JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 10)
Let f(x) = ex – x and g(x) = x2 – x, $$\forall $$ x $$ \in $$ R. Then the set of all x $$ \in $$ R, where the function h(x) = (fog) (x) is increasing, is :
[0, $$\infty $$)
$$\left[ { - 1, - {1 \over 2}} \right] \cup \left[ {{1 \over 2},\infty } \right)$$
$$\left[ { - {1 \over 2},0} \right] \cup \left[ {1,\infty } \right)$$
$$\left[ {0,{1 \over 2}} \right] \cup \left[ {1,\infty } \right)$$
Explanation
f(x) = ex – x, g(x) = x2 – x
f(g(x)) = e(x2 - x) - (x2 - x)
If f(g(x)) is increasing function
(f (g(x)))x = $${e^{\left( {{x^2} - x} \right)}} \times (2x - 1) - 2x + 1$$
$$ \Rightarrow \mathop {(2x - 1)}\limits_A \mathop {[{e^{\left( {{x^2} - x} \right)}} - 1]}\limits_B $$
A & B are either both positive or negative
_10th_April_Morning_Slot_en_10_2.png)
for (f (g(x)))' $$ \ge $$ 0,
$$x \in \left[ {0,{1 \over 2}} \right] \cup \left[ {1,\infty } \right)$$
f(g(x)) = e(x2 - x) - (x2 - x)
If f(g(x)) is increasing function
(f (g(x)))x = $${e^{\left( {{x^2} - x} \right)}} \times (2x - 1) - 2x + 1$$
$$ \Rightarrow \mathop {(2x - 1)}\limits_A \mathop {[{e^{\left( {{x^2} - x} \right)}} - 1]}\limits_B $$
A & B are either both positive or negative
for (f (g(x)))' $$ \ge $$ 0,$$x \in \left[ {0,{1 \over 2}} \right] \cup \left[ {1,\infty } \right)$$
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