Continuity and Differentiability (JEE Advanced)

Question 1

Let f(x)=x/(1+e^(1/x)) for x≠ 0 and f(0)=0. Write f'(0⁻)=lim_(h→ 0⁻)(f(h)−f(0))/h and f'(0⁺)=lim_(h→ 0⁺)(f(h)−f(0))/h. The ordered pair (f'(0⁻), f'(0⁺)) is ______. Exactly one option is correct.

Accepted Answer

B. For h≠0, (f(h)−f(0))/h=1/(1+e^(1/h)).
As h→0⁻, 1/h→−∞, so e^(1/h)→0 and the quotient tends to 1. Hence f'(0⁻)=1.
As h→0⁺, 1/h→+∞, so e^(1/h)→∞ and the quotient tends to 0. Hence f'(0⁺)=0.
(Continuity at 0 holds as well: lvert f(x)rvert≤lvert xrvert.) The pair is (1,0); f is continuous but not differentiable at 0.
Trap: swapping the sides, i.e. thinking e^(1/h) is large for negative h, gives (A).
A

Question 2

Let P(x)=(x−1)(x−3)(x−5)+(x−2)(x−4)(x−6) for real x. Which of the following statements about the real roots of P is correct? Exactly one option is correct.

Accepted Answer

C. P is a cubic with leading coefficient 2, so it has at most three real roots. Evaluate at 1,…,6 (one product vanishes each time): P(1)=0+(−1)(−3)(−5)=−15, P(2)=(1)(−1)(−3)+0=3, P(3)=0+(1)(−1)(−3)=3, P(4)=(3)(1)(−1)+0=−3, P(5)=0+(3)(1)(−1)=−3, P(6)=(5)(3)(1)+0=15. P is a polynomial, hence continuous on every closed interval. Since P(1)<00>P(4) and P(5)<0

Question 3

Let φ(t)=t³−3t. For x∈[0,3] define g(x)=min{φ(t):0≤ t≤ x} and G(x)=max{φ(t):0≤ t≤ x}. Which of the following is correct? Exactly one option is correct.

Accepted Answer

A. φ'(t)=3(t²−1): φ decreases on [0,1] from 0 to −2 and increases on [1,3] from −2 to 18; φ(√3)=0.
Running minimum: for 0≤ x≤1 the minimum on [0,x] is φ(x); for x≥1 it stays φ(1)=−2 (later values exceed −2). So g(x)=x³−3x on [0,1] and g(x)=−2 on [1,3]. At x=1: g'(1⁻)=φ'(1)=0 and g'(1⁺)=0, so g is differentiable at 1, hence on (0,3).
Running maximum: on [0,√3], φ≤0=φ(0), so G(x)=0. For x>√3, φ is incr

Continuity and Differentiability: questions with solutions

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