Paragraph:
Let ψ1:[0,∞)→R,ψ2:[0,∞)→R,f:[0,∞)→R and g:[0,∞)→R be functions such that f(0)=g(0)=0,
ψ1(x)=e−x+x,x≥0,
ψ2(x)=x2−2x−2e−x+2,x≥0,
f(x)=∫−xx(∣t∣−t2)e−t2dt,x>0
and g(x)=∫0x2te−tdt,x>0 Question:
Which of the following statements is TRUE ?
Options
Aψ1(x)≤1 for all x>0
Bψ2(x)≤0, for all x>0
Cf(x)≥1−e−x2−32x3+52x5, for all x∈(0,21)
Dg(x)≤32x3−52x5+71x7 for all x∈(0,21)
Correct answer
D. g(x)≤32x3−52x5+71x7 for all x∈(0,21)
Step-by-step solution
(A) Given
ψ1\left(x\right) = {e}^{- x} + x, x \ge 0</p><p>{ψ}^{'}1\left(x\right) = - {e}^{- x} + 1, x \ge 0</p><p>{ψ}^{'}1\left(x\right)isincreasingfunction.</p><p>{ψ}_{1}\left(x\right) > {ψ}_{1}\left(0\right) \forall x \ge 0</p><p>{ψ}_{1}\left(x\right) \ge 1</p><p>So,</p><p>{e}^{- x} + x < 1forx \in ( 0, \infty )isincorrect.</p><p>LHSisincreasingandunboundedfunction.</p><p>(B)Given{ψ}_{2}(x)=x2−2x−2e−x+2,x≥0
ψ′2(x)=2x−2+2e−x,x≥0
ψ′2(x)=2ψ1(x)−2≥0,x≥0
So, ψ2(x) is increasing function.
x2−2x−2e−x+2<1 for x∈(0,∞) is incorrect because LHS →∞ when x→∞
