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KCET 2026 Mathematics Solved Paper

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Question : 13 of 60

Marks: +1, -0

If

f(x)=\left\{\begin{array}{lll}x^{2}-1 & \text { if } & x \geq 2 \\ x+1 & \text { if } & x<2\end{array}\right.

\lim _{x \rightarrow 1} f(x)+\lim _{x \rightarrow 2} f(x)=

3
5
7
9

Solution:

Concept:

For a piecewise function, use the branch that applies near the given

x

-value.

Explanation:

For

x \to 1

, we have

x < 2

, so use

f(x) = x + 1

.

\lim_{x \to 1} f(x) = \lim_{x \to 1} (x+1) = 1 + 1 = 2

.

For

x \to 2

, check both sides.

Left-hand limit: for

x < 2

,

f(x) = x + 1

, so

\lim_{x \to 2^{-}} f(x) = 2 + 1 = 3

.

Right-hand limit: for

x \geq 2

,

f(x) = x^2 - 1

, so

\lim_{x \to 2^{+}} f(x) = 2^2 - 1 = 4 - 1 = 3

.

Both limits are equal, hence

\lim_{x \to 2} f(x) = 3

.

Therefore,

\lim_{x \to 1} f(x) + \lim_{x \to 2} f(x) = 2 + 3 = 5

.

Answer:

5

, which is Option B.

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