JEE Main 2021 (27 Aug Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: The set of all values of $k > - 1$ , for which the equation $(3 x^{2} + 4 x + 3)^{2} - (k + 1) (3 x^{2} + 4 x + 3) (3 x^{2} + 4 x + 2) + k (3 x^{2} + 4 x + 2)^{2} = 0$ has real roots is :

Q: Why do we have to check the case where $k = 1$ separately?

When we derive a quadratic equation $Ax^2 + Bx + C = 0$, we assume $A \neq 0$. If $k = 1$, the coefficient of $x^2$ becomes zero, turning our quadratic into a linear equation (or in this case, an impossible statement $-1 = 0$). Always check if your leading coefficient vanishes at the boundaries of your interval!

Select Answer:

Visualized Solution

Identifying the Repeating Structure

Observe the repeating quadratic expressions:
$(3 x^{2} + 4 x + 3)^{2} - (k + 1) (3 x^{2} + 4 x + 3) (3 x^{2} + 4 x + 2) + k (3 x^{2} + 4 x + 2)^{2} = 0$

Applying Substitution

Let $a = 3 x^{2} + 4 x + 3$
Let $b = 3 x^{2} + 4 x + 2$
Notice that $a - b = 1$
Substitute $a$ and $b$ into the equation:
$a^{2} - (k + 1) ab + k b^{2} = 0$

Factorizing the Expression

Expand the middle term:
$a^{2} - k ab - ab + k b^{2} = 0$
Group and factorize:
$a (a - k b) - b (a - k b) = 0$
$(a - k b) (a - b) = 0$

Analyzing the Factors

From $(a - k b) (a - b) = 0$ , at least one factor is zero.
Since $a - b = 1$ , it cannot be zero ( $a - b \neq = 0$ ).
Therefore, the other factor must be zero:
$a - k b = 0 \Rightarrow a = k b$

Back Substitution

Substitute the original expressions for $a$ and $b$ back into $a = k b$ :
$3 x^{2} + 4 x + 3 = k (3 x^{2} + 4 x + 2)$

Forming the Quadratic in $x$

Expand and rearrange all terms to one side:
$3 x^{2} + 4 x + 3 = 3 k x^{2} + 4 k x + 2 k$
Group the coefficients of $x^{2}$ , $x$ , and the constant:
$3 (k - 1) x^{2} + 4 (k - 1) x + (2 k - 3) = 0$

Condition for Real Roots

For a quadratic equation to have real roots, its discriminant $D$ must be non-negative:
$D = B^{2} - 4 A C \geq 0$
Here, $A = 3 (k - 1)$ , $B = 4 (k - 1)$ , and $C = 2 k - 3$

Setting up the Discriminant Inequality

Substitute $A$ , $B$ , and $C$ into the discriminant condition:
$[4 (k - 1)]^{2} - 4 [3 (k - 1)] [2 k - 3] \geq 0$

Simplifying the Inequality

Factor out the common term $4 (k - 1)$ :
$16 (k - 1)^{2} - 12 (k - 1) (2 k - 3) \geq 0$
$4 (k - 1) [4 (k - 1) - 3 (2 k - 3)] \geq 0$
$4 (k - 1) (- 2 k + 5) \geq 0$

Solving for $k$ using Wavy Curve

Divide by $- 4$ and flip the inequality sign:
$(k - 1) (2 k - 5) \leq 0$
The critical points are $k = 1$ and $k = \frac{5}{2}$ .
Using the wavy curve method, the negative region is:
$k \in [1, \frac{5}{2}]$

Checking the Boundary Condition

Check the leading coefficient of the quadratic equation:
If $k = 1$ , the equation $3 (k - 1) x^{2} + 4 (k - 1) x + (2 k - 3) = 0$ becomes:
$0 x^{2} + 0 x - 1 = 0 \Rightarrow - 1 = 0$
This is impossible, so $k \neq = 1$ .

Final Answer

Combine the interval with the boundary restriction:
$k \in [1, \frac{5}{2}]$ and $k \neq = 1$
Final range for $k$ :
$k \in (1, \frac{5}{2}]$

00:00 / 00:00

The Sigma Insight: Nature of Roots

Solution Diagram

Analyzing the Setup

The given equation is:

(3 x^{2} + 4 x + 3)^{2} - (k + 1) (3 x^{2} + 4 x + 3) (3 x^{2} + 4 x + 2) + k (3 x^{2} + 4 x + 2)^{2} = 0

To simplify this, we define two variables to represent the repeating blocks: Let

a = 3 x^{2} + 4 x + 3

and

b = 3 x^{2} + 4 x + 2

Note that the difference between these two expressions is constant:

a - b = (3 x^{2} + 4 x + 3) - (3 x^{2} + 4 x + 2) = 1

The Master Equation

Substituting

a

and

b

into the original equation, we obtain a homogeneous quadratic form:

a^{2} - (k + 1) ab + k b^{2} = 0

Expanding the middle term allows us to factor the expression by grouping:

a^{2} - k ab - ab + k b^{2} = 0

a (a - k b) - b (a - k b) = 0

(a - k b) (a - b) = 0

Since we established that

a - b = 1

, the term

(a - b)

cannot be zero. Therefore, we must have:

a - k b = 0 \Rightarrow a = k b

The Descent into the Quadratic

Now, we substitute the original expressions for

a

and

b

back into the equation

a = k b

3 x^{2} + 4 x + 3 = k (3 x^{2} + 4 x + 2)

Rearranging this into the standard quadratic form

A x^{2} + B x + C = 0

3 (k - 1) x^{2} + 4 (k - 1) x + (2 k - 3) = 0

For the roots to be real, the discriminant

D = B^{2} - 4 A C

must satisfy

D \geq 0

[4 (k - 1)]^{2} - 4 [3 (k - 1)] [2 k - 3] \geq 0

Final Calculation

We factor out

4 (k - 1)

to simplify the inequality:

4 (k - 1) [4 (k - 1) - 3 (2 k - 3)] \geq 0

4 (k - 1) [4 k - 4 - 6 k + 9] \geq 0

4 (k - 1) (- 2 k + 5) \geq 0

Multiplying by

- 1

reverses the inequality sign, yielding:

(k - 1) (2 k - 5) \leq 0

Using the wavy curve method, we find the interval

k \in [1, \frac{5}{2}]

. However, we must ensure the equation remains a quadratic by checking the coefficient of

x^{2}

: If

k = 1

, the equation becomes

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

, which is impossible.

Thus, we must exclude

k = 1

from our interval. The final solution is: $k \in (1, \frac{5}{2}]$

Common Traps & FAQs

Wait, why can we just ignore the factor

(a - b) = 0

Great observation! We aren't ignoring it; we are analyzing it. Since

a = 3 x^{2} + 4 x + 3

and

b = 3 x^{2} + 4 x + 2

, their difference

a - b

is exactly

1

. Since

1

is never zero, the factor

(a - b)

can never be zero. Therefore, for the product to be zero, the other factor must be the one that vanishes.

Why do we have to check the case where

k = 1

separately?

When we derive a quadratic equation

A x^{2} + B x + C = 0

, we assume

A \neq = 0

. If

k = 1

, the coefficient of

x^{2}

becomes zero, turning our quadratic into a linear equation (or in this case, an impossible statement

- 1 = 0

). Always check if your leading coefficient vanishes at the boundaries of your interval!

Can I use the quadratic formula directly on the original equation?

Technically, yes, but it would be a nightmare of algebra! The beauty of JEE problems lies in finding the hidden symmetry. By using substitution, we reduce a massive expression into a simple, manageable quadratic. Always look for patterns before diving into brute-force calculation.

How do I know if the discriminant should be

\geq 0

or just

> 0

The problem asks for 'real roots'. Real roots include the case where the two roots are identical (a repeated root). Therefore,

D = 0

is allowed, which is why we use the non-strict inequality

\geq 0

Is the wavy curve method always reliable for these inequalities?

Absolutely! The wavy curve method is your best friend for polynomial inequalities. Just ensure that the coefficient of the highest power of

k

is positive before you start drawing your 'waves'. If it's negative, multiply by

- 1

and remember to flip the inequality sign!

Animated Solution for Mathematics - Quadratic Equations: The set of all values of $k > - 1$ , for which the equation $(3 x^{2} + 4 x + 3)^{2} - (k + 1) (3 x^{2} + 4 x + 3) (3 x^{2} + 4 x + 2) + k (3 x^{2} + 4 x + 2)^{2} = 0$ has real roots is :

Select Answer:

Visualized Solution

Identifying the Repeating Structure

Applying Substitution

Factorizing the Expression

Analyzing the Factors

Back Substitution

Forming the Quadratic in $x$

Condition for Real Roots

Setting up the Discriminant Inequality

Simplifying the Inequality

Solving for $k$ using Wavy Curve

Checking the Boundary Condition

Final Answer

The Sigma Insight: Nature of Roots

Analyzing the Setup

The Master Equation

The Descent into the Quadratic

Final Calculation

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Select Answer:

Visualized Solution

Identifying the Repeating Structure

Applying Substitution

Factorizing the Expression

Analyzing the Factors

Back Substitution

Forming the Quadratic in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Condition for Real Roots

Setting up the Discriminant Inequality

Simplifying the Inequality

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k using Wavy Curve

Checking the Boundary Condition

Final Answer

The Sigma Insight: Nature of Roots

Analyzing the Setup

The Master Equation

The Descent into the Quadratic

Final Calculation

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