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:

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

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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:
\[\begin{align} e^{x+1} - 1 = 0 &\Rightarrow x = - 1 \\ x^3 + x^2 = 0 &\Rightarrow x = 0, - 1. \end{align} \] 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.