Let f : ℝ → ℝ be a function such that:
|f(x) − f(y)| ≤ ½ |x − y|, for all x, y ∈ ℝ
f′(x) ≥ ½, for all x ∈ ℝ, and f(1) = ½
Find the number of points of intersection of the curves:
y = f(x) and y = x² − 2x − 5
From the given:
So assume:
f(x) = (1/2)x + c
Use the point: f(1) = 1/2 ⇒ (1/2)·1 + c = 1/2 ⇒ c = 0
Therefore:
f(x) = (1/2)x
Solve:
(1/2)x = x² − 2x − 5
⇒ x² − (5/2)x − 5 = 0
⇒ 2x² − 5x − 10 = 0
⇒ x = [5 ± √(25 + 80)] / 4
⇒ x = [5 ± √105] / 4
Therefore, there are 2 real points of intersection.
| List-I | List-II | ||
|---|---|---|---|
| (A) | $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ | (I) | $[\frac{1}{3} , 1 ]$ |
| (B) | $(x)=|[x]|,x \in [R$ | (II) | Z |
| (C) | $h(x) = |x - [x]| , x \in [R$ | (III) | W |
| (D) | $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ | (IV) | [0, 1) |
| (V) | { -1, 1} | ||
| List I | List II | ||
|---|---|---|---|
| (A) | $\lambda=8, \mu \neq 15$ | 1. | Infinitely many solutions |
| (B) | $\lambda \neq 8, \mu \in R$ | 2. | No solution |
| (C) | $\lambda=8, \mu=15$ | 3. | Unique solution |