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Question: If we have a function as \[f\left( x \right) = \dfrac{{3{x^2} + ax + a + 1}}{{{x^2} + x - 2}}\] , th...

If we have a function as f(x)=3x2+ax+a+1x2+x−2f\left( x \right) = \dfrac{{3{x^2} + ax + a + 1}}{{{x^2} + x - 2}} , then which of the following can be correct?
(A) lim⁡x→1f(x)\mathop {\lim }\limits_{x \to 1} f\left( x \right) exists ⇒a=−2 \Rightarrow a = - 2
(B) lim⁡x→−2f(x)\mathop {\lim }\limits_{x \to - 2} f\left( x \right) exists ⇒a=13 \Rightarrow a = 13
(C) If the limit exists,lim⁡x→1f(x)=43\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \dfrac{4}{3}
(D) If the limit exists,lim⁡x→−2f(x)=−13\mathop {\lim }\limits_{x \to - 2} f\left( x \right) = - \dfrac{1}{3}

Explanation

Solution

In this question, we have the function ff. We need to find out the limits for each of the options given. For finding that we first need to evaluate the limits at the given points where it is given, then find out the value of a. For the other two options we need to find out the value of function f as x approaches to given numbers.

Complete step-by-step solution:
It is given that the function f is defined as,
f(x)=3x2+ax+a+1x2+x−2f\left( x \right) = \dfrac{{3{x^2} + ax + a + 1}}{{{x^2} + x - 2}}
We need to find out the limits for the given conditions.
For option (A)
As x tends to 11 the denominator tends to 00, so the numerator should also tend to 00.
lim⁡x→13x2+ax+a+1=0\mathop {\lim }\limits_{x \to 1} 3{x^2} + ax + a + 1 = 0
Substituting the limit values we get,
3×12+a×1+a+1=03 \times {1^2} + a \times 1 + a + 1 = 0
Simplifying we get,
3+a+a+1=03 + a + a + 1 = 0
Adding terms,
2a+4=02a + 4 = 0
Hence,
2a=−42a = - 4
a=−42=−2a = \dfrac{{ - 4}}{2} = - 2
Hence, lim⁡x→1f(x)\mathop {\lim }\limits_{x \to 1} f\left( x \right) exists ⇒a=−2 \Rightarrow a = - 2 is true.
For option (B)
As x tends to −2 - 2 the denominator tends to 00, so the numerator should also tend to 00.
lim⁡x→−23x2+ax+a+1=0\mathop {\lim }\limits_{x \to - 2} 3{x^2} + ax + a + 1 = 0
Substituting the limit values we get,
3×(−2)2+a×(−2)+a+1=03 \times {\left( { - 2} \right)^2} + a \times \left( { - 2} \right) + a + 1 = 0
Simplifying we get,
3×4−2a+a+1=03 \times 4 - 2a + a + 1 = 0
Add and subtract the terms,
−a+13=0- a + 13 = 0
Hence,
a=13a = 13
Hence, lim⁡x→−2f(x)\mathop {\lim }\limits_{x \to - 2} f\left( x \right) exists ⇒a=13 \Rightarrow a = 13is true.
For option (C)
We have, lim⁡x→1f(x)\mathop {\lim }\limits_{x \to 1} f\left( x \right) exists ⇒a=−2 \Rightarrow a = - 2
Now, if the limit exists,
lim⁡x→1f(x)=lim⁡x→13x2+(−2)x+(−2)+1x2+x−2\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{3{x^2} + \left( { - 2} \right)x + \left( { - 2} \right) + 1}}{{{x^2} + x - 2}}
Solving we get,
⇒lim⁡x→13x2−2x−1x2+x−2\Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{3{x^2} - 2x - 1}}{{{x^2} + x - 2}}
Using the relation between root and coefficients, $$ \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{3{x^2} - 3x + x - 1}}{{{x^2} + 2x - x - 2}}$$ Simplifying we get, $$ \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{3x\left( {x - 1} \right) + 1\left( {x - 1} \right)}}{{x\left( {x + 2} \right) - 1\left( {x + 2} \right)}}$$ $$ \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {3x + 1} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}$$ Hence, $$ \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{3x + 1}}{{x + 2}}$$ $$ \Rightarrow \dfrac{4}{3}$$ Hence, if the limit exists,$$\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \dfrac{4}{3}$$ is true. For option (D) We have, $$\mathop {\lim }\limits_{x \to - 2} f\left( x \right)$$ exists $$ \Rightarrow a = 13$$ Now, if the limit exists, $$\mathop {\lim }\limits_{x \to - 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{3{x^2} + 13 \times x + 13 + 1}}{{{x^2} + x - 2}}$$ Simplifying we get, $$ \Rightarrow \mathop {\lim }\limits_{x \to - 2} \dfrac{{3{x^2} + 13x + 14}}{{{x^2} + x - 2}}$$ Using the relation between root and coefficients, $$$$
⇒lim⁡x→−23x2+6x+7x+14x2+2x−x−2\Rightarrow \mathop {\lim }\limits_{x \to - 2} \dfrac{{3{x^2} + 6x + 7x + 14}}{{{x^2} + 2x - x - 2}}
Factorize we get,
⇒lim⁡x→−23x(x+2)+7(x+2)x(x+2)−1(x+2)\Rightarrow \mathop {\lim }\limits_{x \to - 2} \dfrac{{3x\left( {x + 2} \right) + 7\left( {x + 2} \right)}}{{x\left( {x + 2} \right) - 1\left( {x + 2} \right)}}
⇒lim⁡x→−2(x+2)(3x+7)(x+2)(x−1)\Rightarrow \mathop {\lim }\limits_{x \to - 2} \dfrac{{\left( {x + 2} \right)\left( {3x + 7} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}}
Hence,
⇒lim⁡x→−23x+7x−1\Rightarrow \mathop {\lim }\limits_{x \to - 2} \dfrac{{3x + 7}}{{x - 1}}
⇒3×(−2)+7−3\Rightarrow \dfrac{{3 \times \left( { - 2} \right) + 7}}{{ - 3}}
Solving the terms,
⇒−6+7−3\Rightarrow \dfrac{{ - 6 + 7}}{{ - 3}}
⇒−13\Rightarrow - \dfrac{1}{3}
Hence, If the limit exists, lim⁡x→−2f(x)=−13\mathop {\lim }\limits_{x \to - 2} f\left( x \right) = - \dfrac{1}{3}is true.
Options (A), (B), (C), and (D) is correct.

Note: We have to know that, the sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient,
(r1+r2)=−ba\left( {{r_1} + {r_2}} \right) = - \dfrac{b}{a}
The product of the roots of a quadratic equation is equal to the constant term (the third term), divided by the leading coefficient,
(r1⋅r2)=ca\left( {{r_1} \cdot {r_2}} \right) = \dfrac{c}{a}