Case I: when x>2−x ⇒x>1 ⇒φ′(x)>0 ∀x∈(1,2)
∴φ(x) is increasing on (1,2)
Case II: when x<2−x ⇒x<1 ⇒φ′(x)<0 ∀x∈(0,1)
∴φ(x) is decreasing on (0,1)

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" /> Case I: when x>2−x ⇒x>1 ⇒φ′(x)>0 ∀x∈(1,2)
∴φ(x) is increasing on (1,2)
Case II: when x<2−x ⇒x<1 ⇒φ′(x)<0 ∀x∈(0,1)
∴φ(x) is decreasing on (0,1)

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devarshi-dt-logo

Question:

Let f:[0,2]→R be a twice differentiable function such that f"(x)>0, for all x∈(0,2). If φ(x)=f(x)+f(2-x), then φ is:

increasing on(0,2)

decreasing on(0,2)

increasing on(0,1)and decreasing on(1,2)

decreasing on(0,1)and increasing on(1,2)

Solution:

Correct option is D. decreasing on(0,1)and increasing on(1,2)
φ(x)=f(x)+f(2−x)
φ′(x)=f′(x)−f′(2−x).. (1)
Since f"(x)>0 ⇒f′(x) is increasing ∀x∈(0,2)
Case I: when x>2−x ⇒x>1 ⇒φ′(x)>0 ∀x∈(1,2)
∴φ(x) is increasing on (1,2)
Case II: when x<2−x ⇒x<1 ⇒φ′(x)<0 ∀x∈(0,1)
∴φ(x) is decreasing on (0,1)