Question

I Challange Brainlians .

Evaluate This Question
related to limits.

class 11th

Jee Mains

If

$\displaystyle \lim_{x \to 0} \: \frac{ {x}^{2} - ax + b}{x - 1} = 5$
Then find out the value of a + b
​

Answer 1

Answer:

ANSWER :-) 7

Solution :-)

Given that

\displaystyle \lim_{x \to 1} \: \frac{ {x}^{2} - ax + b}{x - 1} = 5

x→1

lim

x−1

x

2

−ax+b

=5

As limit exists and is equal to 5 and also denominator is zero at x = 1

So numerator

{x}^{2} - ax + bx

2

−ax+b

Should be zero at x = 1

Hence

\begin{gathered}1 - a + b = 0 \\ \\ a = 1 + b\end{gathered}

1−a+b=0

a=1+b

Now putting the value of a from here to the given limit we get

\begin{gathered}\displaystyle \lim_{x \to 1} \: \frac{ {x}^{2} - ax + b}{x - 1} = 5 \\ \\ \implies \displaystyle \lim_{x \to 1} \: \frac{ {x}^{2} - (1 + b)x + b}{x - 1} = 5 \\ \\ \implies\displaystyle \lim_{x \to 1} \: \frac{ ({x}^{2} - x) - b(x - 1)}{x - 1} = 5 \\ \\ \implies\displaystyle \lim_{x \to 1} \: \frac{ (x - )(x - b)}{x - 1} = 5 \\ \\ \implies\displaystyle \lim_{x \to 1} \:(x - b) = 5 \\ \\ \implies1 - b = 5 \\ \\ \implies \pink{ \boxed{ \boxed{b = - 4}}}\end{gathered}

x→1

lim

x−1

x

2

−ax+b

=5

⟹

x→1

lim

x−1

x

2

−(1+b)x+b

=5

⟹

x→1

lim

x−1

(x

2

−x)−b(x−1)

=5

⟹

x→1

lim

x−1

(x−)(x−b)

=5

⟹

x→1

lim

(x−b)=5

⟹1−b=5

⟹

b=−4

So,

\blue{ \boxed{ \boxed{a = - 3}}}

a=−3

Now,

\red{ \boxed{ \boxed{a + b= - 7}}}

a+b=−7

Which is your required answer .