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Some equations cannot be solved with the algebra you know: x² = 2^x, sin x = x/10, ln x = kx. JEE does not ask you to solve them. It asks how many real roots they have, or for which k there are exactly two. The method is to split the equation into two sides, sketch each, and count crossings. This post sets out the method, works four examples (each checked numerically), and lists the traps.
Where is this in the syllabus?
The NTA Syllabus for JEE (Main) 2026 lists polynomial, trigonometric, logarithmic and exponential functions and "Graphs of simple functions" in Unit 7. The idea starts earlier. The NCERT Class 10 chapter Polynomials (reprint 2026-27) has a section on the geometrical meaning of zeroes: the zeroes of a polynomial are the x-coordinates where its graph meets the x-axis. Counting the roots of f(x) = g(x) is the same idea with a curve in place of the x-axis. The basic graphs are in NCERT Class 11 Relations and Functions, Section 2.4.1.
What is the method?
Split the equation into f(x) = g(x), with two sides you can sketch.
Bound the region: where can roots not be? (|sin x| ≤ 1; ln x needs x > 0.)
Sketch both on the same axes, marking intercepts, peaks and where each side is larger.
Count crossings. Where the curves only touch, use slopes.
Check one or two crossings by substituting simple values.
Why bother with a picture? Larkin and Simon (1987) argued that diagrams display information that is only implicit in equations, which would otherwise have to be computed. Which side is larger on each interval is exactly that kind of information.
Example 1: how many roots does x² = 2^x have?
By inspection x = 2 (4 = 4) and x = 4 (16 = 16). Most students stop there. Now look left of zero. At x = 0, x² = 0 and 2⁰ = 1, so the exponential is higher. At x = −1, x² = 1 and 2⁻¹ = 1/2, so the parabola is higher. They cross between −1 and 0 (near x ≈ −0.767). Between 2 and 4 the parabola is above (at x = 3, 9 > 8). Beyond 4 the exponential stays ahead (at x = 5, 32 > 25). Answer: three real roots.
Example 2: how many roots does sin x = x/10 have?
Bound: |sin x| ≤ 1, so |x| ≤ 10. Both sides are odd, so roots come in ± pairs plus x = 0.
For x > 0:
On (0, π): the sine starts steeper than the line and ends below it (sin π = 0 < π/10), so one crossing.
On (π, 2π): sine is negative, the line positive, so none.
On (2π, 3π): the hump peaks at 1 at x = 5π/2 ≈ 7.85, where the line is only about 0.785, so two crossings.
From 3π ≈ 9.42 to 10: sine is negative, so none.
That is 3 positive roots, 3 negative and x = 0: seven in total. Numerically they are 0, ±2.852, ±7.068 and ±8.423.
Example 3: when the curves only touch
How many roots does e^x = 1 + x have? At x = 0 both sides equal 1 and both have slope 1, so the line is the tangent to e^x there. Since e^x curves upwards, it stays above that tangent everywhere else. Answer: one root. Counting two, because "a line and a curve cross twice", is a common slip.
Example 4: a parameter question
For which k does ln x = kx have exactly two roots? The line y = kx passes through the origin. The tangent from the origin to y = ln x touches it at x = e, where the slope of ln x is 1/e and the line through the origin with slope 1/e reaches (e, 1), a point on the curve. So:
0 < k < 1/e: two roots (for k = 0.2, near x ≈ 1.30 and 12.71);
k = 1/e: one root, at x = e, where they touch;
k > 1/e: none;
k ≤ 0: one root, since ln x − kx then only increases.
Which mistakes cost marks?
Sketching only for x > 0. Example 1's third root is negative.
Treating a touch as a crossing, or the other way round. Compare slopes where the curves meet.
Ignoring the domain. ln x does not exist for x ≤ 0.
Trusting a rough sketch near steep growth. Check values at a few integers.
Choosing hard-to-sketch sides. x² − 2^x = 0 as one curve is harder than two simple curves.
For moving a known graph into position, see our guide to graph transformations for JEE.
A practice routine
Days 1 and 2: count the roots of |x| = cos x (2), ln x = x − 2 (2), e^(−x) = x (1) and x³ = 3^x (2, at x = 3 and near 2.48). Sketch first, then check.
Days 3 and 4: parameter questions. For which k does e^x = kx have two roots? (k > e, from the tangent at x = 1.)
Day 5: a timed mixed set of eight, half with modulus graphs.
After each set: check your counts with a graphing app, as our post on graphing tools for maths suggests, but only after you commit to an answer.
When the shape of an exponential or log curve is the gap, the Fermi Simulation Lab has interactive simulations for the ideas that are easier to see than to read.
Start now: sketch y = x² and y = 2^x on one set of axes and find the negative crossing between −1 and 0.
