Let g(x)=f(x)+f(1-x) and f^{ }(x)

Correct Answer is : g(x)g(x)g(x) decreases on [12,1]≤ft[1/2, 1][21,1] and g(x)g(x)g(x) increases on [0,12]≤ft[0, 1/2][0,21]

g(x) increases on ≤ft[1/2, 1] and g(x) decreases on ≤ft[0, 1/2]

g(x) decreases on ≤ft[1/2, 1] and g(x) increases on ≤ft[0, 1/2]

Correct Answer is : g(x)g(x)g(x) decreases on [12,1]≤ft[1/2, 1][21,1] and g(x)g(x)g(x) increases on [0,12]≤ft[0, 1/2][0,21]

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MHTCET 18 Apr 2026 Shift 2 Paper

Question : 125 of 150

Marks: +1, -0

Let

g(x)=f(x)+f(1-x)

f^{\prime \prime}(x) < 0,0 \leq x \leq 1

, then…

$g(x)$ increases on $\left[\frac{1}{2}, 1\right]$ and $g(x)$ decreases on $\left[0, \frac{1}{2}\right]$
$g(x)$ decreases on [0,1]
$g(x)$ increases on $[0,1]$
$g(x)$ decreases on $\left[\frac{1}{2}, 1\right]$ and $g(x)$ increases on $\left[0, \frac{1}{2}\right]$

Solution:

Concept:

f''(x)<0

means

f'(x)

is a decreasing function.

Explanation:

Given

g(x)=f(x)+f(1-x)

Differentiating,

g'(x)=f'(x)-f'(1-x)

Since

f''(x)<0

[0,1]

f'(x)

is decreasing.

For

0 \le x < \frac{1}{2}

, we have

x < 1-x

f'

is decreasing,

f'(x)>f'(1-x)

Therefore

g'(x)=f'(x)-f'(1-x)>0

, so

g(x)

increases on

[0,\frac{1}{2}]

For

\frac{1}{2} < x \le 1

, we have

x > 1-x

Hence

f'(x) < f'(1-x)

, giving

g'(x)<0

g(x)

decreases on

[\frac{1}{2},1]

Answer:

Option D:

g(x)

decreases on

[\frac{1}{2},1]

and increases on

[0,\frac{1}{2}]

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