JEE Main 2022 (25 July Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Quadratic Equations: Let $f (x)$ be a quadratic polynomial with leading coefficient 1 such that $f (0) = p, p \neq = 0$ and $f (1) = \frac{1}{3}$ . If the equation $f (x) = 0$ and $f (f (f (f (x)))) = 0$ have a common real root, then $f (- 3)$ is equal to

Q: Why did we divide by $p$ in the equation $\frac{p^2}{9} + \frac{bp}{3} + p = 0$? What if $p=0$?

The problem statement explicitly gives us the condition $p \neq 0$. In mathematics, we must always respect the constraints provided in the prompt. Since $p$ is guaranteed to be non-zero, dividing by it is a perfectly safe and elegant way to simplify our algebraic expression.

Enter Numerical Value:

Visualized Solution

Defining the Polynomial $f (x)$

Let the quadratic polynomial be $f (x) = x^{2} + b x + c$ .
Given $f (0) = p$ , we substitute $x = 0$ : $f (0) = 0^{2} + b (0) + c = p ⟹ c = p$ .
Thus, $f (x) = x^{2} + b x + p$ where $p \neq = 0$ .

Applying the Condition $f (1) = \frac{1}{3}$

Given $f (1) = \frac{1}{3}$ .
Substitute $x = 1$ into $f (x) = x^{2} + b x + p$ :
$1^{2} + b (1) + p = \frac{1}{3}$
$1 + b + p = \frac{1}{3} ⟹ b + p = - \frac{2}{3}$ (Equation 1)

Identifying the Common Root $α$

Let $α$ be the common real root.
Since $α$ is a root of $f (x) = 0$ , we have $f (α) = 0$ .
Since $α$ is also a root of $f (f (f (f (x)))) = 0$ , we have $f (f (f (f (α)))) = 0$ .

Simplifying the Nested Composition

Substitute $f (α) = 0$ into the nested equation:
$f (f (f (f (α)))) = f (f (f (0))) = 0$
Substitute $f (0) = p$ :
$f (f (p)) = 0$

Interpreting $f (f (p)) = 0$

The equation $f (f (p)) = 0$ implies that the value $f (p)$ is a root of the polynomial $f (x)$ .
Let the roots of $f (x) = 0$ be $α$ and $β$ .
Then $f (p) \in {α, β}$ , which means $f (p)$ acts as an input that makes the function zero.

Calculating $f (p)$ in terms of $p$

Calculate $f (p)$ using $f (x) = x^{2} + b x + p$ :
$f (p) = p^{2} + b p + p = p (p + b + 1)$
From Equation 1, $b + p = - \frac{2}{3}$ . Substitute this:
$f (p) = p (- \frac{2}{3} + 1) = p (\frac{1}{3}) = \frac{p}{3}$

Setting up the Root Equation $f (\frac{p}{3}) = 0$

Since $f (p) = \frac{p}{3}$ is a root of $f (x) = 0$ , we have $f (\frac{p}{3}) = 0$ .
$(\frac{p}{3})^{2} + b (\frac{p}{3}) + p = 0$
$\frac{p ^{2}}{9} + \frac{b p}{3} + p = 0$
Divide by $p$ (since $p \neq = 0$ ): $\frac{p}{9} + \frac{b}{3} + 1 = 0$

Forming the System of Equations

We have a system of two linear equations:
1) $b + p = - \frac{2}{3}$
2) $\frac{p}{9} + \frac{b}{3} + 1 = 0 ⟹ 3 b + p = - 9$

Solving for $b$ and $p$

Subtract Equation 1 from Equation 2:
$(3 b + p) - (b + p) = - 9 - (- \frac{2}{3})$
$2 b = - 9 + \frac{2}{3} = - \frac{25}{3} ⟹ b = - \frac{25}{6}$
Substitute $b$ into Equation 1:
$- \frac{25}{6} + p = - \frac{2}{3} ⟹ p = \frac{25}{6} - \frac{4}{6} = \frac{21}{6} = \frac{7}{2}$

Finding the Final Polynomial $f (x)$

The polynomial is $f (x) = x^{2} - \frac{25}{6} x + \frac{7}{2}$ .
We need to find $f (- 3)$ .

Calculating $f (- 3)$

Substitute $x = - 3$ into $f (x)$ :
$f (- 3) = (- 3)^{2} - \frac{25}{6} (- 3) + \frac{7}{2}$
$f (- 3) = 9 + \frac{25}{2} + \frac{7}{2}$
$f (- 3) = 9 + \frac{32}{2} = 9 + 16 = 25$

Summary and Key Takeaway

Final Answer: $f (- 3) = 25$
Key Takeaway: For any polynomial $f (x)$ , if $f (f (k)) = 0$ , then $f (k)$ must be one of the roots of $f (x) = 0$ .
Next Challenge: What if the common root was between $f (x) = 0$ and $f (f (x)) = 0$ ? How would that change the constraints on $p$ ?

00:00 / 00:00

The Sigma Insight: Relation Between Roots and Coefficients

Solution Diagram

Analyzing the Setup

Imagine you are standing before a complex, multi-layered puzzle. You have a quadratic polynomial,

f (x) = x^{2} + b x + p

, and you are told that it shares a root with its own four-fold composition,

f (f (f (f (x)))) = 0

We define our polynomial as

f (x) = x^{2} + b x + p

. We know

f (0) = p

, which is our constant term. We are also given

f (1) = 1/3

Substituting

x = 1

, we get

1 + b + p = 1/3

, which simplifies to the following anchor equation:

b + p = - \frac{2}{3}

The Collapse of Complexity

Consider the common root

α

. By definition,

f (α) = 0

. When we look at the nested equation

f (f (f (f (α)))) = 0

, we can substitute from the inside out.

Since

f (α) = 0

, the expression becomes

f (f (f (0))) = 0

. Because

f (0) = p

, this simplifies further to

f (f (p)) = 0

The entire sixteen-degree complexity has vanished. We are left with the simple statement that

f (p)

must be a root of

f (x) = 0

The Algebraic Bridge

f (p)

is a root, then

f (f (p)) = 0

. Let us calculate

f (p)

explicitly:

f (p) = p^{2} + b p + p = p (p + b + 1)

Using our earlier anchor,

b + p = - 2/3

, we substitute this into our expression for

f (p)

f (p) = p (- \frac{2}{3} + 1) = \frac{p}{3}

Since

f (p) = p /3

is a root of

f (x) = 0

, it must satisfy the equation

f (p /3) = 0

. Substituting

p /3

into our quadratic form:

(\frac{p}{3})^{2} + b (\frac{p}{3}) + p = 0

Dividing by

p

(assuming $p eq 0$), we get

\frac{p}{9} + \frac{b}{3} + 1 = 0

. This simplifies to the linear constraint:

3 b + p = - 9

The Final Resolution

We now have a system of two linear equations:

b + p = - \frac{2}{3}

3 b + p = - 9

Subtracting the first from the second, we find

2 b = - 9 + 2/3 = - 25/3

, which yields

b = - 25/6

. Substituting this back into our first equation, we find

p = 7/2

Our polynomial is now fully revealed:

f (x) = x^{2} - \frac{25}{6} x + \frac{7}{2}

. To find

f (- 3)

, we substitute:

f (- 3) = (- 3)^{2} - \frac{25}{6} (- 3) + \frac{7}{2} = 9 + \frac{25}{2} + \frac{7}{2} = 9 + \frac{32}{2} = 9 + 16

The final result is:

f (- 3) = 25

Common Traps & FAQs

Wait, why can we just assume

f (p)

is a root of

f (x) = 0

? Is it possible that

f (p)

is not a root?

That is a brilliant question! The logic follows directly from the definition of a root. If

f (f (p)) = 0

, we are essentially saying that the value

f (p)

is an input that, when fed into the function

f

, yields zero. By definition, any input that yields zero is a root of the polynomial. Thus,

f (p)

must be a root of

f (x) = 0

Why did we divide by

p

in the equation

\frac{p ^{2}}{9} + \frac{b p}{3} + p = 0

? What if

p = 0

The problem statement explicitly gives us the condition

p \neq = 0

. In mathematics, we must always respect the constraints provided in the prompt. Since

p

is guaranteed to be non-zero, dividing by it is a perfectly safe and elegant way to simplify our algebraic expression.

I got confused with the nested function

f (f (f (f (x))))

. Do I need to expand that entire polynomial?

Absolutely not! Expanding a nested quadratic four times would lead to a polynomial of degree 16, which is a nightmare. The beauty of this problem lies in the 'common root' property. By using the fact that

f (α) = 0

, the entire nested structure collapses like a house of cards into a much simpler expression.

How do I know if

α

is a real root? Does the problem guarantee it?

The problem states that the equations have a 'common real root'. This is a powerful hint. It ensures that the discriminant of our quadratic will be non-negative, and it justifies our algebraic manipulation. We are working within the domain of real numbers, so we don't need to worry about complex roots here.

Is there a faster way to find

f (- 3)

without solving for

b

and

p

individually?

While you could potentially look for shortcuts, solving for the coefficients

b

and

p

is the most robust method. It ensures you have the complete definition of the polynomial, which prevents errors. Once you have

b

and

p

, finding

f (- 3)

is just a simple substitution!

Animated Solution for Mathematics - Quadratic Equations: Let $f (x)$ be a quadratic polynomial with leading coefficient 1 such that $f (0) = p, p \neq = 0$ and $f (1) = \frac{1}{3}$ . If the equation $f (x) = 0$ and $f (f (f (f (x)))) = 0$ have a common real root, then $f (- 3)$ is equal to

Enter Numerical Value:

Visualized Solution

Defining the Polynomial $f (x)$

Applying the Condition $f (1) = \frac{1}{3}$

Identifying the Common Root $α$

Simplifying the Nested Composition

Interpreting $f (f (p)) = 0$

Calculating $f (p)$ in terms of $p$

Setting up the Root Equation $f (\frac{p}{3}) = 0$

Forming the System of Equations

Solving for $b$ and $p$

Finding the Final Polynomial $f (x)$

Calculating $f (- 3)$

Summary and Key Takeaway

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Collapse of Complexity

The Algebraic Bridge

The Final Resolution

Similar Questions

Let $f (x)$ be a quadratic polynomial such that $f (- 2) + f (3) = 0$ . If one of the roots of $f (x) = 0$ is $- 1$ , then the sum of the roots of $f (x) = 0$ is equal to:

Let $f (x)$ be a quadratic polynomial such that $f (- 1) + f (2) = 0$ . If one of the roots of $f (x) = 0$ is 3, then its other root lies in :

Let $λ \neq = 0$ be a real number. Let $α, β$ be the roots of the equation $14 x^{2} - 31 x + 3 λ = 0$ and $α, γ$ be the roots of the equation $35 x^{2} - 53 x + 4 λ = 0$ . Then $\frac{3 α}{β}$ and $\frac{4 α}{γ}$ are the roots of the equation :

Let $p$ and $q$ be real numbers such that $p \neq = 0, p^{3} \neq = - q$ . If $α$ and $β$ are nonzero complex numbers satisfying $α + β = - p$ and $α^{3} + β^{3} = q$ , then a quadratic equation having $α / β$ and $β / α$ as its roots is

If f(x) is a quadratic expression such that $f (1) + f (2) = 0$ , and -1 is a root of $f (x) = 0$ , then the other root of $f (x) = 0$ is :-

If one root of the quadratic equation $a x^{2} + b x + c = 0$ is equal to the $n$ -th power of the other, then show that $(a c^{n})^{1/ (n + 1)} + (a^{n} c)^{1/ (n + 1)} + b = 0$

Let $α$ and $β$ be the roots of $x^{2} + 3 x - 16 = 0$ , and $γ$ and $δ$ be the roots of $x^{2} + 3 x - 1 = 0$ . If $P_{n} = α^{n} + β^{n}$ and $Q_{n} = γ^{n} + δ^{n}$ , then $\frac{P _{25} + 3 P _{24}}{2 P _{23}} + \frac{Q _{25} - Q _{23}}{Q _{24}}$ is equal to

If one root is square of the other root of the equation $x^{2} + p x + q = 0$ , then the relation between $p$ and $q$ is

For the equation $3 x^{2} + p x + 3 = 0, p > 0$ , if one of the root is square of the other, then $p$ is equal to

Let $a, b \in R, a \neq = 0$ be such that the equation, $a x^{2} - 2 b x + 5 = 0$ has a repeated root $α$ , which is also a root of the equation, $x^{2} - 2 b x - 10 = 0$ . If $β$ is the other root of this equation, then $α^{2} + β^{2}$ is equal to:

Enter Numerical Value:

Visualized Solution

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

Applying the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)=31​

Identifying the Common Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Simplifying the Nested Composition

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(p) in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Setting up the Root Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(3p​)=0

Forming the System of Equations

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Finding the Final Polynomial .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−3)

Summary and Key Takeaway

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Collapse of Complexity

The Algebraic Bridge

The Final Resolution

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a quadratic polynomial such that f(−2)+f(3)=0. If one of the roots of f(x)=0 is −1, then the sum of the roots of f(x)=0 is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) be a quadratic polynomial such that f(−1)+f(2)=0. If one of the roots of f(x)=0 is 3, then its other root lies in :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x) is a quadratic expression such that f(1)+f(2)=0, and -1 is a root of f(x)=0, then the other root of f(x)=0 is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one root of the quadratic equation ax2+bx+c=0 is equal to the n-th power of the other, then show that (acn)1/(n+1)+(anc)1/(n+1)+b=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If one root is square of the other root of the equation x2+px+q=0, then the relation between p and q is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the equation 3x2+px+3=0,p>0, if one of the root is square of the other, then p is equal to

Defining the Polynomial $f (x)$

Applying the Condition $f (1) = \frac{1}{3}$

Identifying the Common Root $α$

Interpreting $f (f (p)) = 0$

Calculating $f (p)$ in terms of $p$

Setting up the Root Equation $f (\frac{p}{3}) = 0$

Solving for $b$ and $p$

Finding the Final Polynomial $f (x)$

Calculating $f (- 3)$

Let $f (x)$ be a quadratic polynomial such that $f (- 2) + f (3) = 0$ . If one of the roots of $f (x) = 0$ is $- 1$ , then the sum of the roots of $f (x) = 0$ is equal to:

Let $f (x)$ be a quadratic polynomial such that $f (- 1) + f (2) = 0$ . If one of the roots of $f (x) = 0$ is 3, then its other root lies in :

If f(x) is a quadratic expression such that $f (1) + f (2) = 0$ , and -1 is a root of $f (x) = 0$ , then the other root of $f (x) = 0$ is :-

If one root of the quadratic equation $a x^{2} + b x + c = 0$ is equal to the $n$ -th power of the other, then show that $(a c^{n})^{1/ (n + 1)} + (a^{n} c)^{1/ (n + 1)} + b = 0$

If one root is square of the other root of the equation $x^{2} + p x + q = 0$ , then the relation between $p$ and $q$ is

For the equation $3 x^{2} + p x + 3 = 0, p > 0$ , if one of the root is square of the other, then $p$ is equal to