JEE Advanced 1998

LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: Let $f (x) = A x^{2} + B x + C$ where $A, B, C$ are real numbers. Prove that if $f (x)$ is an integer whenever $x$ is an integer, then the numbers $2 A, A + B$ and $C$ are all integers. Conversely, prove that if the numbers $2 A, A + B$ and $C$ are all integers then $f (x)$ is an integer whenever $x$ is an integer.

Q: Is the term $\frac{x(x-1)}{2}$ always an integer?

Yes, absolutely! The expression $x(x-1)$ represents the product of two consecutive integers. In any pair of consecutive integers, one of them must be even. Since an even number is divisible by $2$, their product $x(x-1)$ is always divisible by $2$. Therefore, the fraction is always an integer.

Visualized Solution

The Quadratic Function $f (x)$

Given function: $f (x) = A x^{2} + B x + C$
Condition: $f (x) \in Z$ for all $x \in Z$
Goal (Part 1): Prove $2 A, A + B, C \in Z$

Finding the Constant Term $C$

Substitute $x = 0$ into $f (x)$ :
$f (0) = A (0)^{2} + B (0) + C = C$
Since $f (0)$ is an integer, $C$ must be an integer.

Finding $A + B$

Substitute $x = 1$ into $f (x)$ :
$f (1) = A (1)^{2} + B (1) + C = A + B + C$
Rearranging: $A + B = f (1) - C$
Since $f (1)$ and $C$ are integers, $A + B$ is an integer.

Evaluating at $x = - 1$

Substitute $x = - 1$ into $f (x)$ :
$f (- 1) = A (- 1)^{2} + B (- 1) + C = A - B + C$
This gives us another integer expression.

Finding $2 A$

Add $f (1)$ and $f (- 1)$ :
$f (1) + f (- 1) = (A + B + C) + (A - B + C)$
$f (1) + f (- 1) = 2 A + 2 C$
Rearranging: $2 A = f (1) + f (- 1) - 2 C$
Since $f (1), f (- 1), C$ are integers, $2 A$ is an integer.

The Converse Statement

Given: $2 A, A + B, C \in Z$
Goal (Part 2): Prove $f (x) \in Z$ for all $x \in Z$

Algebraic Manipulation of $f (x)$

Start with $f (x) = A x^{2} + B x + C$
We need to express this using $2 A$ and $A + B$ .
Add and subtract $A x$ :
$f (x) = A x^{2} - A x + A x + B x + C$

Grouping the Terms

Group the terms:
$f (x) = A (x^{2} - x) + (A + B) x + C$
Factor out $x$ from the first term:
$f (x) = A x (x - 1) + (A + B) x + C$

Creating the $2 A$ Term

Multiply and divide the first term by $2$ :
$f (x) = \frac{2 A}{2} x (x - 1) + (A + B) x + C$
Rewrite as:
$f (x) = 2 A (\frac{x ( x - 1 )}{2}) + (A + B) x + C$

The Product of Consecutive Integers

Look at the expression $\frac{x ( x - 1 )}{2}$
For any integer $x$ , $x$ and $x - 1$ are consecutive integers.
One of them must be even.
Therefore, their product $x (x - 1)$ is always divisible by $2$ .

Concluding the Proof

$\frac{x ( x - 1 )}{2} \in Z$
$2 A \in Z$ , $(A + B) \in Z$ , $C \in Z$
$f (x) = \in Z 2 A \cdot \in Z \frac{x ( x - 1 )}{2} + \in Z (A + B) \cdot \in Z x + \in Z C$
Sum of integers is an integer, so $f (x) \in Z$ .

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The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

The machine is defined by the quadratic rule

f (x) = A x^{2} + B x + C

. We are given that for any integer

x

, the output

f (x)

is also an integer. Our objective is to determine the necessary and sufficient conditions for the coefficients

A, B,

and

C

Testing the Boundaries

Let us begin by testing the simplest inputs to constrain the coefficients. Substituting

x = 0

into the function, we get:

f (0) = A (0)^{2} + B (0) + C = C

Since

f (0)

must be an integer, we conclude that $C$ is an integer.

Next, we test

x = 1

and

x = - 1

f (1) = A + B + C

f (- 1) = A - B + C

Since

f (1)

and

f (- 1)

are integers, their sum and difference must also be integers. Adding these two equations yields:

f (1) + f (- 1) = 2 A + 2 C

Rearranging for

2 A

, we find:

2 A = f (1) + f (- 1) - 2 C

Because

f (1)

f (- 1)

, and

C

are integers, it follows that $2 A$ must be an integer. Furthermore, since

A + B + C

is an integer and

C

is an integer,

A + B

must also be an integer.

The Algebraic Surgery

Now, we verify the converse: if

2 A

A + B

, and

C

are integers, is

f (x)

always an integer? We rewrite the function

f (x) = A x^{2} + B x + C

to isolate these known integer components.

By adding and subtracting

A x

, we obtain:

f (x) = A x (x - 1) + (A + B) x + C

To introduce the

2 A

term, we multiply and divide the first term by

2

f (x) = 2 A (\frac{x ( x - 1 )}{2}) + (A + B) x + C

Final Conclusion

In the expression above,

2 A

(A + B)

, and

C

are integers by our hypothesis. We must only confirm that

\frac{x ( x - 1 )}{2}

is an integer.

Since

x

and

x - 1

are consecutive integers, one of them must be even. Therefore, their product

x (x - 1)

is always divisible by

2

, ensuring that

\frac{x ( x - 1 )}{2}

is an integer.

Because

f (x)

is the sum of products of integers,

f (x)

is guaranteed to be an integer for any integer

x

. Thus, the necessary and sufficient conditions are that $2 A$ , $A + B$ , and $C$ must all be integers.

Common Traps & FAQs

Wait, does this mean

A, B,

and

C

must be integers themselves?

Not necessarily! That is the classic trap. The condition is that

f (x)

is an integer for all integer

x

. This implies that specific combinations like

2 A

A + B

, and

C

must be integers, but

A

itself could be a fraction like

1/2

. For example, if

f (x) = \frac{1}{2} x^{2} + \frac{1}{2} x

, then

f (x)

is always an integer, even though the coefficients are not.

Why did we choose

x = 0, 1, - 1

to test the function?

We chose these values because they are the simplest integers that allow us to create a system of linear equations.

x = 0

immediately isolates

C

x = 1

and

x = - 1

create equations involving

A

and

B

that are easy to manipulate and solve, allowing us to isolate the required terms

A + B

and

2 A

Is the term

\frac{x ( x - 1 )}{2}

always an integer?

Yes, absolutely! The expression

x (x - 1)

represents the product of two consecutive integers. In any pair of consecutive integers, one of them must be even. Since an even number is divisible by

2

, their product

x (x - 1)

is always divisible by

2

. Therefore, the fraction is always an integer.

Visualized Solution

The Quadratic Function $f (x)$

Finding the Constant Term $C$

Finding $A + B$

Evaluating at $x = - 1$

Finding $2 A$

The Converse Statement

Algebraic Manipulation of $f (x)$

Grouping the Terms

Creating the $2 A$ Term

The Product of Consecutive Integers

Concluding the Proof

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

Testing the Boundaries

The Algebraic Surgery

Final Conclusion

Similar Questions

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Visualized Solution

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

Finding the Constant Term .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A+B

Evaluating at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2A

The Converse Statement

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

Grouping the Terms

Creating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2A Term

The Product of Consecutive Integers

Concluding the Proof

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

Analyzing the Setup

Testing the Boundaries

The Algebraic Surgery

Final Conclusion

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The Quadratic Function $f (x)$

Finding the Constant Term $C$

Finding $A + B$

Evaluating at $x = - 1$

Finding $2 A$

Algebraic Manipulation of $f (x)$

Creating the $2 A$ Term

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Let $f (x)$ be a quadratic expression which is positive for all the real values of $x$ . If $g (x) = f (x) + f^{'} (x) + f^{''} (x)$ , then for any real $x$ ,

The integer ' $k$ ', for which the inequality $x^{2} - 2 (3 k - 1) x + 8 k^{2} - 7 > 0$ is valid for every $x$ in $R$ is :

If the roots of the equation $b x^{2} + c x + a = 0$ be imaginary, then for all real values of $x$ , the expression $3 b^{2} x^{2} + 6 b c x + 2 c^{2}$ is

For real $x$ , the function $\frac{( x - a ) ( x - b )}{x - c}$ will assume all real values provided

Let $f (x) = (1 + b^{2}) x^{2} + 2 b x + 1$ and let $m (b)$ be the minimum value of $f (x)$ . As $b$ varies, the range of $m (b)$ is

Let $f : R - {2, 6} \to R$ be real valued function defined as $f (x) = \frac{x ^{2} + 2 x + 1}{x ^{2} - 8 x + 12}$ . Then range of $f$ is

If $a^{2} + b^{2} + c^{2} = 1$ , then $ab + b c + c a$ lies in the interval