Let $f(x) = e^x - x$ and $g(x) = x^2 - x, x \in R$. Then the set of all $x \in R$, where the function $h(x) = (fog)(x)$ is increasing, is : Options: (A) $[-1/2, 0] \cup [1, \infty)$ (B) $[0, 1/2] \cup [1, \infty)$ (C) $[-1/2, 0] \cup [1, \infty)$ (D) $[0, \infty)$

### Analyzing the Setup We are given two functions: the outer function $f(x) = e^x - x$ and the inner function $g(x) = x^2 - x$. The composite function is defined as $h(x) = f(g(x))$. Substituting $g(x)$ into $f(x)$, we obtain the explicit form of the composite function: $$h(x) = e^{x^2-x} - (x^2-x)$$ ### Applying the Chain Rule To determine where the function is increasing, we must find the derivative $h'(x)$ and identify where $h'(x) \ge 0$. According to the **Chain Rule**, the derivative is given by: $$h'(x) = f'(g(x)) \cdot g'(x)$$ Given $f'(x) = e^x - 1$ and $g'(x) = 2x - 1$, we substitute these into the formula to get: $$h'(x) = (e^{x^2-x} - 1) \cdot (2x - 1)$$ ### Finding Critical Points To solve the inequality $h'(x) \ge 0$, we first identify the critical points by setting each factor to zero. For the first factor, $2x - 1 = 0$, we find: $$x = \frac{1}{2}$$ For the second factor, $e^{x^2-x} - 1 = 0$, we note that $e^u = 1$ implies $u = 0$. Therefore, we solve: $$x^2 - x = 0 \Rightarrow x(x - 1) = 0$$ This yields the critical points $x = 0$ and $x = 1$. ### Sign Analysis and Conclusion We now have three critical points: $0$, $1/2$, and $1$. These points divide the real number line into four intervals. We test the sign of $h'(x)$ in each region: * For $x > 1$: Both $(2x-1) > 0$ and $(e^{x^2-x}-1) > 0$, so $h'(x) > 0$. * For $1/2 0$ but $(e^{x^2-x}-1) 0$. * For $x 0$, so $h'(x) < 0$. By observing where the derivative is non-negative, we conclude that the function $h(x)$ is increasing on the intervals: **$[0, 1/2] \cup

JEE Main 2019 (10 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Differentiation: Let $f (x) = e^{x} - x$ and $g (x) = x^{2} - x, x \in R$ . Then the set of all $x \in R$ , where the function $h (x) = (f o g) (x)$ is increasing, is :

Q: Why is the interval $[0, 1/2]$ included? I thought it would be negative.

It is a common mistake to assume that because $x$ is small, the whole expression is negative. In the interval $(0, 1/2)$, the first factor $(2x-1)$ is negative, and the second factor $(e^{x^2-x}-1)$ is also negative. A negative times a negative is a positive! Always check the sign of each factor individually.

Select Answer:

Progressive Hints

Hint 1

Think of the composite function

h (x) = f (g (x))

as a nested structure. To determine where it is increasing, you do not need to graph the entire function. You need to know the 'velocity' of the function, which is its derivative. What is the fundamental condition for a function to be increasing? Recall that

h^{'} (x) \geq 0

Hint 2

Apply the Chain Rule carefully. Remember that

\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)

. Once you differentiate, do not rush to expand everything. Look for common factors—they are the keys to unlocking the inequality. The derivative of

e^{u}

e^{u} \cdot u^{'}

, which will help you factor the expression.

Hint 3

You should arrive at the inequality

(2 x - 1) (e^{x^{2} - x} - 1) \geq 0

. Now, treat this like a Wavy Curve problem. Find the roots of each factor separately:

2 x - 1 = 0

and

e^{x^{2} - x} - 1 = 0

. Remember that

e^{u} - 1

has the same sign as

u

because

e^{u}

is an increasing function.

Visualized Solution

Defining the Composite Function $h (x)$

Given: $f (x) = e^{x} - x$ and $g (x) = x^{2} - x$
We need to find where $h (x) = f (g (x))$ is increasing.

Substituting $g (x)$ into $f (x)$

Substitute $g (x)$ into $f (x)$ : $h (x) = e^{g (x)} - g (x)$
$h (x) = e^{x^{2} - x} - (x^{2} - x)$

Condition for Increasing Function

For a function to be increasing, its first derivative must be non-negative.
Condition: $h^{'} (x) \geq 0$

Calculating the Derivative $h^{'} (x)$

Differentiate $h (x)$ using the chain rule.
$h^{'} (x) = \frac{d}{d x} [e^{x^{2} - x}] - \frac{d}{d x} [x^{2} - x]$
$h^{'} (x) = e^{x^{2} - x} (2 x - 1) - (2 x - 1)$

Factoring the Derivative

Notice the common factor $(2 x - 1)$ .
$h^{'} (x) = (2 x - 1) (e^{x^{2} - x} - 1)$
We need $(2 x - 1) (e^{x^{2} - x} - 1) \geq 0$

Identifying Critical Points

Set each factor to zero to find critical points.
$2 x - 1 = 0 ⟹ x = \frac{1}{2}$
$e^{x^{2} - x} - 1 = 0 ⟹ e^{x^{2} - x} = e^{0}$
$x^{2} - x = 0 ⟹ x (x - 1) = 0 ⟹ x = 0, 1$

Plotting Critical Points

The critical points are $0$ , $\frac{1}{2}$ , and $1$ .
Plot these points on the real number line to divide it into intervals.

Sign Analysis for $x > 1$

Interval: $(1, \infty)$
Let $x = 2$ : $2 x - 1 = 3 > 0$
$e^{2^{2} - 2} - 1 = e^{2} - 1 > 0$
Product is $(+) (+) = +$

Sign Analysis for $\frac{1}{2} < x < 1$

Interval: $(\frac{1}{2}, 1)$
$2 x - 1 > 0$ (Positive)
$x^{2} - x < 0 ⟹ e^{x^{2} - x} < 1 ⟹ e^{x^{2} - x} - 1 < 0$ (Negative)
Product is $(+) (-) = -$

Sign Analysis for $0 < x < \frac{1}{2}$

Interval: $(0, \frac{1}{2})$
$2 x - 1 < 0$ (Negative)
$x^{2} - x < 0 ⟹ e^{x^{2} - x} - 1 < 0$ (Negative)
Product is $(-) (-) = +$

Sign Analysis for $x < 0$

Interval: $(- \infty, 0)$
$2 x - 1 < 0$ (Negative)
$x^{2} - x > 0 ⟹ e^{x^{2} - x} - 1 > 0$ (Positive)
Product is $(-) (+) = -$

Final Solution

$h^{'} (x) \geq 0$ in intervals where the sign is positive ( $+$ ).
These intervals are $[0, \frac{1}{2}]$ and $[1, \infty)$ .
Final Answer: $x \in [0, \frac{1}{2}] \cup [1, \infty)$

00:00 / 00:00

The Sigma Insight: Monotonicity

Solution Diagram

Analyzing the Setup

We are given two functions: the outer function

f (x) = e^{x} - x

and the inner function

g (x) = x^{2} - x

. The composite function is defined as

h (x) = f (g (x))

Substituting

g (x)

into

f (x)

, we obtain the explicit form of the composite function:

h (x) = e^{x^{2} - x} - (x^{2} - x)

Applying the Chain Rule

To determine where the function is increasing, we must find the derivative

h^{'} (x)

and identify where

h^{'} (x) \geq 0

. According to the Chain Rule, the derivative is given by:

h^{'} (x) = f^{'} (g (x)) \cdot g^{'} (x)

Given

f^{'} (x) = e^{x} - 1

and

g^{'} (x) = 2 x - 1

, we substitute these into the formula to get:

h^{'} (x) = (e^{x^{2} - x} - 1) \cdot (2 x - 1)

Finding Critical Points

To solve the inequality

h^{'} (x) \geq 0

, we first identify the critical points by setting each factor to zero.

For the first factor,

2 x - 1 = 0

, we find:

x = \frac{1}{2}

For the second factor,

e^{x^{2} - x} - 1 = 0

, we note that

e^{u} = 1

implies

u = 0

. Therefore, we solve:

x^{2} - x = 0 \Rightarrow x (x - 1) = 0

This yields the critical points

x = 0

and

x = 1

Sign Analysis and Conclusion

We now have three critical points:

0

1/2

, and

1

. These points divide the real number line into four intervals. We test the sign of

h^{'} (x)

in each region:

For $x > 1$ : Both $(2 x - 1) > 0$ and $(e^{x^{2} - x} - 1) > 0$ , so $h^{'} (x) > 0$ . For

1/2 < x < 1

(2 x - 1) > 0

but

(e^{x^{2} - x} - 1) < 0

, so

h^{'} (x) < 0

. For $0 < x < 1/2$ : Both $(2 x - 1) < 0$ and $(e^{x^{2} - x} - 1) < 0$ , so $h^{'} (x) > 0$ . For

x < 0

(2 x - 1) < 0

but

(e^{x^{2} - x} - 1) > 0

, so

h^{'} (x) < 0

By observing where the derivative is non-negative, we conclude that the function

h (x)

is increasing on the intervals:

$[0, 1/2] \cup

Common Traps & FAQs

Why do we include the points where the derivative is zero in our final answer?

This is a classic JEE trap! A function is defined as 'increasing' (or non-decreasing) if its derivative is

\geq 0

. Even if the derivative is zero at isolated points, the function is still considered increasing there because it does not decrease. We only exclude points where the derivative is undefined, which is not the case here.

I tried to expand the exponential term and got stuck. Is there a shortcut?

Never expand if you can factor! The beauty of this problem lies in the structure

(e^{u} - 1)

. Since

e^{u}

is an increasing function,

e^{u} - 1

behaves exactly like

u

regarding its sign. By keeping it factored, you avoid messy algebra and see the critical points immediately.

Why is the interval

[0, 1/2]

included? I thought it would be negative.

It is a common mistake to assume that because

x

is small, the whole expression is negative. In the interval

(0, 1/2)

, the first factor

(2 x - 1)

is negative, and the second factor

(e^{x^{2} - x} - 1)

is also negative. A negative times a negative is a positive! Always check the sign of each factor individually.

Animated Solution for Mathematics - Differentiation: Let $f (x) = e^{x} - x$ and $g (x) = x^{2} - x, x \in R$ . Then the set of all $x \in R$ , where the function $h (x) = (f o g) (x)$ is increasing, is :

Select Answer:

Visualized Solution

Defining the Composite Function $h (x)$

Substituting $g (x)$ into $f (x)$

Condition for Increasing Function

Calculating the Derivative $h^{'} (x)$

Factoring the Derivative

Identifying Critical Points

Plotting Critical Points

Sign Analysis for $x > 1$

Sign Analysis for $\frac{1}{2} < x < 1$

Sign Analysis for $0 < x < \frac{1}{2}$

Sign Analysis for $x < 0$

Final Solution

The Sigma Insight: Monotonicity

Analyzing the Setup

Applying the Chain Rule

Finding Critical Points

Sign Analysis and Conclusion

Similar Questions

The function $f (x) = x e^{x (1 - x)}, x \in R$ , is

Let $f : (0, 1) \to R$ be a function defined by $f (x) = \frac{1}{1 - e ^{- x}}$ , and $g (x) = (f (- x) - f (x))$ . Consider two statements (I) $g$ is an increasing function in $(0, 1)$ (II) $g$ is one-one in $(0, 1)$ Then,

Let $f$ be a real valued function, defined on $R - {- 1, 1}$ and given by $f (x) = 3 lo g_{e} \frac{x - 1}{x + 1} - \frac{2}{x - 1}$ . Then in which of the following intervals, function $f (x)$ is increasing?

If $f (x) = x e^{x (1 - x)}$ , then $f (x)$ is

The interval in which the function $f (x) = x^{x}, x > 0$ , is strictly increasing is

The function defined by $f (x) = (x + 2) e^{- x}$ is

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

Let $f : R \to R$ and $g : R \to R$ be two functions defined by $f (x) = lo g_{e} (x^{2} + 1) - e^{- x} + 1$ and $g (x) = \frac{1 - 2 e ^{2 x}}{e ^{x}}$ . Then, for which of the following range of $α$ , the inequality $f (g (\frac{( α - 1 ) ^{2}}{3})) > f (g (α - \frac{5}{3}))$ holds?

The function, $f (x) = (3 x - 7) x^{2/3}$ , $x \in R$ , is increasing for all $x$ lying in :

If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=ex−x and g(x)=x2−x,x∈R. Then the set of all x∈R, where the function h(x)=(fog)(x) is increasing, is :

Select Answer:

Visualized Solution

Defining the Composite Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(x)

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x) into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Condition for Increasing Function

Calculating the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h′(x)

Factoring the Derivative

Identifying Critical Points

Plotting Critical Points

Sign Analysis for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>1

Sign Analysis for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 21​<x<1

Sign Analysis for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0<x<21​

Sign Analysis for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0

Final Solution

The Sigma Insight: Monotonicity

Analyzing the Setup

Applying the Chain Rule

Finding Critical Points

Sign Analysis and Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f(x)=xex(1−x),x∈R, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=xex(1−x), then f(x) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The interval in which the function f(x)=xx,x>0, is strictly increasing is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function defined by f(x)=(x+2)e−x is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined as f(x)={−34​x3+2x2+3x3xex​,x>0,x≤0​. Then f is increasing function in the interval

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function, f(x)=(3x−7)x2/3, x∈R, is increasing for all x lying in :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=∫x2x2+1​e−t2dt, then f(x) increases in

Animated Solution for Mathematics - Differentiation: Let $f (x) = e^{x} - x$ and $g (x) = x^{2} - x, x \in R$ . Then the set of all $x \in R$ , where the function $h (x) = (f o g) (x)$ is increasing, is :

Defining the Composite Function $h (x)$

Substituting $g (x)$ into $f (x)$

Calculating the Derivative $h^{'} (x)$

Sign Analysis for $x > 1$

Sign Analysis for $\frac{1}{2} < x < 1$

Sign Analysis for $0 < x < \frac{1}{2}$

Sign Analysis for $x < 0$

The function $f (x) = x e^{x (1 - x)}, x \in R$ , is

If $f (x) = x e^{x (1 - x)}$ , then $f (x)$ is

The interval in which the function $f (x) = x^{x}, x > 0$ , is strictly increasing is

The function defined by $f (x) = (x + 2) e^{- x}$ is

Let $f : R \to R$ be defined as $f (x) = {- \frac{4}{3} x^{3} + 2 x^{2} + 3 x 3 x e^{x}, x > 0, x \leq 0$ . Then $f$ is increasing function in the interval

The function, $f (x) = (3 x - 7) x^{2/3}$ , $x \in R$ , is increasing for all $x$ lying in :

If $f (x) = \int_{x^{2}}^{x^{2} + 1} e^{- t^{2}} d t$ , then $f (x)$ increases in