Question

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Evaluate this question of Limits

Class 11th Jee Mains

Question

If

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

Answer 1

ANSWER :-) 7

Solution :-)

Given that

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

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

So numerator

${x}^{2} - ax + b$

Should be zero at x = 1

Hence

$1 - a + b = 0 \\ \\ a = 1 + b$

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

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

So,

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

Now,

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

Which is your required answer .

Answer 2

that

As limit exists

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

and is equal to 5

Also at x = 1

denominator is zero .

to this limit to be exist

numerator

${x}^{2} - ax + b$

have to be zero when x = 1

Means

$1 - a + b = 0 \\ \\ a = 1 + b$

Put this value of a in given lmt.

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

So

a = -3 ( if we put value of be in a = 1 + b )

Hence

a + b = -7