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QJEEM19Maca4a: JEE Main 2019 Mathematics - Class 11 JEEM19Maca4a Solution: Let f(x) = e x – x and g(x) = x 2 – x, x R. Then the set of all …

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 :

Multiple Choice Options

  • A [0, ∞\infty )
  • B [−1,−12]∪[12,∞)\left[ { - 1, - {1 \over 2}} \right] \cup \left[ {{1 \over 2},\infty } \right)
  • C [−12,0]∪[1,∞)\left[ { - {1 \over 2},0} \right] \cup \left[ {1,\infty } \right)
  • D [0,12]∪[1,∞)\left[ {0,{1 \over 2}} \right] \cup \left[ {1,\infty } \right)

Step-by-Step Verified Solution

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(x2−x)×(2x−1)−2x+1{e^{\left( {{x^2} - x} \right)}} \times (2x - 1) - 2x + 1

⇒(2x−1)A[e(x2−x)−1]B\Rightarrow \mathop {(2x - 1)}\limits_A \mathop {[{e^{\left( {{x^2} - x} \right)}} - 1]}\limits_B

A & B are either both positive or negative

JEE Main 2019 (Online) 10th April Morning Slot Mathematics - Functions Question 111 English Explanation for (f (g(x)))' ≥\ge 0,

x∈[0,12]∪[1,∞)x \in \left[ {0,{1 \over 2}} \right] \cup \left[ {1,\infty } \right)

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