If φ(x)=f(x)+f(2 a-x) and f prime prime(x)0, a0,0 ≤ x ≤ 2 a then

Question

If φ(x)=f(x)+f(2 a-x) and f prime prime(x)>0, a>0,0 ≤ x ≤ 2 a then (A) φ( x ) increases in ( a , 2 a ) (B) φ(x) increases in (0, a) (C)  φ( x ) decreases in (0, a ) (D) φ(x) decreases in (a, 2 a)

Tardigrade Admin · Accepted Answer

φ(x)=f(x)+f(2 a-x) φ prime(x)=f prime(x)+f prime(2 a-x) text Since f prime prime(x)>0 ⇒ f prime(x) text is an increasing function text hence f prime(x)>f φ(2 a-x) ⇒ x>2 a-x ⇒ x>a ∴ f prime(x)>0 if x>a ⇒ f(x) is uparrow in (a, 2 a) | | text ly f ( x ) is downarrow in (0, a )

Q. If $ϕ (x) = f (x) + f (2 a - x)$ and $f^{''} (x) > 0, a > 0, 0 \leq x \leq 2 a$ then

$ϕ (x)$ increases in $(a, 2 a)$

$ϕ (x)$ increases in $(0, a)$

$ϕ (x)$ decreases in $(0, a)$

$ϕ (x)$ decreases in $(a, 2 a)$

Q. If ϕ(x)=f(x)+f(2a−x) and f′′(x)>0,a>0,0≤x≤2a then

ϕ(x) increases in (a,2a)

ϕ(x) increases in (0,a)

ϕ(x) decreases in (0,a)

ϕ(x) decreases in (a,2a)

ϕ(x)=f(x)+f(2a−x) ϕ′(x)=f′(x)+f′(2a−x) Since f′′(x)>0⇒f′(x) is an increasing function hence f′(x)>fϕ(2a−x)⇒x>2a−x⇒x>a ∴f′(x)>0 if x>a⇒f(x) is ↑ in (a,2a) ∥∥ly f(x) is ↓ in (0,a)

Q. If $ϕ (x) = f (x) + f (2 a - x)$ and $f^{''} (x) > 0, a > 0, 0 \leq x \leq 2 a$ then

$ϕ (x)$ increases in $(a, 2 a)$

$ϕ (x)$ increases in $(0, a)$

$ϕ (x)$ decreases in $(0, a)$

$ϕ (x)$ decreases in $(a, 2 a)$