Prove which way a function moves, then count crossings without guessing their locations.
JEE Main · JEE Advanced60 min study guide9 indexed sections
BY THE END, YOU CAN
Use the mean value theorem to justify derivative-sign arguments.
Separate existence from uniqueness when counting real roots.
Handle parameters, excluded points, repeated roots and endpoint equality.
Use Rolle’s theorem as an upper-bound tool without claiming it locates roots.
BEFORE YOU START
Differentiation, continuity, interval notation and sign charts.
Polynomial factorization and elementary logarithms.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. Two questions behind every root count
An equation f(x) = 0 asks where a graph meets the horizontal axis. Finding an exact expression for each meeting point is one possible goal, but counting the meetings is often much easier and more informative. A polynomial of high degree or an equation involving logarithms may have roots with no convenient closed form. Continuity and the direction of travel of the graph can still determine exactly how many there are.
Keep two questions separate. Existence asks whether a root occurs in an interval. Uniqueness asks whether two different roots could occur there. Opposite signs at two endpoints prove existence when the function is continuous between them; they do not establish uniqueness. Strict monotonicity establishes at most one root on an interval; it does not establish that a root actually occurs. A complete exactly-one argument needs both parts.
For example, the function x³ + x increases strictly on the whole real line. This tells us that a horizontal line can meet its graph at most once. The limits −∞ and +∞ at the two ends tell us that every horizontal line does meet it. These are logically different facts. If we instead consider eˣ, strict increase remains true, but the line y = −1 never meets the graph.
Always write the domain first. A rational expression with a pole, a logarithm with a restricted argument, or a square root with a boundary can split the domain into separate intervals. A proof on each interval is not automatically a proof on their union. The direction of motion within each piece must be combined with the function values across the gaps.
02 / UNDERSTAND & APPLY
2. What increasing actually means
A function is strictly increasing on a set if x₁ < x₂ in that set always implies f(x₁) < f(x₂). It is nondecreasing when the final inequality is allowed to be equality. Strictly decreasing and nonincreasing reverse the comparison. These are statements comparing two function values. They are not definitions involving derivatives, and so they also apply to functions that are not differentiable.
The mean value theorem supplies the bridge. Suppose f is continuous on [u,v] and differentiable on (u,v). Then for some c between u and v, f(v) − f(u) = f′(c)(v − u). If f′ is positive throughout an interval, this difference is positive for any two points in the interval, and f is strictly increasing there. If f′ is nonnegative, the same proof gives nondecrease.
A zero derivative at a single point does not stop strict increase. The function x³ has derivative 3x², which vanishes at zero, yet x³ is strictly increasing. One convenient sufficient refinement is that f′ ≥ 0 everywhere in an interval and is not identically zero on any nontrivial subinterval. The mean value theorem first gives nondecrease; if two different endpoints had equal values, nondecrease would force f to be constant between them, contradicting the refinement.
Do not reverse a sufficient criterion carelessly. A differentiable increasing function has nonnegative derivative wherever that derivative exists, but its derivative need not be strictly positive at every point. Also, f′ = 0 at one point means only that the tangent is horizontal there. To conclude that f is constant, the derivative must vanish everywhere on a connected interval.
f(v) − f(u) = f′(c)(v − u), u < c < v
The mean value theorem needs continuity on the closed segment and differentiability in its interior.
03 / UNDERSTAND & APPLY
3. Build a sign chart on the actual domain
The useful breakpoints of a derivative sign chart are its zeros, points where it is undefined, and boundaries or holes of the original function. Factor the derivative when possible. Record the sign of each factor on the resulting open intervals. An even power does not change sign across its zero; an odd power does. A squared denominator is positive where defined, but its zero still marks an excluded point if it came from the original denominator.
If f is continuous at a breakpoint, the strict increase or decrease established on adjacent open pieces can often be extended to include the breakpoint. For instance, x³ is increasing on each side of zero and continuity plus value comparisons join the pieces into one increasing interval. By contrast, f(x) = −1/x has f′(x) = 1/x² > 0 on both components of its domain, but is not increasing on the entire punctured real line: −1 < 1 while f(−1) = 1 > −1 = f(1).
Draw the sign chart before naming intervals. An interval of increase is a set of x-values, not a set of y-values. A local turning value belongs in a separate row because it will later tell us how many horizontal lines cross each branch. A sketch need only show branch direction, endpoints and heights. It does not need artistic accuracy or a numerical plot.
ORIGINAL WORKED EXAMPLE
A stationary point that does not turn
Find maximal intervals of strict increase and decrease of f(x) = x⁵/5 − x³/3.
The domain is all real numbers. Differentiate: f′(x) = x⁴ − x² = x²(x − 1)(x + 1).
The factor x² is positive away from zero. The remaining factor x² − 1 is positive for |x| > 1 and negative for |x| < 1.
Therefore the graph rises before −1, falls from −1 to 1, and rises after 1. The derivative vanishes at zero without a sign change.
Within [−1,1], the derivative is nonpositive and is not zero on any interval of positive length. Hence the entire middle interval is strictly decreasing; it need not be split at zero.
Answer Strictly increasing on (−∞,−1] and [1,∞); strictly decreasing on [−1,1].
04 / UNDERSTAND & APPLY
4. Sign changes, tangencies and missing roots
The intermediate value theorem says that a continuous function on [a,b] takes every value between f(a) and f(b). If f(a)f(b) < 0, zero lies strictly between these endpoint values, so at least one root lies in (a,b). An endpoint with f(a) = 0 must be recorded separately. A sign test that uses a pole instead of a continuous interval is invalid: 1/x changes sign across zero but has no zero.
Same-sign endpoints do not prove absence. The function (x − 1)² is positive at 0 and 2 but vanishes at 1. It touches the axis without crossing. Even when roots do cross, an interval can contain two crossings that restore the original sign. This is why splitting into monotone pieces is so valuable: each branch has a controlled shape, so endpoint data become decisive.
To prove exactly one root on an unbounded interval, replace a finite endpoint value by an appropriate limit and, if necessary, choose sufficiently large finite endpoints for the intermediate value theorem. A limit of zero at infinity is not a root located at infinity. If a branch approaches zero from above and stays above it, no zero is attained.
ORIGINAL WORKED EXAMPLE
An equation with no useful closed-form root
Show that ln x + x − 2 = 0 has exactly one positive real solution and locate it between 1 and 2.
The logarithm requires x > 0. On this domain f′(x) = 1/x + 1 > 0, so there is at most one root.
The function is continuous on [1,2], with f(1) = −1 and f(2) = ln 2 > 0.
The intermediate value theorem gives a root strictly between 1 and 2. Strict increase excludes every other positive root.
No nonpositive value is in the original domain. Thus the domain-wide count is exactly one.
Answer Exactly one real solution, in (1,2).
05 / UNDERSTAND & APPLY
5. Move the horizontal line instead of the graph
An equation g(x) = k compares one fixed graph with a moving horizontal line. Once the turning values and end behavior of g are known, many parameter cases can be handled together. The critical parameter values are usually heights at local extrema, values at included endpoints, and limiting heights at excluded endpoints. Sorting these heights creates the correct case split.
Count distinct x-values unless the problem explicitly asks for multiplicity. At the height of a local maximum or minimum, two crossings can merge into a single tangent contact. For a polynomial that contact may be a repeated root, but it is still one distinct solution. Conversely, a root shared by two closed monotonic intervals is the same root and must not be counted twice.
For a diagram, mark the two turning points, put their heights on the vertical axis, and draw horizontal lines above, between and below those heights. The picture suggests the answer; the derivative signs, endpoint limits and intermediate value theorem certify it. This combination is safer than estimating intersections from a graphing window.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
g(x) = x³ − 3x on a displayed part of ℝ; branch behavior extends beyond the window.
ORIGINAL WORKED EXAMPLE
A complete parameter count
Count the distinct real solutions of x³ − 3x = k for every real k.
Let g(x) = x³ − 3x. Then g′(x) = 3(x − 1)(x + 1), so g rises on (−∞,−1], falls on [−1,1], and rises on [1,∞).
The turning heights are g(−1) = 2 and g(1) = −2. The far-left and far-right limits are −∞ and +∞.
If −2 < k < 2, each of the three open branches contains exactly one solution, producing three distinct roots.
If k = 2, x³ − 3x − 2 = (x − 2)(x + 1)², so the roots are −1 and 2. If k = −2, x³ − 3x + 2 = (x + 2)(x − 1)², so the roots are −2 and 1.
For k > 2 only the right branch reaches that height; for k < −2 only the left branch does.
Answer Three roots for |k| < 2; two distinct roots for |k| = 2; one root for |k| > 2.
06 / UNDERSTAND & APPLY
6. Rolle’s theorem turns roots into derivative roots
If f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), Rolle’s theorem gives at least one point c inside with f′(c) = 0. Imagine leaving a height and returning to it without a jump or corner: some intermediate tangent must be horizontal. The theorem does not say where that tangent occurs, and it does not require the root itself to be a turning point.
If f has n distinct roots in one interval, applying Rolle’s theorem between consecutive roots gives at least n − 1 distinct derivative roots. Consequently, if f′ has at most m roots there, f has at most m + 1. Repeating the argument with higher derivatives can make an upper bound easy. These are upper bounds only: a positive quadratic has no real roots even though its derivative has one.
The hypotheses matter particularly for absolute values and domain gaps. The function |x| − 1 has two roots but has no zero derivative anywhere it is differentiable; the corner at zero prevents a single Rolle argument across the roots. Likewise, do not apply Rolle between roots on opposite sides of a pole. Use piecewise analysis instead.
ORIGINAL WORKED EXAMPLE
Two roots certified without solving an exponential equation
Prove that eˣ − 2x − 2 = 0 has exactly two real roots.
Let f(x) = eˣ − 2x − 2. It is continuous and differentiable on all real numbers, with f′(x) = eˣ − 2.
The derivative has exactly one zero, x = ln 2. Rolle’s theorem therefore limits f to at most two distinct roots.
Now f(−1) = e⁻¹ > 0 and f(0) = −1 < 0, so one root lies in (−1,0). Also f(2) = e² − 6 > 0, so another root lies in (0,2).
The intervals are disjoint and the upper bound is two. Thus these are all the roots. The inequality e² > 6 can be checked from its positive series terms: 1 + 2 + 2 + 8/6 > 6.
Answer Exactly two real roots: one in (−1,0) and one in (0,2).
07 / UNDERSTAND & APPLY
7. Use monotonicity to prove a comparison
To compare two expressions, subtract them and study the difference. A known value at a convenient anchor, often 0 or 1, converts derivative information into an inequality. Keep track of the direction when moving left of the anchor: if a function increases and f(1) = 0, then f(x) < 0 for x < 1 and f(x) > 0 for x > 1.
Parameters inside the derivative lead to a different kind of question: for which parameters is the derivative nonnegative throughout the required interval? This becomes an algebraic inequality over a domain. A condition sufficient on all real numbers may be unnecessarily restrictive on a smaller interval. Conversely, checking only the endpoints of a quadratic derivative is unsafe unless you know where its minimum occurs.
When equality is allowed in the derivative condition, inspect whether the resulting function remains strictly increasing. Isolated zeros generally do not destroy strict increase, whereas a whole interval of zero derivative creates a constant segment. State the distinction explicitly rather than replacing every non-strict derivative inequality by a strict one.
ORIGINAL WORKED EXAMPLE
A parameter condition with an equality boundary
For which real a is f(x) = x³ + ax strictly increasing on the whole real line?
Differentiate: f′(x) = 3x² + a. If a ≥ 0, this is nonnegative for every real x.
For a > 0 the derivative is everywhere positive. For a = 0 it vanishes only at x = 0 and x³ is still strictly increasing.
If a < 0, then f′(0) = a < 0. By continuity of the derivative, f decreases on a small interval around zero, contradicting strict increase.
Thus the boundary case a = 0 belongs in the answer; requiring f′ > 0 everywhere would wrongly exclude it.
Answer Exactly a ≥ 0.
08 / UNDERSTAND & APPLY
8. Original practice with complete checks
Attempt each item by stating a domain and a justification, not just a root count. For count questions, label the part that proves existence and the part that proves uniqueness. These exercises are original practice, not past-paper reproductions.
ORIGINAL PRACTICE · 1
Determine the intervals of increase and decrease of x³ − 6x² + 9x.
Reveal answer & reasoning
Increasing on (−∞,1] and [3,∞); decreasing on [1,3].
The derivative is 3(x − 1)(x − 3), positive outside the two roots and negative between them. Continuity allows the endpoints to be included.
ORIGINAL PRACTICE · 2
How many real roots does x³ + x + 7 have?
Reveal answer & reasoning
Exactly one.
Its derivative 3x² + 1 is positive everywhere. Its limits at −∞ and +∞ have opposite infinite signs, so continuity supplies a root and strict increase makes it unique.
ORIGINAL PRACTICE · 3
How many positive solutions does x − ln x = 2 have?
Reveal answer & reasoning
Exactly two.
On x > 0, the derivative is 1 − 1/x. The function decreases to its minimum value 1 at x = 1 and then increases. It tends to +∞ as x approaches 0 from the right and as x tends to +∞. Each branch therefore crosses height 2 once.
ORIGINAL PRACTICE · 4
Does f′(x) > 0 wherever defined prove f is increasing on a disconnected domain?
Reveal answer & reasoning
No.
The counterexample f(x) = −1/x has positive derivative on both domain components but f(−1) = 1 > −1 = f(1). The mean value theorem cannot cross the missing point zero.
ORIGINAL PRACTICE · 5
Count distinct roots of x⁴ − 2x² = k.
Reveal answer & reasoning
Four for −1 < k < 0; three for k = 0; two for k = −1 or k > 0; none for k < −1.
Write (x² − 1)² = k + 1. For −1 < k < 0, both x² = 1 ± √(k + 1) are positive. At k = −1 only x² = 1 remains. At k = 0 the values are x² = 0 and 2. For k > 0 only the plus value is nonnegative. For k < −1 a square cannot equal k + 1.
ORIGINAL PRACTICE · 6
Prove ln x ≤ x − 1 for x > 0 and identify equality.
Reveal answer & reasoning
Equality only at x = 1.
Set h(x) = x − 1 − ln x. Its derivative (x − 1)/x is negative on (0,1) and positive on (1,∞), so its minimum is h(1) = 0.
ORIGINAL PRACTICE · 7
A differentiable function on ℝ has exactly two zeros of its derivative. What is the largest possible number of distinct real zeros of the function?
Reveal answer & reasoning
Three.
Four distinct zeros would force at least three distinct derivative zeros by Rolle’s theorem. The cubic x³ − x attains the bound of three function zeros and has two derivative zeros.
ORIGINAL PRACTICE · 8
For f(x) = x³ − 3x, count roots of f(x) = 2 restricted to [−1,1].
Reveal answer & reasoning
One distinct root, x = −1.
On [−1,1] the function is strictly decreasing from 2 to −2. The endpoint −1 is the only allowed point with value 2; the other unrestricted root x = 2 lies outside the interval.
09 / UNDERSTAND & APPLY
9. Revision: turn a sketch into a proof
The most common incomplete answer says “the derivative is positive, hence one root.” Replace it with “at most one root,” then supply a sign change or range argument. The second common error is to look only for sign changes and miss tangent roots. The third is to count a shared turning-point root twice when two adjacent branches include it.
A reliable written solution has four compact pieces: the allowed domain, a derivative-sign table, the image or endpoint behavior of each monotonic piece, and the final count with equality cases. If a parameter is present, list the critical heights in increasing order before making cases. If the original question restricts x to an interval, intersect every candidate branch with that interval first.
As a final independent check, compare the result with algebraic degree when the function is a polynomial, substitute easily identified boundary roots, and test one representative parameter from each region. These checks can reveal arithmetic slips, but they do not replace continuity and monotonicity arguments. A graph is excellent evidence for a conjecture; your theorem hypotheses explain why the count is certain.
KEEP THE BIG IDEAS
Quick revision
A sign change plus continuity proves existence; strict monotonicity proves at most one.
Derivative tests apply on intervals and must respect holes and endpoints.
Parameter thresholds are branch heights, and repeated roots count once when counting distinct solutions.
Rolle’s theorem converts bounds on derivative roots into bounds on function roots.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.