Zaymiey

📐 Mathematics  ·  Continuity and Differentiability  ·  JEE

For what value of k is f(x) = {kx+1 if x<=5, 3x-5 if x>5} continuous at x = 5?

Answer: k = 9/5.

  • A k = 1
  • B k = 9/5
  • C k = 5
  • D k = 3

Correct answer: B. k = 9/5

Explanation: Continuity requires left limit = right limit = f(5). Left: lim(x→5⁻) = 5k+1. Right: lim(x→5⁺) = 3(5)−5 = 10. Setting 5k+1 = 10 gives 5k = 9, so k = 9/5.

f(x) = |x|: Continuous but NOT Differentiable at 0x=0 (sharp "kink")LHD = -1RHD = +1LHD ≠ RHD at the kink → not differentiable, even though the curve has no gap (continuous)

f(x)=|x| is perfectly continuous at x=0 (no break, no jump), but the curve has a sharp corner there: approaching from the left gives slope -1 and from the right gives slope +1 - since these one-sided derivatives disagree, the function is not differentiable at that point.

Concept context

When a function has no breaks (continuity) and when it has a well-defined slope (differentiability), plus rules for differentiating implicit, inverse trig, exponential, log, and parametric functions.

Read the full Continuity and Differentiability notes →