁡x→1(x+1−a)=5 ⇒1+1−a=5 \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - ax + b}}{{x - 1}} = 5 \\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{(x - 1)(x + 1 - a)}}{{x - 1}} = 5 \\\ \Rightarrow \mathop {\lim }\limits_{x \to 1} (x + 1 - a) = 5 \\\ \Rightarrow 1 + 1 - a = 5 \\\

Substituting this value of a, in equation (1) we get ,

⇒1+3+b=0 ⇒b=−4 \Rightarrow 1 + 3 + b = 0 \\\ \Rightarrow b = - 4 \\\

Now we have to find a+ba + b .
a+b=−3+(−4)=−7a + b = - 3 + ( - 4) = - 7

Hence (A) is the correct option.

Note: 1) Factor Theorem : If x−ax - a is a factor of any polynomial p(x), then p(a)=0.
2) Remainder Theorem : In order to obtain the remainder when a polynomial p (x) is divided by x−ax - a, substitute x=ax = a in the polynomial. Thus the remainder will be p(a).
3) The limit of a function f (x) when x tends to a is a value towards which f(x) approaches as x tends to a form either side. For a limit of a function to exist, the left hand limit (L . H . L) should be equal to the right hand limit (R . H . L). If the left hand limit is not equal to the right hand limit, we say the limit of function does not exist.
4) In order to evaluate algebraic limits we should use factorization.