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  • Answer please! - Limit , continuity and differentiability - JEE Main-18

If \lim_{x\rightarrow 1}\frac{x^{2}-ax+b}{x-1}=5 , then a+b is equal to : 

  • Option 1)

    -4

  • Option 2)

    5

  • Option 3)

    -7

  • Option 4)

    1

Answers (1)

\lim_{x\rightarrow 1}\frac{x^{2}-ax+b}{x-1}=5

As x\rightarrow 1 denominator will become 0

=> for finite limit numerator must also approach to zero as x\rightarrow 1.

So, L' Hospital Law will be applicable.

1-a+b=0........................(1)

Now,

\lim_{x\rightarrow 1}\frac{x^{2}-ax+b}{x-1}=\frac{0}{0}form

\lim_{x\rightarrow 1}\frac{{2}x-a}{1}=5

2-a=5  => a=-3

and 

b=a-1 =>b=-3-1=-4

So,

a+b=-3-4=-7

option(3) is correct.


Option 1)

-4

Option 2)

5

Option 3)

-7

Option 4)

1

Posted by

Vakul

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