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KCET 2026 Mathematics Solved Paper

Question : 14 of 60

Marks: +1, -0

\lim _{x \rightarrow 3}\left(\frac{x^{2}-a x-3 b}{x-3}\right)=5

a+b=

Solution:

Concept:

For a finite limit when the denominator tends to

0

, the numerator must also tend to

0

x = 3

Explanation:

Given:

\lim_{x \to 3} \frac{x^{2} - ax - 3b}{x - 3} = 5

x = 3

, the denominator becomes

3 - 3 = 0

For the limit to exist as a finite value, the numerator must also be

0

x = 3

Substitute

x = 3

in the numerator:

3^{2} - a(3) - 3b = 0

9 - 3a - 3b = 0

Divide the entire equation by

3

3 - a - b = 0

a + b = 3

Answer:

a + b = 3

Correct option: C. 3

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