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Exercise 5.1 · Q6

Q.Find all points of discontinuity of ff, where ff is defined by f(x)={2x+3,if x≤22x−3,if x>2f(x) = \begin{cases} 2x+3, & \text{if } x \leq 2 \\ 2x-3, & \text{if } x > 2 \end{cases}

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The function ff is defined by two linear pieces that meet at x=2x=2. The left-hand limit at 22 is 77, the right-hand limit is 11, and since these are not equal, the function is discontinuous at x=2x=2. It is continuous everywhere else.

The Core Idea: Continuity at a Point

A function is continuous at a point x=ax = a if three things happen together:

  1. The function is defined at aa — f(a)f(a) exists.
  2. The limit of f(x)f(x) as xx approaches aa exists (both sides agree).
  3. That limit equals f(a)f(a).

For a piecewise function, the only place where things can go wrong is at the "seam" — the point where the definition changes. Here, that seam is x=2x = 2. Everywhere else, each piece is a straight line, and straight lines are continuous everywhere. So the entire question boils down to: what happens at x=2x = 2?


Step-by-Step Analysis

1. Check the function value at x=2x = 2

Since x=2x = 2 falls in the first case (x≤2x \leq 2), we use f(x)=2x+3f(x) = 2x + 3.

f(2)=2(2)+3=4+3=7f(2) = 2(2) + 3 = 4 + 3 = 7

So f(2)=7f(2) = 7. That's fine — the function is defined.

2. Compute the left-hand limit as x→2−x \to 2^-

When xx approaches 22 from the left (values like 1.9,1.99,1.9991.9, 1.99, 1.999), we are still in the region x≤2x \leq 2, so we use the same piece: 2x+32x + 3.

lim⁡x→2−f(x)=lim⁡x→2−(2x+3)=2(2)+3=7\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x + 3) = 2(2) + 3 = 7

The left-hand limit is 77, which matches f(2)f(2).

3. Compute the right-hand limit as x→2+x \to 2^+

When xx approaches 22 from the right (values like 2.1,2.01,2.0012.1, 2.01, 2.001), we are in the region x>2x > 2, so we use the other piece: 2x−32x - 3.

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

The right-hand limit is 11.

Watch out

A common mistake is to assume that because both pieces are linear, the function must be continuous. But the two lines have different yy-values at x=2x=2 — one gives 77, the other gives 11. The function literally jumps from 77 to 11 as you cross x=2x=2.

4. Compare the two one-sided limits

Left-hand limit: 77

Right-hand limit: 11

Since 7≠17 \neq 1, the two-sided limit lim⁡x→2f(x)\displaystyle \lim_{x \to 2} f(x) does not exist.

5. Apply the continuity condition …

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