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Exercise 7.3 · Q1

Q.Explain why Rolle's theorem is not applicable to the following functions in the respective intervals.

(i) f(x)=∣1x∣, x∈[−1,1]f(x)=\left|\dfrac1x\right|,\ x\in[-1,1]
(ii) f(x)=tan⁡x, x∈[0,π]f(x)=\tan x,\ x\in[0,\pi]
(iii) f(x)=x−2log⁡x, x∈[2,7]f(x)=x-2\log x,\ x\in[2,7]
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✓ Free question

Rolle's theorem needs all three of: continuity on [a,b][a,b], differentiability on (a,b)(a,b), and f(a)=f(b)f(a)=f(b). Each part fails exactly one of these.

Step 1 (i). f(x)=∣1x∣, x∈[−1,1]f(x)=\left|\dfrac1x\right|,\ x\in[-1,1].

ff is not defined at x=0x=0 (division by zero), and x=0∈[−1,1]x=0\in[-1,1]. So ff is not continuous on [−1,1][-1,1] (it isn't even defined there) — even though f(−1)=1=f(1)f(-1)=1=f(1). Rolle's theorem does not apply.

Step 2 (ii). f(x)=tan⁡x, x∈[0,π]f(x)=\tan x,\ x\in[0,\pi].

tan⁡x\tan x has an infinite discontinuity at x=π2∈(0,π)x=\dfrac{\pi}{2}\in(0,\pi) (tan⁡π2\tan\tfrac{\pi}{2} is undefined). So ff is not continuous on [0,π][0,\pi], even though f(0)=0=f(π)f(0)=0=f(\pi). Rolle's theorem does not apply.

Step 3 (iii). f(x)=x−2log⁡x, x∈[2,7]f(x)=x-2\log x,\ x\in[2,7].

ff is continuous on [2,7][2,7] and differentiable on (2,7)(2,7) (no discontinuity or corner anywhere on this interval). But f(2)=2−2log⁡2≈0.614f(2)=2-2\log2\approx0.614 and f(7)=7−2log⁡7≈3.108f(7)=7-2\log7\approx3.108, so f(2)≠f(7)f(2)\ne f(7) — the hypothesis f(a)=f(b)f(a)=f(b) fails. Rolle's theorem does not apply.

✓Final answer

(i) ff is undefined at the interior point x=0x=0, so continuity fails. (ii) tan⁡x\tan x is discontinuous at x=π/2x=\pi/2 inside [0,π][0,\pi], so continuity fails. (iii) f(2)≠f(7)f(2)\ne f(7), so the equal-endpoints hypothesis fails, even though ff is continuous and differentiable throughout.

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