Basics & LogarithmshardFree

Basics & Logarithms: Minimum Value Range

JEE Maths question with a full step-by-step solution.

Question
If λ\lambda is the minimum value of ∣x−p∣+∣x−15∣+∣x−p−15∣|x-p| + |x-15| + |x-p-15| for xx in the range p≤x≤15p \leq x \leq 15 where 0<p<150 < p < 15, then λ5\dfrac{\lambda}{5} is
Solution
Answer: 3
Step 1: For p≤x≤15p \leq x \leq 15: ∣x−p∣=x−p|x-p| = x-p (since x≥px \geq p) ∣x−15∣=15−x|x-15| = 15-x (since x≤15x \leq 15) ∣x−(p+15)∣=(p+15)−x|x-(p+15)| = (p+15)-x (since p+15>15≥xp+15 > 15 \geq x) Step 2: The expression simplifies to:
(x−p)+(15−x)+(p+15−x)=30−x(x-p) + (15-x) + (p+15-x) = 30 - x
 Step 3: Since E=30−xE = 30-x is decreasing, EE is minimised at x=15x = 15:
λ=30−15=15\lambda = 30 - 15 = 15
 Step 4: λ5=155=3\dfrac{\lambda}{5} = \dfrac{15}{5} = 3. Answer: 3
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