Question 1Level 2 · ExamSingle correct
Let a be a real parameter and f(x)=3a2−1x3+(a−1)x2+2x+1 for x∈R. The set of all values of a for which f is strictly increasing on R is ______. Exactly one option is correct.
A(−∞,−3)∪(1,∞)
B(−∞,−3]∪(1,∞)
C(−∞,−3]∪[1,∞)
D(−∞,−1)∪[1,∞)
Show answer and solution
Answer: C
f′(x)=(a2−1)x2+2(a−1)x+2. A differentiable f is strictly increasing on R iff f′≥0 everywhere and f′ does not vanish on any interval.
a=1: f′=2>0, works. a=−1: f′=−4x+2<0 for x>21, fails. a2<1: leading coefficient negative, f′→−∞, fails.
a2>1: need discriminant ≤0: 4(a−1)2−8(a2−1)≤0⟺(a−1)(a−1−2a−2)≤0⟺(a−1)(a+3)≥0⟺a≤−3 or a≥1. With a2>1 this gives a≤−3 or a>1.
Equality case a=−3: f′=8x2−8x+2=2(2x−1)2, zero only at x=21, so f is still strictly increasing.
Hence a∈(−∞,−3]∪[1,∞).
Answer: (C).
Difficulty: HardQuestion 3Level 2 · ExamSingle correct
Exactly one of the following functions satisfies all the hypotheses of Rolle's theorem (continuity on the closed interval, differentiability on the open interval, equal values at the end points) on the interval shown. Here ⌊t⌋ denotes the greatest integer not exceeding t. That function is ______. Exactly one option is correct.
Af(x)=x2/3 on [−1,1]
Bf(x)=x−⌊x⌋ on [0,1]
Df(x)=x2sinx1 for x=0, f(0)=0, on [−π1,π1]
Show answer and solution
Answer: D
(A) f′(x)=32x−1/3 does not exist at x=0∈(−1,1). Fails.
(B) f(0)=f(1)=0, but f(x)=x→1 as x→1−, so f is not continuous at 1. Fails.
(C) f(±1)=43 and f is continuous, but at x=±21 the one-sided derivatives are ±1 and ∓1 (a corner), so f is not differentiable on (−1,1). Fails.
(D) f(±π1)=π21sin(±π)=0. ∣f(x)∣≤x2 gives continuity at 0; for x=0 f is differentiable, and f′(0)=limh→0hsinh1=0. All hypotheses hold.
Answer: (D).
Difficulty: Hard