Let -x-fx-f2 a-x and f-x-0 for all x -0- a- Then- -xa increases on -0- a-b decreases on -0- a-c increases on -a- 0-d decreases on -a- 2 a-

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Standard XII

Mathematics

Monotonicity in an Interval

Let ϕx=fx+f2 ...

Question

Let ϕ(x) = f(x) + f(2a − x) and f"(x) > 0 for all x ∈ [0, a]. Then, ϕ (x)
(a) increases on [0, a]
(b) decreases on [0, a]
(c) increases on [−a, 0]
(d) decreases on [a, 2a]

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Solution

Given: ϕ(x) = f(x) + f(2a − x)

Differentiating above equation with respect to x we get,

ϕ'(x) = f'(x) − f(2a − x) .....(1)

Since, f''(x) > 0, f'(x) is an increasing function.

Now,

when
$x \in [0, a]$
$x \leq 2 a - x f' (x) \leq f (2 a - x) . . . . . (2)$

Considering equation (1) and (2) we get,

ϕ'(x) ≤ 0

⇒ ϕ'(x) is decreasing in [0, a]

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Let ϕ(x) = f(x) + f(2a − x) and f"(x) > 0 for all x ∈ [0, a]. Then, ϕ (x) (a) increases on [0, a] (b) decreases on [0, a] (c) increases on [−a, 0] (d) decreases on [a, 2a]

Let ϕ(x) = f(x) + f(2a − x) and f"(x) > 0 for all x ∈ [0, a]. Then, ϕ (x)
(a) increases on [0, a]
(b) decreases on [0, a]
(c) increases on [−a, 0]
(d) decreases on [a, 2a]