JEE Quadratic Expressions

This page will teach you how to master JEE quadratic expressions by highlighting the main concepts and providing a list of examples to work through. Establish your mastery by solving the attached problems.

A quadratic expression is a polynomial in any independent variable (let's say $x$) having degree 2. An expression of the form $ax^2+bx+c$, where $a \neq 0$ and $a,b,c \in \mathbb{R}$, is an example of the quadratic expression.

In order to confidently answer JEE problems, you should know the following concepts. Click through on the links if you are unsure.

Graphs:

• Parabola upward facing/downward facing
• Extreme point and extreme value at the vertex of parabola $\left(-\frac{b}{2a},-\frac{D}{4a}\right)$
• Graph transformation : $a ( x-b)^2 + c$
• The graph of $y=ax^2+bx+c$ is as follows:

simple graph of quadratic equations

Rational Expressions:
A rational function of $x$ is defined as the ratio of two polynomials of $x$, say $P(x)$ and $Q(x)$, where $Q(x) \neq 0$. We can write that as $\frac{P(x)}{Q(x)}$.

• Domain of rational functions
• Range of rational functions

Sign of Leading Term:

• $a(x-a)(x-b) > 0 \Leftrightarrow x< a \text{ or } x > b$
• $(x-a ) (x-b) < 0 \Leftrightarrow a < x < b$
• Wavy curve method and its applications

Range:

• If $a > 0$, then the range of $ax^2 + bx + c$ is $\left[-\frac{D}{4a}, \infty\right)$.
• If $a < 0$, then the range of $ax^2 + bx + c$ is $\left( -\infty,-\frac{D}{4a}\right]$.
• Range of $ax^2+bx+c$ when the domain is restricted

In 2 Variables:

• $ax^2 + 2hxy + by^2 + 2gx + 2fy + c$
• The discriminant is $\Delta=abc+2fgh-af^2-bg^2-ch^2$.
• $ax^2 + 2hxy + by^2 + 2gx + 2fy + c$ can be resolved into two linear factors, if $\Delta=0$.

Find the range of $x^2+x+1$ when $-100 \leq x \leq 100$.

$$\begin{array} { llllll } \text{A) } [ -9901, 10101] &&&&&&& &\text{B) }[ 9901, 10101 ] \\ \text{C) } \left[\frac{3}{4},10101 \right] &&&&&&& &\text{D) } \left[\frac{3}{4}, 9901\right] \end{array}$$

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Concepts tested: range, range over restricted domain, graph

Answer: C) $\left[\frac{3}{4},10101 \right]$

Solution 1:
The vertex $\text{at } x=-\frac{1}{2}$ of the parabola $x^2+x+1$ lies in the given domain $-100 \leq x \leq 100$.
Thus, we calculate three values $f(-100)=9901, f\left(-\frac{1}{2}\right)=\frac{3}{4},$ and $f(100)=10101$.
Among these, the minimum is $\frac{3}{4}$ and the maximum is $10101$, implying the range of $x^2+x+1$ is $\left[\frac{3}{4},10101\right].$

Solution 2:
Sketching the graph of $x^2+x+1$ and looking over the values of $y$ when $-100 \leq x \leq 100$, we have the following:

Now, from the graph, it's clear that $(x^2+x+1) \in \left[\frac{3}{4},10101\right]$.

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Common mistakes:

• If you take the values only at the extreme points, i.e. $f(-100)=9901$ and $f(100)=10101$, you'll miss many values above the vertex.

For real $x$ satisfying $\dfrac{6x^2-5x-3}{x^2-2x+6} \leq 4$, the least and the highest values of $4x^2$ are $A$ and $B$, respectively. Find the value of $A-B$.

$$\begin{array} { l l } \text{A) } 81 &&&&&&& &\text{B) }\frac{81}{4} \\ \text{C) } -\frac{81}{4} &&&&&&& &\text{D) } -81 \end{array}$$

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Concepts tested: range, rational expression, sign

Answer: D) $-81$

Solution :
Since the denominator ${ x }^{ 2 }-2x+6={ (x-1) }^{ 2 }+5$ is always positive, we can multiply both sides of the given inequality by it without affecting the inequality sign:

$$\begin{aligned} 6{ x }^{ 2 } -5x-3 &\le 4x^ 2 -8x+24\\ 2{ x }^{ 2 }+3x-27 &\le 0\\ (2x+9)(x-3) &\le 0\\ x &\in \left[ \frac { -9 }{ 2 }, 3 \right] \\ x &\in \left[-\frac{9}{2}, 0\right] \cup (0,3] \\ x^2 &\in \left[0,\frac{81}{4}\right] \cup (0,9] \\ x^2 &\in \left[0,\frac{81}{4}\right]\\ 4x^2 &\in [0, 81]\\ \Rightarrow A-B &=0-81\\&=-81. \end{aligned}$$

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Common mistakes:

• After you get $x \in \left[ \cfrac { -9 }{ 2 } ,3 \right]$, don't just square, because it will result in $x^2 \in \left[9,\frac{81}{4}\right]$.

Given that the following graph represents $y=ax^2+bx+c$, predict the signs of $a,b,c,$ and $D=b^2-4ac$.

$$\begin{array} { l l } \text{A) } a>0, ~b<0, ~c>0, ~D>0 &&&&& &\text{B) } a>0, ~b<0, ~c<0, ~D>0 \\ \\ \text{C) } a>0, ~b>0, ~c>0, ~D>0 &&&&& &\text{D) } a>0, ~b>0, ~c<0, ~D>0 \end{array}$$

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Concepts tested: graph

Answer : B) $a>0,b<0,c<0,D>0$

Solution :
$a>0$, since the parabola opens upward.
$b<0$, since $a>0$ and the $x$-coordinate of the vertex is $-\frac{b}{2a}>0$.
$c<0$, since the $y$-intercept is negative.
$D>0$, since there exist two distinct real roots as in the graph.

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Common mistakes:

• As it requires to know the sign of $a$ to get the sign of $b$, if you tried to find the sign of $b$ before finding the sign of $a$, then you won't get any information about the sign of $b$.

Find the range of $f(x)=3(x^2+x+1)^2+2(x^2+x+1)-3$ over all real values $x$.

$$\begin{array} { l l } \text{A) } \left[ \frac{3}{16}, \infty\right) &&&&& &\text{B) } [ -3, \infty ) \\ \\ \text{C) } [2, \infty) &&&&& &\text{D) } ( 0, \infty ) \end{array}$$

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Concepts tested: range, range over restricted domain, graph

Answer: A) $\left[ \frac{3}{16}, \infty \right)$.

Solution 1:
Step 1: Simplify the expression into $f(x) = h(g(x))$.
We see that $f(x)=3(x^2+x+1)^2+2(x^2+x+1)-3$ is a quadratic expression in terms of $x^2+x+1$.
We can simplify the calculations by setting $g(x) = x^2+x+1$ and $h(x) = 3x^2 + 2x - 3$.
Then, $f(x) = h (g (x) )$.

Step 2: Find the range of $g(x)$.
Since $x^2 + x + 1 = \left( x + \frac{1}{2} \right) ^2 + \frac{3}{4}$, it follows that $g(x) \in \left[ \frac{3}{4} , \infty \right)$.

Step 3: Given range of $g(x)$, find the range of $h(g(x) )$.
Consider the graph of the parabola $h(x) = 3x ^2 + 2x -3$.
The vertex occurs at $x = - \frac{2}{3}$, which is not in $\left[ \frac{3}{4}, \infty \right)$.
Hence, the minimum of $h(g(x))$ will occur at $g\left( \frac{3}{4} \right)$, which is equal to $\frac{3}{16}$.
Clearly, as $x$ tends to infinity, $h(g(x) )$ tend to infinity.

Conclusion: Thus, the range of $f(x)$ is $\left[ \frac{3}{16} , \infty \right)$.

Solution 2:
Let $x^2+x+1=p$, then we have to find the range of $f(p)=3p^2+2p-3,$ where $p \in \left[\frac{3}{4},\infty\right)$.
Let's draw the graph of $y=f(p)=3p^2+2p-3$, which looks like the following:

So, as $p \in \left[ \dfrac{3}{4},\infty\right)$, we can see that $f(p) \in \left[ \dfrac{3}{16},\infty\right).$

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Common mistakes:

• If we tried to expand out $f(x)$ completely, we would obtain a 4$^\text{th}$ degree polynomial. However, we don’t know how to easily solve range problems of 4$^\text{th}$ degree polynomials.

• If you tried substituting in integer values to find the range, you will get some integral values, and miss out on the point $\left(- \frac{3}{4}, \frac{3}{16}\right)$.

Let $p_1,p_2,p_2, \ldots, p_n$ be $n$ real values of $p$ for which the least value of $4x^2-4px+p^2-2p+2$ on the interval $0 \leq x \leq 2$ is equal to 3.

Find the value of $\frac{p_1+p_2+p_3+\cdots+p_n}{n}.$

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Concepts Tested : range, range over restricted domain

Answer : $3.874$

Solution :
Step 1: Figure out possible cases.
Drawing graphs of possible scenarios will be helpful in determining all possible cases.
We have 3 possibilities depending on where the vertex $x = \frac{p}{2}$ is located.

• Case 1: $\frac{p}{2}<0$
• Case 2: $\frac{p}{2}>2$
• Case 3: $0\leq\frac{p}{2}\leq 2$

Step 2: Analyze each case.

• $\frac{p}{2}<0:$ In this case, the minimum will be at $x=0,$ giving $p=1-\sqrt{2}.$

• $\frac{p}{2}>2:$ In this case, the minimum will be at $x=2,$ giving $p=5+\sqrt{10}.$

• $0\leq\frac{p}{2}\leq 2:$ In this case, the minimum will be at $x=\frac{p}{2},$ giving no value of $p.$
  This is because at $x=\frac{p}{2}$ the minimum value will be $2-2p$, and $2-2p=3\implies p=-\frac{1}{2} \notin [0,4]$.

Step 3: Summing all the values $$\frac{\displaystyle\sum_{i=1}^n p_{n}}{n}=\frac{1-\sqrt{2}+5+\sqrt{10}}{2}=3.874.$$

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Common mistakes:

• If you don't consider all the cases, you're gonna miss out.

Find the values of $x$ for which $\dfrac{\left(e^{x+1}-1\right)\left(x^3+x^2\right)}{2x-2} \geq 0$.

$$\begin{array} { l l } \text{A) } [-1,0] \cup (1,\infty) &&&&& &\text{B) } \{-1,0\} \cup (1,\infty) \\ \text{C) } (-1,0) \cup (1,\infty) &&&&& &\text{D) } (1,\infty ) \\ \end{array}$$

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Concepts Tested : sign, wavy curve method

Answer : B) $\{-1,0\} \cup (1,\infty)$

Solution :
We use the wavy curve method.

Step 1: Identify the points where the curve touches the $x$-axis, or there is a point of discontinuity.
Curve touches the $x$-axis when the numerator is 0:
$$\begin{aligned} e^{x+1} - 1 = 0 &\Rightarrow x = - 1 \\ x^3 + x^2 = 0 &\Rightarrow x = 0, - 1. \end{aligned}$$ Curve has a vertical asymptote when the denominator is 0:
$$2x - 2 = 0 \Rightarrow x = 1.$$ Thus, the points to consider are $x = -1, 0, 1$.

Step 2: Plot the points and test the domains.
Testing $x < -1$, use $x = -2$. The expression is $\frac{ (-) (-) } { (-) } = -$.
Testing $-1 < x < 0$, use $x = -0.5$. The expression is $\frac{ (+)(+)} { (-) } = -$.
Testing $0 < x < 1$, use $x = 0.5$. The expression is $\frac{ (+)(+) } { (-1) } = -$.
Testing $1 < x$, use $x = 2$. The expression is $\frac{ (+)(+) } { (+)} = +$.

Step 3: Draw the graph, find the positive regions.
From the above steps, we obtain the following graph:

Hence, the positive region is $(1,\infty)$.

Step 4: Consider the cases of equality.
The numerator equals to $0$, if $x=0$ or $x=-1$.

Hence, the set of the values of $x$ satisfying $\frac{(e^{x+1}-1)(x^3+x^2)}{2x-2} \geq 0$ is $\{-1,0\} \cup (1,\infty)$.

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Common mistakes:

• It is not true that when the curve touches the $x$-axis, the sign must change. Instead, the curve could bounce back as in the case of $y = x^2$.

Once you are confident about quadratic expressions, move on to JEE Quadratic Equations or JEE Complex Numbers.