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    JEE Main Revision Plan 2026 - Subject- Wise, JEE Notes, Preparation Tips

    Quadratic Inequalities - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

    Quick Facts

    • 15 Questions around this concept.

    Solve by difficulty

    Solution of (x+1)(x-9)<0 is

    Solution of x(x-4)(x+5)>0

    The least integer satisfying, $49.4- \frac{(27-y)}{10} \: < \: 47.4-\left ( \frac{27-9y}{10} \right )$ is

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    Solutions of (x - 1) (x - 2) = (x - 1) are

    Let $f(x)=x^2+3 x+2$, then which of the following is correct

    Find the set of values of ' $a$ ' for which the given condition is true $(a-1)(a-3)(a+5)>0$

    Concepts Covered - 1

    Wavy Curve Method

    The wavy curve method is used to solve the inequality of the type 

    $\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}>0$ or $(<, \leq, \geq)$

    We use the following steps in the wavy curve method to solve a question (start by getting 0 on one side of inequality)

    1: Factorize the numerator and denominator into linear factors.

    2: Make coefficients of x positive in all linear factors.

    3: Equate each linear factor to zero and find the values of x in each case. The values are called critical points. Do not include the linear factors with even power while finding critical points.

    4: Identify distinct critical points on the real number line. The “n” numbers of distinct critical points divide real number lines in (n+1) sub-intervals.

    5: The sign of rational function in the rightmost interval is positive. Alternate sign in adjoining intervals on the left.

    6. Check for each critical point and points from even powered linear factors, if these are to be included in the answer or not.

    For Example: using the  wavy curve method to find the interval of x for the inequality given :

    $\frac{x}{x-1} \geq 0$

    Steps

    0 on the right-hand side (already there)

    all linear factors (already present)

    Critical points are: x=0,1

    The critical points are marked on the real number line. Starting with a positive sign in the rightmost interval, we denote signs of adjacent intervals by alternating signs.

    Hence, $\mathrm{x} \in(-\infty, 0] \cup(1, \infty)$

    Note that 1 cannot be taken in the answer as at x=1, the denominator becomes 0 and hence expression is not defined.

     

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    Wavy Curve Method

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