JEE Advanced 1987

LEVELJEE Main

Animated Solution for Mathematics - Functions: Let $f$ and $g$ be increasing and decreasing functions, respectively from $[0, \infty)$ to $[0, \infty)$ . Let $h (x) = f (g (x))$ . If $h (0) = 0$ , then $h (x) - h (1)$ is

Q: Wait, why do we reject the possibility of $h(x)$ being negative? The inequality $h(x) \le 0$ allows for negative values, right?

That is a brilliant observation! While the inequality $h(x) \le 0$ suggests values less than or equal to zero, we must look at the co-domain. The problem states $f: [0, \infty) \to [0, \infty)$. This means the output of $f$ is strictly non-negative. Since $h(x) = f(g(x))$, $h(x)$ must be $\ge 0$. We are trapped between $\le 0$ and $\ge 0$, which forces $h(x)$ to be exactly zero.

Select Answer:

Progressive Hints

Hint 1

Think about the behavior of composite functions. If you have an increasing function

f

and a decreasing function

g

, what happens to the output of

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

x

increases? Visualize the 'direction' of the mapping and how the monotonicity of

f

interacts with the monotonicity of

g

Hint 2

Recall the formal definitions:

g

is decreasing means for

x_{1} \leq x_{2}

g (x_{1}) \geq g (x_{2})

f

is increasing means for

y_{1} \leq y_{2}

f (y_{1}) \leq f (y_{2})

. Combine these two inequalities carefully to see what happens to

h (x)

relative to

h (0)

Hint 3

Start with the domain constraint

x \geq 0

. Since

g

is decreasing, we know

g (x) \leq g (0)

. Now, apply the increasing function

f

to both sides of this inequality. What does the resulting inequality tell you about

h (x)

compared to

h (0)

? Remember to check the co-domain of

f

to see if

h (x)

can actually be negative.

Visualized Solution

Introduction to Functions $f$ and $g$

Given functions: $f, g : [0, \infty) \to [0, \infty)$
Function $f$ is increasing.
Function $g$ is decreasing.
Composite function: $h (x) = f (g (x))$

Analyzing Function $g (x)$

Since $g$ is a decreasing function on $[0, \infty)$ :
For any $x \geq 0$ , we have $g (x) \leq g (0)$

Property of Increasing Function $f$

Since $f$ is an increasing function:
If $a \leq b$ , then $f (a) \leq f (b)$

Applying $f$ to the Inequality

Applying $f$ to $g (x) \leq g (0)$ :
$f (g (x)) \leq f (g (0))$
This simplifies to: $h (x) \leq h (0)$

Using the Given Condition $h (0) = 0$

Given condition: $h (0) = 0$
Substituting this into $h (x) \leq h (0)$ :
$h (x) \leq 0$

Considering the Range of $f$

Range of $f$ is $[0, \infty)$
So, $f (t) \geq 0$ for all $t$
Therefore, $h (x) = f (g (x)) \geq 0$

Concluding $h (x) = 0$

From $h (x) \leq 0$ and $h (x) \geq 0$ :
We must have $h (x) = 0$ for all $x \geq 0$

Final Calculation and Answer

Since $h (x) = 0$ for all $x \geq 0$
$h (1) = 0$
$h (x) - h (1) = 0 - 0 = 0$
The value is always zero.

00:00 / 00:00

The Sigma Insight: Classification of Functions

Solution Diagram

Analyzing the Functional Constraints

We are given two functions,

f : [0, \infty) \to [0, \infty)

and

g : [0, \infty) \to [0, \infty)

The function

f

is strictly increasing, while

g

is strictly decreasing. We define the composite function as

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

, with the specific anchor point

h (0) = 0

The Monotonicity Argument

Since

g

is a decreasing function on the interval

[0, \infty)

, for any

x \geq 0

, we must have:

g (x) \leq g (0)

Because

f

is an increasing function, applying

f

to both sides of this inequality preserves the direction of the inequality sign:

f (g (x)) \leq f (g (0))

Substituting the definition of

h (x)

, this simplifies to:

h (x) \leq h (0)

The Range Constraint

We are given that

h (0) = 0

, which leads us to the inequality

h (x) \leq 0

However, we must also consider the co-domain of the functions. Since

f : [0, \infty) \to [0, \infty)

, the output of

f

is always non-negative.

Consequently, for any

x \geq 0

h (x) = f (g (x)) \geq 0

Final Conclusion

We are now constrained by two simultaneous conditions:

h (x) \leq 0

and

h (x) \geq 0

The only value that satisfies both conditions is

h (x) = 0

for all

x \geq 0

Since

h (x)

is identically zero, it follows that

h (1) = 0

. Therefore, the expression

h (x) - h (1)

evaluates to:

h (x) - h (1) = 0 - 0 = 0

Common Traps & FAQs

Wait, why do we reject the possibility of

h (x)

being negative? The inequality

h (x) \leq 0

allows for negative values, right?

That is a brilliant observation! While the inequality

h (x) \leq 0

suggests values less than or equal to zero, we must look at the co-domain. The problem states

f : [0, \infty) \to [0, \infty)

. This means the output of

f

is strictly non-negative. Since

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

h (x)

must be

\geq 0

. We are trapped between

\leq 0

and

\geq 0

, which forces

h (x)

to be exactly zero.

Does this mean

h (x)

is a constant function?

Exactly! When a function is squeezed between being less than or equal to zero and greater than or equal to zero, it has no choice but to be zero everywhere. It is a beautiful example of a 'sandwich' or 'squeeze' argument in function analysis. It is not just constant; it is the zero function.

Why can't we just use energy conservation or some other physics principle here?

While physics intuition is powerful, this is a pure mathematical analysis of monotonicity. In JEE Advanced, you must rely on the rigorous definitions of functions. Energy conservation doesn't apply here because we are dealing with abstract mappings, not physical systems. Stick to the definitions of increasing and decreasing functions—they are your best tools!

Animated Solution for Mathematics - Functions: Let $f$ and $g$ be increasing and decreasing functions, respectively from $[0, \infty)$ to $[0, \infty)$ . Let $h (x) = f (g (x))$ . If $h (0) = 0$ , then $h (x) - h (1)$ is

Select Answer:

Visualized Solution

Introduction to Functions $f$ and $g$

Analyzing Function $g (x)$

Property of Increasing Function $f$

Applying $f$ to the Inequality

Using the Given Condition $h (0) = 0$

Considering the Range of $f$

Concluding $h (x) = 0$

Final Calculation and Answer

The Sigma Insight: Classification of Functions

Analyzing the Functional Constraints

The Monotonicity Argument

The Range Constraint

Final Conclusion

Similar Questions

Let $g (x) = 1 + x - [x]$ and $f (x) = ⎩ ⎨ ⎧ - 1, 0, 1, x < 0 x = 0 x > 0$ . Then for all $x$ , $f (g (x))$ is equal to

Let $f, g : R \to R$ be defined as : $f (x) = ∣ x - 1∣$ and $g (x) = {e^{x}, x + 1, x \geq 0 x \leq 0$ . Then the function $f (g (x))$ is

If the functions $f (x)$ and $g (x)$ are defined on $R \to R$ such that $f (x) = {0, x, x \in rational x \in irrational$ ; $g (x) = {0, x, x \in irrational x \in rational$ then $(f - g) (x)$ is

List-I

List-II

Let $f, g : N \to N$ such that $f (n + 1) = f (n) + f (a)$ $\forall n \in N$ and $g$ be any arbitrary function. Which of the following statements is NOT true?

List-I

List-II

Let $f : R \to R$ be defined as $f (x) = x - 1$ and $g : R - {1, - 1} \to R$ be defined as $g (x) = \frac{x ^{2}}{x ^{2} - 1}$ . Then the function $f o g$ is :

Let $f (x) = {- a x + a if - a \leq x \leq 0 if 0 < x \leq a$ where $a > 0$ and $g (x) = (f (∣ x ∣) - ∣ f (x) ∣) /2$ . Then the function $g : [- a, a] \to [- a, a]$ is

A real valued function $f (x)$ satisfies the functional equation $f (x - y) = f (x) f (y) - f (a - x) f (a + y)$ where $a$ is a given constant and $f (0) = 1, f (2 a - x)$ is equal to

Select Answer:

Visualized Solution

Introduction to Functions .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g

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

Property of Increasing Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f to the Inequality

Using the Given Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h(0)=0

Considering the Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f

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

Final Calculation and Answer

The Sigma Insight: Classification of Functions

Analyzing the Functional Constraints

The Monotonicity Argument

The Range Constraint

Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let g(x)=1+x−[x] and f(x)=⎩⎨⎧​−1,0,1,​x<0x=0x>0​. Then for all x, f(g(x)) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f,g:R→R be defined as : f(x)=∣x−1∣ and g(x)={ex,x+1,​x≥0x≤0​. Then the function f(g(x)) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the functions f(x) and g(x) are defined on R→R such that f(x)={0,x,​x∈rationalx∈irrational​; g(x)={0,x,​x∈irrationalx∈rational​ then (f−g)(x) is

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f,g:N→N such that f(n+1)=f(n)+f(a) ∀n∈N and g be any arbitrary function. Which of the following statements is NOT true?

List-I

List-II

.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)=x−1 and g:R−{1,−1}→R be defined as g(x)=x2−1x2​. Then the function fog is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)={−ax+a​if −a≤x≤0if 0<x≤a​ where a>0 and g(x)=(f(∣x∣)−∣f(x)∣)/2. Then the function g:[−a,a]→[−a,a] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A real valued function f(x) satisfies the functional equation f(x−y)=f(x)f(y)−f(a−x)f(a+y) where a is a given constant and f(0)=1,f(2a−x) is equal to

Introduction to Functions $f$ and $g$

Analyzing Function $g (x)$

Property of Increasing Function $f$

Applying $f$ to the Inequality

Using the Given Condition $h (0) = 0$

Considering the Range of $f$

Concluding $h (x) = 0$

Let $g (x) = 1 + x - [x]$ and $f (x) = ⎩ ⎨ ⎧ - 1, 0, 1, x < 0 x = 0 x > 0$ . Then for all $x$ , $f (g (x))$ is equal to

Let $f, g : R \to R$ be defined as : $f (x) = ∣ x - 1∣$ and $g (x) = {e^{x}, x + 1, x \geq 0 x \leq 0$ . Then the function $f (g (x))$ is

If the functions $f (x)$ and $g (x)$ are defined on $R \to R$ such that $f (x) = {0, x, x \in rational x \in irrational$ ; $g (x) = {0, x, x \in irrational x \in rational$ then $(f - g) (x)$ is

Let $f, g : N \to N$ such that $f (n + 1) = f (n) + f (a)$ $\forall n \in N$ and $g$ be any arbitrary function. Which of the following statements is NOT true?

Let $f : R \to R$ be defined as $f (x) = x - 1$ and $g : R - {1, - 1} \to R$ be defined as $g (x) = \frac{x ^{2}}{x ^{2} - 1}$ . Then the function $f o g$ is :

Let $f (x) = {- a x + a if - a \leq x \leq 0 if 0 < x \leq a$ where $a > 0$ and $g (x) = (f (∣ x ∣) - ∣ f (x) ∣) /2$ . Then the function $g : [- a, a] \to [- a, a]$ is

A real valued function $f (x)$ satisfies the functional equation $f (x - y) = f (x) f (y) - f (a - x) f (a + y)$ where $a$ is a given constant and $f (0) = 1, f (2 a - x)$ is equal to