25 original JEE Main-style maths questions to sit in one go, then mark. It is set out like the maths section of JEE Main Paper 1: multiple choice first, then numerical-value questions.
If \(f(x) = \dfrac{x - 1}{x + 1}\), \(x \ne -1\), then \(f(f(x))\) (where defined) equals
\(f(f(x)) = \dfrac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1} = \dfrac{(x - 1) - (x + 1)}{(x - 1) + (x + 1)} = \dfrac{-2}{2x} = -\dfrac1x\).
If \(|z - 3 - 4i| = 2\), the greatest value of \(|z|\) is
\(z\) lies on the circle with centre \(3 + 4i\) (distance 5 from the origin) and radius 2. The farthest point is \(5 + 2 = 7\) from the origin.
If \(\alpha\) and \(\beta\) are the roots of \(x^2 + x + 1 = 0\), then \(\alpha^{10} + \beta^{10}\) equals
The roots are \(\omega\) and \(\omega^2\). \(\omega^{10} + \omega^{20} = \omega + \omega^2 = -1\).
If \(A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\) and \(\theta = \dfrac{\pi}{8}\), then \(A^4\) is
\(A\) is a rotation through \(\theta\), so \(A^4\) is a rotation through \(4\theta = \dfrac{\pi}{2}\): \(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\).
\(\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}\) equals
Subtract row 1 from rows 2 and 3 and take out \((b - a)\) and \((c - a)\): the determinant is \((b - a)(c - a)(c - b)\), which equals \((a - b)(b - c)(c - a)\).
Five boys and three girls sit in a row so that no two girls are next to each other. The number of arrangements is
Seat the boys: \(5! = 120\). They make 6 gaps; choose and fill 3 of them with the girls: \(^6P_3 = 120\).
\(120 \times 120 = 14400\).
The middle term in the expansion of \(\left(2x - \dfrac1x\right)^{10}\) is
Eleven terms, so the middle one is \(T_6 = \binom{10}{5}(2x)^5\left(-\dfrac1x\right)^5 = 252 \times 32 \times (-1) = -8064\).
The sum of the first 20 terms of the arithmetic progression \(3, 7, 11, \dots\) is
\(S_{20} = \dfrac{20}{2}\left[2(3) + 19(4)\right] = 10 \times 82 = 820\).
\(\displaystyle\lim_{x \to 0} \frac{e^{3x} - 1}{\sin 2x}\) equals
\(\dfrac{e^{3x} - 1}{3x} \cdot \dfrac{2x}{\sin 2x} \cdot \dfrac{3x}{2x} \to 1 \cdot 1 \cdot \dfrac32\).
\(f(x) = \dfrac{x^3 - 8}{x - 2}\) for \(x \ne 2\) and \(f(2) = k\). If \(f\) is continuous at 2, then \(k\) is
For \(x \ne 2\), \(f(x) = x^2 + 2x + 4 \to 4 + 4 + 4 = 12\). So \(k = 12\).
\(\dfrac{d}{dx}\left[\log(\operatorname{cosec} x - \cot x)\right]\) equals
\(\dfrac{-\operatorname{cosec} x\cot x + \operatorname{cosec}^2 x}{\operatorname{cosec} x - \cot x} = \dfrac{\operatorname{cosec} x(\operatorname{cosec} x - \cot x)}{\operatorname{cosec} x - \cot x} = \operatorname{cosec} x\).
Which points of the parabola \(y = \dfrac{x^2}{2}\) are closest to \((0, 5)\)?
Squared distance \(D = x^2 + \left(\tfrac{x^2}{2} - 5\right)^2\). With \(t = x^2 \ge 0\): \(D = \tfrac{t^2}{4} - 4t + 25\), least at \(t = 8\).
So \(x = \pm 2\sqrt2\), \(y = 4\).
\(\displaystyle\int \frac{dx}{x^2 - 2x + 10}\) equals
Complete the square: \(x^2 - 2x + 10 = (x - 1)^2 + 9\). Then \(\displaystyle\int\frac{dx}{(x - 1)^2 + 3^2} = \frac13\tan^{-1}\frac{x - 1}{3} + C\).
\(\displaystyle\int_0^{\pi} x\sin x\,dx\) equals
By parts: \(\left[-x\cos x\right]_0^{\pi} + \displaystyle\int_0^{\pi}\cos x\,dx = \pi + 0 = \pi\).
The area of the region in the first quadrant between the curve \(y = x^3\) and the line \(y = 4x\) is
They meet where \(x^3 = 4x\): \(x = 0\) or \(x = 2\) in the first quadrant, and the line is above the curve between them.
Area \(= \displaystyle\int_0^2 (4x - x^3)\,dx = \Big[2x^2 - \frac{x^4}{4}\Big]_0^2 = 8 - 4 = 4\).
Trap: Integrating the curve minus the line gives \(-4\): an area is never negative, so take the upper function minus the lower one.
The integrating factor of \(\dfrac{dy}{dx} + \dfrac{2}{x}y = x\), \(x > 0\), is
IF \(= e^{\int 2/x\,dx} = e^{2\log x} = x^2\).
The line through \((2, 3)\) perpendicular to \(3x - 4y + 5 = 0\) is
A perpendicular line has the form \(4x + 3y = c\). Through \((2, 3)\): \(c = 8 + 9 = 17\).
The centre and radius of the circle \(2x^2 + 2y^2 - 8x + 12y - 6 = 0\) are
Divide by 2: \(x^2 + y^2 - 4x + 6y - 3 = 0\). Centre \((2, -3)\), radius \(\sqrt{4 + 9 + 3} = 4\).
Trap: Read the centre from the equation only after making the \(x^2\) and \(y^2\) coefficients 1.
The angle between the planes \(2x - y + z = 1\) and \(x + y + 2z = 3\) is
Normals \((2, -1, 1)\) and \((1, 1, 2)\): \(\cos\theta = \dfrac{2 - 1 + 2}{\sqrt6\sqrt6} = \dfrac12\). The angle is \(\dfrac{\pi}{3}\).
Three fair coins are tossed. Given that at least one shows a head, the probability that exactly two show heads is
\(P(\text{at least one head}) = \tfrac78\), \(P(\text{exactly two}) = \tfrac38\). Conditional probability \(= \dfrac{3/8}{7/8} = \dfrac37\).
Find the number of terms in the expansion of \((x + y + z)^6\).
Terms \(x^ay^bz^c\) with \(a + b + c = 6\): \(\binom{6 + 2}{2} = 28\).
The sum of the coefficients in the expansion of \((3x - 1)^n\) is 256. Find \(n\).
Put \(x = 1\): \(2^n = 256\), so \(n = 8\).
If \(\displaystyle L = \lim_{x \to 0} \frac{\tan 3x - \sin 3x}{x^3}\), find \(2L\).
\(\tan u - \sin u = \tan u(1 - \cos u)\), and with \(u = 3x\): \(\dfrac{\tan 3x}{3x} \cdot \dfrac{1 - \cos 3x}{(3x)^2} \cdot 27 \to 1 \cdot \dfrac12 \cdot 27\).
\(L = \dfrac{27}{2}\), so \(2L = 27\).
Find the number of real solutions of \(|x|^2 - 3|x| + 2 = 0\).
\((|x| - 1)(|x| - 2) = 0\), so \(|x| = 1\) or \(2\): \(x = \pm 1, \pm 2\). Four solutions.
Ten observations have mean 15 and variance 4. Each observation is multiplied by 3. Find the new variance.
Multiplying every value by 3 multiplies the variance by \(3^2 = 9\): \(4 \times 9 = 36\).
All questions are original, written by CBSE Math Revision in the style of each exam; none are taken from NTA, IIT, CBSE or NCERT papers. Every answer was checked twice: by a computer re-solve and by hand. No calculator needed. Spotted a mistake? Tell us.
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