Let \psi _1:[0,\infty ) \to R , \psi _2:[0,\infty ) \to R , f : (0, \infty ) \to R and g : [0, \infty ) \to R be functions such that f(0) = g(0) = 0, \psi _1(x) = e^ - x + x,x \ge 0 , \psi _2(x) = x^2 - 2x - 2e^ - x + 2,x \ge 0 , f(x) = \int_ - x^x (|t| - t^2)e^ - t^2dt,x \gt 0 and g(x) = \int_0^x^2 \sqrt t e^ - tdt,x \gt 0 . Which of the following statements is TRUE?

g(x) \le 2 \over 3x^3 - 2 \over 5x^5 + 1 \over 7x^7 , for all x \in ( 0,1 \over 2 )

f(x) \ge 1 - e^ - x^2 - 2 \over 3x^3 + 2 \over 5x^5 , for all x \in ( 0,1 \over 2 )

Definite Integration PYQ — JEE Advanced 2021 · Shift 2 · Q35 with Solution

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Correct answer: D

The user wants me to determine which of the four given statements is true.

Let's analyze each statement one by one.

Statement A: ${\psi _1}(x) \le 1$ , for all x > 0

The function is given by ${\psi _1}(x) = {e^{ - x}} + x$ for $x \ge 0$ .

To analyze its behavior, we find its derivative: ${\psi' _1}(x) = \frac{d}{dx}(e^{-x} + x) = -e^{-x} + 1$
We check the sign of the derivative for $x > 0$ . For $x > 0$ , we have $-x < 0$ , which implies $e^{-x} < e^0 = 1$ . Therefore, $1 - e^{-x} > 0$ . So, ${\psi' _1}(x) > 0$ for all $x > 0$ .
This means that ${\psi _1}(x)$ is a strictly increasing function on the interval $(0, \infty)$ .
The value of the function at $x=0$ is ${\psi _1}(0) = e^0 + 0 = 1$ .
Since the function is strictly increasing for $x > 0$ , for any $x > 0$ , we have ${\psi _1}(x) > {\psi _1}(0) = 1$ .
The statement ${\psi _1}(x) \le 1$ for all $x > 0$ is therefore FALSE.

Statement B: ${\psi _2}(x) \le 0$ , for all x > 0

The function is given by ${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2$ for $x \ge 0$ .

We find its derivative: ${\psi' _2}(x) = \frac{d}{dx}({x^2} - 2x - 2{e^{ - x}} + 2) = 2x - 2 + 2e^{-x} = 2(x - 1 + e^{-x})$
Let's analyze the sign of the term $h(x) = x - 1 + e^{-x}$ . Its derivative is $h'(x) = 1 - e^{-x}$ .
As shown for statement A, $1 - e^{-x} > 0$ for $x > 0$ . So, $h(x)$ is strictly increasing for $x > 0$ .
The value at $x=0$ is $h(0) = 0 - 1 + e^0 = 0$ .
Since $h(x)$ is strictly increasing, for any $x > 0$ , we have $h(x) > h(0) = 0$ . This implies ${\psi' _2}(x) = 2h(x) > 0$ for all $x > 0$ .
So, ${\psi _2}(x)$ is a strictly increasing function on $(0, \infty)$ .
The value at $x=0$ is ${\psi _2}(0) = 0^2 - 2(0) - 2e^0 + 2 = -2 + 2 = 0$ .
Since ${\psi _2}(x)$ is strictly increasing for $x > 0$ , for any $x > 0$ , we have ${\psi _2}(x) > {\psi _2}(0) = 0$ .
The statement ${\psi _2}(x) \le 0$ for all $x > 0$ is therefore FALSE.

Statement C: $f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

The function is $f(x) = \int_{ - x}^x {(|t| - {t^2}){e^{ - {t^2}}}dt}$ .

The integrand $F(t) = (|t| - t^2)e^{-t^2}$ is an even function, since $F(-t) = (|-t| - (-t)^2)e^{-(-t)^2} = (|t| - t^2)e^{-t^2} = F(t)$ .
Therefore, $f(x) = 2 \int_0^x {(|t| - {t^2}){e^{ - {t^2}}}dt}$ . For $t \ge 0$ , $|t| = t$ , so $f(x) = 2 \int_0^x {(t - {t^2}){e^{ - {t^2}}}dt}$ .
Let's define a function $\phi(x) = f(x) - (1 - e^{ - {x^2}} - \frac{2}{3}{x^3} + \frac{2}{5}{x^5})$ . We want to check if $\phi(x) \ge 0$ .
We find the derivative of $\phi(x)$ . Using the Leibniz rule for $f'(x)$ : $f'(x) = 2(x - x^2)e^{-x^2} = 2xe^{-x^2} - 2x^2e^{-x^2}$ The derivative of the second part is: $\frac{d}{dx} \left(1 - e^{ - {x^2}} - \frac{2}{3}{x^3} + \frac{2}{5}{x^5}\right) = -e^{-x^2}(-2x) - 2x^2 + 2x^4 = 2xe^{-x^2} - 2x^2 + 2x^4$
So, ${\phi'}(x) = f'(x) - (2xe^{-x^2} - 2x^2 + 2x^4) = (2xe^{-x^2} - 2x^2e^{-x^2}) - (2xe^{-x^2} - 2x^2 + 2x^4) = -2x^2e^{-x^2} + 2x^2 - 2x^4 = 2x^2(1 - e^{-x^2} - x^2)$ .
To determine the sign of ${\phi'}(x)$ , we analyze the term $k(u) = 1 - e^{-u} - u$ for $u = x^2 > 0$ . $k'(u) = e^{-u} - 1$ . For $u > 0$ , $e^{-u} < 1$ , so $k'(u) < 0$ . This means $k(u)$ is a decreasing function for $u > 0$ . Since $k(0) = 1 - e^0 - 0 = 0$ , we have $k(u) < 0$ for all $u > 0$ .
Thus, for $x > 0$ , $1 - e^{-x^2} - x^2 < 0$ . Since $2x^2 > 0$ , we have ${\phi'}(x) < 0$ for $x > 0$ .
This means $\phi(x)$ is a strictly decreasing function for $x > 0$ .
Let's check the value at $x=0$ : $\phi(0) = f(0) - (1 - e^0 - 0 + 0) = 0 - (1-1) = 0$ .
Since $\phi(0)=0$ and $\phi(x)$ is strictly decreasing, we must have $\phi(x) < 0$ for all $x > 0$ . This implies $f(x) < 1 - e^{-x^2} - \frac{2}{3}x^3 + \frac{2}{5}x^5$ .
The statement $f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}$ is therefore FALSE.

Statement D: $g(x) \le {2 \over 3}{x^3} - {2 \over 5}{x^5} + {1 \over 7}{x^7}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

The function is $g(x) = \int_0^{{x^2}} {\sqrt t {e^{ - t}}dt}$ .

We use the Taylor series for $e^{-t}$ : $e^{-t} = 1 - t + \frac{t^2}{2!} - \frac{t^3}{3!} + \dots$ .
For $t > 0$ , this is a convergent alternating series with terms that decrease in magnitude. For an alternating series, the sum is bounded by its partial sums. In particular, the sum is less than or equal to any partial sum ending on a positive term. $e^{-t} \le 1 - t + \frac{t^2}{2}$ . This inequality holds for all $t \ge 0$ .
The integrand involves $\sqrt{t} e^{-t}$ . Since $\sqrt{t} \ge 0$ for $t \ge 0$ , we can multiply the inequality by $\sqrt{t}$ : $\sqrt{t} e^{-t} \le \sqrt{t} \left(1 - t + \frac{t^2}{2}\right) = t^{1/2} - t^{3/2} + \frac{1}{2}t^{5/2}$
Now, we integrate both sides from $0$ to $x^2$ . Since the inequality holds for all $t$ in the integration interval, the inequality will hold for the integrals as well. $g(x) = \int_0^{x^2} \sqrt{t} e^{-t} dt \le \int_0^{x^2} \left(t^{1/2} - t^{3/2} + \frac{1}{2}t^{5/2}\right) dt$
We compute the integral on the right-hand side: $\int_0^{x^2} \left(t^{1/2} - t^{3/2} + \frac{1}{2}t^{5/2}\right) dt = \left[ \frac{t^{3/2}}{3/2} - \frac{t^{5/2}}{5/2} + \frac{1}{2}\frac{t^{7/2}}{7/2} \right]_0^{x^2}$ $= \left[ \frac{2}{3}t^{3/2} - \frac{2}{5}t^{5/2} + \frac{1}{7}t^{7/2} \right]_0^{x^2}$ $= \left( \frac{2}{3}(x^2)^{3/2} - \frac{2}{5}(x^2)^{5/2} + \frac{1}{7}(x^2)^{7/2} \right) - (0)$ $= \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7$
So, we have shown that $g(x) \le \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7$ . This inequality holds for all $x \ge 0$ , and therefore it must hold for $x \in (0, 1/2)$ .
The statement is therefore TRUE.

Conclusion

Based on the analysis of all four options, only statement D is true.

More from Definite Integration

View written solutionFree

Correct answer: D

The user wants me to determine which of the four given statements is true.

Let's analyze each statement one by one.

Statement A: ${\psi _1}(x) \le 1$ , for all x > 0

The function is given by ${\psi _1}(x) = {e^{ - x}} + x$ for $x \ge 0$ .

To analyze its behavior, we find its derivative: ${\psi' _1}(x) = \frac{d}{dx}(e^{-x} + x) = -e^{-x} + 1$
We check the sign of the derivative for $x > 0$ . For $x > 0$ , we have $-x < 0$ , which implies $e^{-x} < e^0 = 1$ . Therefore, $1 - e^{-x} > 0$ . So, ${\psi' _1}(x) > 0$ for all $x > 0$ .
This means that ${\psi _1}(x)$ is a strictly increasing function on the interval $(0, \infty)$ .
The value of the function at $x=0$ is ${\psi _1}(0) = e^0 + 0 = 1$ .
Since the function is strictly increasing for $x > 0$ , for any $x > 0$ , we have ${\psi _1}(x) > {\psi _1}(0) = 1$ .
The statement ${\psi _1}(x) \le 1$ for all $x > 0$ is therefore FALSE.

Statement B: ${\psi _2}(x) \le 0$ , for all x > 0

The function is given by ${\psi _2}(x) = {x^2} - 2x - 2{e^{ - x}} + 2$ for $x \ge 0$ .

We find its derivative: ${\psi' _2}(x) = \frac{d}{dx}({x^2} - 2x - 2{e^{ - x}} + 2) = 2x - 2 + 2e^{-x} = 2(x - 1 + e^{-x})$
Let's analyze the sign of the term $h(x) = x - 1 + e^{-x}$ . Its derivative is $h'(x) = 1 - e^{-x}$ .
As shown for statement A, $1 - e^{-x} > 0$ for $x > 0$ . So, $h(x)$ is strictly increasing for $x > 0$ .
The value at $x=0$ is $h(0) = 0 - 1 + e^0 = 0$ .
Since $h(x)$ is strictly increasing, for any $x > 0$ , we have $h(x) > h(0) = 0$ . This implies ${\psi' _2}(x) = 2h(x) > 0$ for all $x > 0$ .
So, ${\psi _2}(x)$ is a strictly increasing function on $(0, \infty)$ .
The value at $x=0$ is ${\psi _2}(0) = 0^2 - 2(0) - 2e^0 + 2 = -2 + 2 = 0$ .
Since ${\psi _2}(x)$ is strictly increasing for $x > 0$ , for any $x > 0$ , we have ${\psi _2}(x) > {\psi _2}(0) = 0$ .
The statement ${\psi _2}(x) \le 0$ for all $x > 0$ is therefore FALSE.

Statement C: $f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

The function is $f(x) = \int_{ - x}^x {(|t| - {t^2}){e^{ - {t^2}}}dt}$ .

The integrand $F(t) = (|t| - t^2)e^{-t^2}$ is an even function, since $F(-t) = (|-t| - (-t)^2)e^{-(-t)^2} = (|t| - t^2)e^{-t^2} = F(t)$ .
Therefore, $f(x) = 2 \int_0^x {(|t| - {t^2}){e^{ - {t^2}}}dt}$ . For $t \ge 0$ , $|t| = t$ , so $f(x) = 2 \int_0^x {(t - {t^2}){e^{ - {t^2}}}dt}$ .
Let's define a function $\phi(x) = f(x) - (1 - e^{ - {x^2}} - \frac{2}{3}{x^3} + \frac{2}{5}{x^5})$ . We want to check if $\phi(x) \ge 0$ .
We find the derivative of $\phi(x)$ . Using the Leibniz rule for $f'(x)$ : $f'(x) = 2(x - x^2)e^{-x^2} = 2xe^{-x^2} - 2x^2e^{-x^2}$ The derivative of the second part is: $\frac{d}{dx} \left(1 - e^{ - {x^2}} - \frac{2}{3}{x^3} + \frac{2}{5}{x^5}\right) = -e^{-x^2}(-2x) - 2x^2 + 2x^4 = 2xe^{-x^2} - 2x^2 + 2x^4$
So, ${\phi'}(x) = f'(x) - (2xe^{-x^2} - 2x^2 + 2x^4) = (2xe^{-x^2} - 2x^2e^{-x^2}) - (2xe^{-x^2} - 2x^2 + 2x^4) = -2x^2e^{-x^2} + 2x^2 - 2x^4 = 2x^2(1 - e^{-x^2} - x^2)$ .
To determine the sign of ${\phi'}(x)$ , we analyze the term $k(u) = 1 - e^{-u} - u$ for $u = x^2 > 0$ . $k'(u) = e^{-u} - 1$ . For $u > 0$ , $e^{-u} < 1$ , so $k'(u) < 0$ . This means $k(u)$ is a decreasing function for $u > 0$ . Since $k(0) = 1 - e^0 - 0 = 0$ , we have $k(u) < 0$ for all $u > 0$ .
Thus, for $x > 0$ , $1 - e^{-x^2} - x^2 < 0$ . Since $2x^2 > 0$ , we have ${\phi'}(x) < 0$ for $x > 0$ .
This means $\phi(x)$ is a strictly decreasing function for $x > 0$ .
Let's check the value at $x=0$ : $\phi(0) = f(0) - (1 - e^0 - 0 + 0) = 0 - (1-1) = 0$ .
Since $\phi(0)=0$ and $\phi(x)$ is strictly decreasing, we must have $\phi(x) < 0$ for all $x > 0$ . This implies $f(x) < 1 - e^{-x^2} - \frac{2}{3}x^3 + \frac{2}{5}x^5$ .
The statement $f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}$ is therefore FALSE.

Statement D: $g(x) \le {2 \over 3}{x^3} - {2 \over 5}{x^5} + {1 \over 7}{x^7}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

The function is $g(x) = \int_0^{{x^2}} {\sqrt t {e^{ - t}}dt}$ .

We use the Taylor series for $e^{-t}$ : $e^{-t} = 1 - t + \frac{t^2}{2!} - \frac{t^3}{3!} + \dots$ .
For $t > 0$ , this is a convergent alternating series with terms that decrease in magnitude. For an alternating series, the sum is bounded by its partial sums. In particular, the sum is less than or equal to any partial sum ending on a positive term. $e^{-t} \le 1 - t + \frac{t^2}{2}$ . This inequality holds for all $t \ge 0$ .
The integrand involves $\sqrt{t} e^{-t}$ . Since $\sqrt{t} \ge 0$ for $t \ge 0$ , we can multiply the inequality by $\sqrt{t}$ : $\sqrt{t} e^{-t} \le \sqrt{t} \left(1 - t + \frac{t^2}{2}\right) = t^{1/2} - t^{3/2} + \frac{1}{2}t^{5/2}$
Now, we integrate both sides from $0$ to $x^2$ . Since the inequality holds for all $t$ in the integration interval, the inequality will hold for the integrals as well. $g(x) = \int_0^{x^2} \sqrt{t} e^{-t} dt \le \int_0^{x^2} \left(t^{1/2} - t^{3/2} + \frac{1}{2}t^{5/2}\right) dt$
We compute the integral on the right-hand side: $\int_0^{x^2} \left(t^{1/2} - t^{3/2} + \frac{1}{2}t^{5/2}\right) dt = \left[ \frac{t^{3/2}}{3/2} - \frac{t^{5/2}}{5/2} + \frac{1}{2}\frac{t^{7/2}}{7/2} \right]_0^{x^2}$ $= \left[ \frac{2}{3}t^{3/2} - \frac{2}{5}t^{5/2} + \frac{1}{7}t^{7/2} \right]_0^{x^2}$ $= \left( \frac{2}{3}(x^2)^{3/2} - \frac{2}{5}(x^2)^{5/2} + \frac{1}{7}(x^2)^{7/2} \right) - (0)$ $= \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7$
So, we have shown that $g(x) \le \frac{2}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7$ . This inequality holds for all $x \ge 0$ , and therefore it must hold for $x \in (0, 1/2)$ .
The statement is therefore TRUE.

Conclusion

Based on the analysis of all four options, only statement D is true.

Definite Integration question

Definite Integration question

Statement A: ${\psi _1}(x) \le 1$ , for all x > 0

Statement B: ${\psi _2}(x) \le 0$ , for all x > 0

Statement C: $f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

Statement D: $g(x) \le {2 \over 3}{x^3} - {2 \over 5}{x^5} + {1 \over 7}{x^7}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

Conclusion

More from Definite Integration

Definite Integration question

Definite Integration question

Statement A: ${\psi _1}(x) \le 1$ , for all x > 0

Statement B: ${\psi _2}(x) \le 0$ , for all x > 0

Statement C: $f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

Statement D: $g(x) \le {2 \over 3}{x^3} - {2 \over 5}{x^5} + {1 \over 7}{x^7}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

Conclusion

More from Definite Integration

Statement A: ψ1(x)≤1{\psi _1}(x) \le 1ψ1​(x)≤1, for all x > 0

Statement B: ψ2(x)≤0{\psi _2}(x) \le 0ψ2​(x)≤0, for all x > 0

Statement C: f(x)≥1−e−x2−23x3+25x5f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}f(x)≥1−e−x2−32​x3+52​x5, for all x∈(0,12)x \in \left( {0,{1 \over 2}} \right)x∈(0,21​)

Statement D: g(x)≤23x3−25x5+17x7g(x) \le {2 \over 3}{x^3} - {2 \over 5}{x^5} + {1 \over 7}{x^7}g(x)≤32​x3−52​x5+71​x7, for all x∈(0,12)x \in \left( {0,{1 \over 2}} \right)x∈(0,21​)

Conclusion

More from Definite Integration

Statement A: ψ1(x)≤1{\psi _1}(x) \le 1ψ1​(x)≤1, for all x > 0

Statement B: ψ2(x)≤0{\psi _2}(x) \le 0ψ2​(x)≤0, for all x > 0

Statement C: f(x)≥1−e−x2−23x3+25x5f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}f(x)≥1−e−x2−32​x3+52​x5, for all x∈(0,12)x \in \left( {0,{1 \over 2}} \right)x∈(0,21​)

Statement D: g(x)≤23x3−25x5+17x7g(x) \le {2 \over 3}{x^3} - {2 \over 5}{x^5} + {1 \over 7}{x^7}g(x)≤32​x3−52​x5+71​x7, for all x∈(0,12)x \in \left( {0,{1 \over 2}} \right)x∈(0,21​)

Conclusion

More from Definite Integration

Statement A: ${\psi _1}(x) \le 1$ , for all x > 0

Statement B: ${\psi _2}(x) \le 0$ , for all x > 0

Statement C: $f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

Statement D: $g(x) \le {2 \over 3}{x^3} - {2 \over 5}{x^5} + {1 \over 7}{x^7}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

Statement A: ${\psi _1}(x) \le 1$ , for all x > 0

Statement B: ${\psi _2}(x) \le 0$ , for all x > 0

Statement C: $f(x) \ge 1 - {e^{ - {x^2}}} - {2 \over 3}{x^3} + {2 \over 5}{x^5}$ , for all $x \in \left( {0,{1 \over 2}} \right)$

Statement D: $g(x) \le {2 \over 3}{x^3} - {2 \over 5}{x^5} + {1 \over 7}{x^7}$ , for all $x \in \left( {0,{1 \over 2}} \right)$