Continuity & Differentiability: Real-World Scenarios (4)

easy 2 min read

Question

A taxi charges ₹5050 for the first 2 km2\ \text{km}, then ₹1515 per additional km. Express the fare as a function of distance xx, and check whether it is continuous and differentiable at x=2x = 2.

Solution — Step by Step

f(x)={50,0x250+15(x2),x>2f(x) = \begin{cases} 50, & 0 \leq x \leq 2 \\ 50 + 15(x - 2), & x > 2 \end{cases}

Left-hand limit: limx2f(x)=50\lim_{x \to 2^-} f(x) = 50.

Right-hand limit: limx2+f(x)=50+15(0)=50\lim_{x \to 2^+} f(x) = 50 + 15(0) = 50.

Function value: f(2)=50f(2) = 50. All three are equal, so ff is continuous at x=2x = 2.

Left-hand derivative: f(x)=0f'(x) = 0 for x<2x < 2, so LHD =0= 0.

Right-hand derivative: f(x)=15f'(x) = 15 for x>2x > 2, so RHD =15= 15.

LHD \neq RHD, so ff is not differentiable at x=2x = 2.

The fare graph has a sharp corner at x=2 kmx = 2\ \text{km} — the rate of change jumps from 00 to 1515 rupees per km. Continuous (no break in the price), but not smooth (sudden change in slope).

Final answer: ff is continuous at x=2x = 2 but not differentiable there.

Why This Works

Continuity asks “is there a break in the graph?” Differentiability asks “is there a corner?” Real-world piecewise pricing functions are usually continuous (otherwise you’d see ₹8080 for 1.99 km1.99\ \text{km} and ₹5050 for 2.00 km2.00\ \text{km}, which makes no business sense), but corners are common at price-tier boundaries.

This is the same reason absolute value x|x| is continuous but not differentiable at 00.

Alternative Method

Plot the function. A V-shape, kink, or corner means continuous but not differentiable. A jump means not continuous (and hence also not differentiable).

For CBSE 4-mark questions, always check three things at the boundary: LHL, RHL, f(c)f(c) for continuity; LHD and RHD for differentiability. Tabulate them — examiners give partial marks for the framework even if arithmetic slips.

Common Mistake

Assuming “continuous implies differentiable”. The reverse is true (differentiable implies continuous), but not this direction. The standard counterexample is x|x| at x=0x = 0.

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