JEE MAIN - Mathematics (2019 - 8th April Morning Slot - No. 23)
Let ƒ : [0, 2] $$ \to $$ R be a twice differentiable
function such that ƒ''(x) > 0, for all x $$ \in $$ (0, 2).
If $$\phi $$(x) = ƒ(x) + ƒ(2 – x), then $$\phi $$ is :
decreasing on (0, 2)
decreasing on (0, 1) and increasing on (1, 2)
increasing on (0, 2)
increasing on (0, 1) and decreasing on (1, 2)
Explanation
$$\phi $$(x) = ƒ(x) + ƒ(2 – x)
$$ \Rightarrow $$ $$\phi $$'(x) = ƒ'(x) - ƒ'(2 – x)
Since ƒ''(x) > 0 for all x $$ \in $$ (0, 2)
$$ \Rightarrow $$ ƒ'(x) is an increasing function for all x $$ \in $$ (0, 2).
Case 1 : When $$\phi $$(x) is increasing function
So $$\phi $$'(x) > 0
$$ \Rightarrow $$ ƒ'(x) - ƒ'(2 – x) > 0
$$ \Rightarrow $$ ƒ'(x) > ƒ'(2 – x)
$$ \Rightarrow $$ x > 2 – x
$$ \Rightarrow $$ x > 1
$$ \therefore $$ $$\phi $$(x) is increasing on (1, 2).
Case 2 : When $$\phi $$(x) is decreasing function
So $$\phi $$'(x) < 0
$$ \Rightarrow $$ ƒ'(x) - ƒ'(2 – x) < 0
$$ \Rightarrow $$ ƒ'(x) < ƒ'(2 – x)
$$ \Rightarrow $$ x < 2 – x
$$ \Rightarrow $$ x < 1
$$ \therefore $$ $$\phi $$(x) is decreasing on (0, 1).
$$ \Rightarrow $$ $$\phi $$'(x) = ƒ'(x) - ƒ'(2 – x)
Since ƒ''(x) > 0 for all x $$ \in $$ (0, 2)
$$ \Rightarrow $$ ƒ'(x) is an increasing function for all x $$ \in $$ (0, 2).
Case 1 : When $$\phi $$(x) is increasing function
So $$\phi $$'(x) > 0
$$ \Rightarrow $$ ƒ'(x) - ƒ'(2 – x) > 0
$$ \Rightarrow $$ ƒ'(x) > ƒ'(2 – x)
$$ \Rightarrow $$ x > 2 – x
$$ \Rightarrow $$ x > 1
$$ \therefore $$ $$\phi $$(x) is increasing on (1, 2).
Case 2 : When $$\phi $$(x) is decreasing function
So $$\phi $$'(x) < 0
$$ \Rightarrow $$ ƒ'(x) - ƒ'(2 – x) < 0
$$ \Rightarrow $$ ƒ'(x) < ƒ'(2 – x)
$$ \Rightarrow $$ x < 2 – x
$$ \Rightarrow $$ x < 1
$$ \therefore $$ $$\phi $$(x) is decreasing on (0, 1).
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