JEE (Main) · 2025
JEE (Main) 2025 — 4 April, Shift 1
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
25 Q
- Limits, Continuity and Differentiability3
- Sets, Relations and Functions2
- Sequences and Series2
4 easy · 20 medium · 1 hard
Physics
6 Q
- Oscillations and Waves1
- Physical World and Measurement1
- Motion of System of Particles and Rigid Body1
5 easy · 1 medium · 0 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Set $D$: solve $x^2 + y^2 = 25$ and $x^2 + 9y^2 = 144$ together. Subtracting, $8y^2 = 119$, so $y^2 = \frac{119}{8}$ and $x^2 = 25 - \frac{119}{8} = \frac{81}{8}$. Both are positive, so $x = \pm\frac{9}{2\sqrt2}$, $y = \pm\sqrt{\frac{119}{8}}$: four points. $n(D) = 4$.
Set $C$: integer points with $x^2 + y^2 \le 4$.
$x = 0$: $y = -2, -1, 0, 1, 2$ (5 points).
$x = \pm1$: $y^2 \le 3$, $y = -1, 0, 1$ (3 each, 6 points).
$x = \pm2$: $y = 0$ (2 points).
$n(C) = 13$.
One-one functions from a 4-element set to a 13-element set: $13 \times 12 \times 11 \times 10 = 17160$.
Answer: (B) $17160$. | ||||
| 2 | Sets, Relations and FunctionsComposition of functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on composition of functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions $f(g(x)) = \frac{2g + 3}{5g + 2}$ with $g = \frac{2-3x}{1-x}$.
$2g + 3 = \frac{4 - 6x + 3 - 3x}{1-x} = \frac{7 - 9x}{1-x}$, and $5g + 2 = \frac{10 - 15x + 2 - 2x}{1-x} = \frac{12 - 17x}{1-x}$.
So $fog(x) = \frac{7 - 9x}{12 - 17x} = \frac{9x - 7}{17x - 12}$.
Its derivative is $\frac{9(17x-12) - 17(9x-7)}{(17x-12)^2} = \frac{11}{(17x-12)^2} > 0$, so it increases on $[2, 4]$.
$\alpha = fog(2) = \frac{11}{22} = \frac12$, $\beta = fog(4) = \frac{29}{56}$.
$\beta - \alpha = \frac{29}{56} - \frac{28}{56} = \frac{1}{56}$, so $\frac{1}{\beta - \alpha} = 56$.
Answer: (C) $56$. | ||||
| 3 | Quadratic EquationsNature of roots | Class 11: Complex Numbers and Quadratic Equations | Single correctEasy | |
A easy single correct question on nature of roots, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations The roots are $x = -2 \pm \sqrt{4 + n}$. They are integers exactly when $4 + n$ is a perfect square $k^2$ (then the roots are $-2 \pm k$).
$n \in [20, 100]$ gives $4 + n \in [24, 104]$. The squares there: $25, 36, 49, 64, 81, 100$, i.e. $n = 21, 32, 45, 60, 77, 96$.
That is 6 values.
Answer: (B) $6$. | ||||
| 4 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series $A$: $1 + 5(k-1)$, $k = 1, \ldots, 2025$; the last term is $1 + 5 \cdot 2024 = 10121$.
$B$: $9 + 7(k-1)$; the last term is $9 + 7 \cdot 2024 = 14177$.
Common terms leave remainder 1 on division by 5 and 2 on division by 7. The first is 16, and they repeat every $\text{lcm}(5, 7) = 35$: $16, 51, 86, \ldots$
They must not exceed 10121 (the end of $A$): $16 + 35m \le 10121 \Rightarrow m \le 288.7$, so $m = 0, 1, \ldots, 288$: 289 common terms.
$n(A \cup B) = 2025 + 2025 - 289 = 3761$.
Answer: (B) $3761$. | ||||
| 5 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctMedium | |
A medium single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series The terms are the odd numbers $1, 3, 5, 7, 9, \ldots$, with every term in an odd position squared: $1^2, 3, 5^2, 7, 9^2, 11, \ldots$
In 40 terms there are 20 squared terms $(4k-3)^2$ and 20 plain terms $4k - 1$, $k = 1, \ldots, 20$.
$\sum_{k=1}^{20}(4k-3)^2 = 16\sum k^2 - 24\sum k + 9 \cdot 20 = 16(2870) - 24(210) + 180 = 45920 - 5040 + 180 = 41060$.
$\sum_{k=1}^{20}(4k-1) = 4(210) - 20 = 820$.
Total $= 41060 + 820 = 41880$.
Answer: (D) $41880$. | ||||
| 6 | Binomial TheoremGeneral and middle term | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on general and middle term, from Binomial Theorem. NCERT: Class 11: Binomial Theorem Let $a = 2^{1/3}$, $b = 3^{-1/3}$.
15th term from the beginning: ${}^nC_{14}a^{n-14}b^{14}$. 15th term from the end is the 15th term of $(b + a)^n$: ${}^nC_{14}b^{n-14}a^{14}$.
Ratio $= \left(\frac{a}{b}\right)^{n-28} = (2^{1/3}\cdot3^{1/3})^{n-28} = 6^{(n-28)/3}$.
$6^{(n-28)/3} = 6^{-1} \Rightarrow n - 28 = -3 \Rightarrow n = 25$.
${}^{25}C_3 = \frac{25\cdot24\cdot23}{6} = 2300$.
Answer: (B) $2300$. | ||||
| 7 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $(x+y)^{2n-3}$ has $2n - 2$ coefficients, whose sum is $2^{2n-3}$.
Mean: $\frac{2^{2n-3}}{2n-2} = 16$. Try $n = 5$: $\frac{2^7}{8} = 16$. (The left side increases with $n$ for $n \ge 3$, and $n = 2, 3, 4$ give $1, 2, 5.33$, so $n = 5$ is the only solution.)
P $= (9, 5)$. Distance from $x + y - 8 = 0$: $\frac{|9 + 5 - 8|}{\sqrt2} = \frac{6}{\sqrt2} = 3\sqrt2$.
Answer: (C) $3\sqrt2$. | ||||
| 8 | ProbabilityProbability of an event | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability of an event, from Probability. NCERT: Class 11: Probability; Class 12: Probability There are 16 persons and a committee of 12 leaves out 4. Total committees: ${}^{16}C_{12} = {}^{16}C_4 = 1820$.
"At least 3 of 4 engineers and at least 1 of 2 doctors in" means at most 1 engineer and at most 1 doctor are left out. Count the 4 left out:
0 engineers, 0 doctors: ${}^{10}C_4 = 210$.
1 engineer, 0 doctors: $4 \cdot {}^{10}C_3 = 480$.
0 engineers, 1 doctor: $2 \cdot {}^{10}C_3 = 240$.
1 engineer, 1 doctor: $4 \cdot 2 \cdot {}^{10}C_2 = 360$.
Favourable $= 1290$. Probability $= \frac{1290}{1820} = \frac{129}{182}$.
Answer: (A) $\frac{129}{182}$. | ||||
| 9 | ProbabilityProbability distribution of a random variable | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on probability distribution of a random variable, from Probability. NCERT: Class 11: Probability; Class 12: Probability The 2 pens are drawn together (without replacement). ${}^{10}C_2 = 45$.
$P(X=0) = \frac{{}^7C_2}{45} = \frac{21}{45}$, $P(X=1) = \frac{3\cdot7}{45} = \frac{21}{45}$, $P(X=2) = \frac{{}^3C_2}{45} = \frac{3}{45}$.
$E(X) = \frac{21 + 6}{45} = \frac{27}{45} = \frac35$.
$E(X^2) = \frac{21 + 12}{45} = \frac{33}{45} = \frac{11}{15}$.
$\text{Var}(X) = \frac{11}{15} - \frac{9}{25} = \frac{55 - 27}{75} = \frac{28}{75}$.
Answer: (D) $\frac{28}{75}$. | ||||
| 10 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | Single correctEasy | |
A easy single correct question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines Slopes: $4x - 7y + 10 = 0$ has slope $\frac47$; $7x + 4y = 15$ has slope $-\frac74$. Their product is $-1$, so these two sides are perpendicular, and the orthocentre of a right triangle is the vertex at the right angle.
Solve $4x - 7y = -10$ and $7x + 4y = 15$: $16x - 28y = -40$ and $49x + 28y = 105$ add to $65x = 65$, so $x = 1$, $y = 2$. Orthocentre $(1, 2)$.
The triangle formed by $x = 0$, $y = 0$, $x + y = 1$ is right-angled at the origin, so its orthocentre is $(0, 0)$.
Distance $= \sqrt{1 + 4} = \sqrt5$.
Answer: (C) $\sqrt5$. | ||||
| 11 | Conic SectionsEllipse | Class 11: Conic Sections | Single correctEasy | |
A easy single correct question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Distance between the foci $= 2ae = 8$, so $ae = 4$. With $e = \frac45$, $a = 5$.
$b^2 = a^2(1 - e^2) = 25\left(1 - \frac{16}{25}\right) = 9$.
Latus rectum $= \frac{2b^2}{a} = \frac{18}{5}$.
Answer: (A) $\frac{18}{5}$. | ||||
| 12 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Let $t = \sin^2\theta$, so $\cos^2\theta = 1 - t$.
$10t^2 + 15(1-t)^2 = 6 \Rightarrow 25t^2 - 30t + 9 = 0 \Rightarrow (5t - 3)^2 = 0 \Rightarrow t = \frac35$.
So $\sin^2\theta = \frac35$, $\cos^2\theta = \frac25$.
$27\,\text{cosec}^6\theta = 27\cdot\frac{125}{27} = 125$; $8\sec^6\theta = 8\cdot\frac{125}{8} = 125$; $16\sec^8\theta = 16\cdot\frac{625}{16} = 625$.
Value $= \frac{250}{625} = \frac25$.
Answer: (C) $\frac25$. | ||||
| 13 | TrigonometryInverse trigonometric functions and their properties | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctMedium | |
A medium single correct question on inverse trigonometric functions and their properties, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions Put $x = \sin\phi$ with $\phi = \sin^{-1}x$. Since $-\frac12 < x < \frac1{\sqrt2}$, $-\frac{\pi}{6} < \phi < \frac{\pi}{4}$, and $\sqrt{1-x^2} = \cos\phi$.
$\frac{\sqrt3}{2}\sin\phi + \frac12\cos\phi = \sin\phi\cos\frac{\pi}{6} + \cos\phi\sin\frac{\pi}{6} = \sin\left(\phi + \frac{\pi}{6}\right)$.
$\phi + \frac{\pi}{6}$ lies in $\left(0, \frac{5\pi}{12}\right)$, inside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $\sin^{-1}\sin\left(\phi + \frac{\pi}{6}\right) = \phi + \frac{\pi}{6}$.
Value $= \frac{\pi}{6} + \sin^{-1}x$.
Answer: (B). | ||||
| 14 | Three Dimensional GeometrySkew lines and shortest distance | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on skew lines and shortest distance, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry Line 1 passes through $(3, \alpha, 3)$ with direction $\vec b_1 = (3, -1, 1)$; line 2 through $(-3, -7, \beta)$ with $\vec b_2 = (-3, 2, 4)$.
$\vec b_1 \times \vec b_2 = \hat i(-4 - 2) - \hat j(12 + 3) + \hat k(6 - 3) = (-6, -15, 3)$, magnitude $\sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}$.
Joining vector: $(-3 - 3, -7 - \alpha, \beta - 3) = (-6, -7-\alpha, \beta-3)$.
Dot product with $(-6, -15, 3)$: $36 + 15(7 + \alpha) + 3(\beta - 3) = 132 + 15\alpha + 3\beta$.
Shortest distance $= \frac{|132 + 15\alpha + 3\beta|}{3\sqrt{30}} = 3\sqrt{30} \Rightarrow |132 + 15\alpha + 3\beta| = 270$.
$15\alpha + 3\beta = 138$ or $-402$, i.e. $5\alpha + \beta = 46$ or $-134$. The positive value is 46.
Answer: (C) $46$. | ||||
| 15 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry A general point of $L$: $(6 + 3t, 7 + 2t, 7 - 2t)$.
Foot from $P(1, 2, 3)$: the vector $(5 + 3t, 5 + 2t, 4 - 2t)$ must be perpendicular to $(3, 2, -2)$:
$15 + 9t + 10 + 4t - 8 + 4t = 17 + 17t = 0 \Rightarrow t = -1$. Foot $F = (3, 5, 9)$.
The unit direction is $\frac{(3, 2, -2)}{\sqrt{17}}$, so $A, B = F \pm 2\sqrt{17}\cdot\frac{(3,2,-2)}{\sqrt{17}} = F \pm (6, 4, -4)$:
$A = (9, 9, 5)$, $B = (-3, 1, 13)$.
$\overrightarrow{OA}\cdot\overrightarrow{OB} = -27 + 9 + 65 = 47$.
Answer: (B) $47$. | ||||
| 16 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec u\cdot\vec v = 6 - 1 = 5$, $|\vec u| = \sqrt{10}$, $|\vec v| = \sqrt{5 + \lambda^2}$.
$\frac{5}{\sqrt{10}\sqrt{5+\lambda^2}} = \frac{\sqrt5}{2\sqrt7}$. Squaring: $\frac{25}{10(5+\lambda^2)} = \frac{5}{28} \Rightarrow 700 = 50(5 + \lambda^2) \Rightarrow \lambda^2 = 9$, $\lambda = 3$.
Since $\vec v_1 \perp \vec v_2$, $|\vec v|^2 = |\vec v_1|^2 + |\vec v_2|^2$ (Pythagoras).
$|\vec v|^2 = 4 + 1 + 9 = 14$.
(Check: $|\vec v_1| = \frac{\vec u\cdot\vec v}{|\vec u|} = \frac{5}{\sqrt{10}}$, $|\vec v_1|^2 = 2.5$, $|\vec v_2|^2 = 11.5$; sum 14.)
Answer: (A) $14$. | ||||
| 17 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Put $t = x - 1 \to 0^+$; $\sin(1-x) = -\sin t$. The numerator is $t(6 + \lambda\cos t) - \mu\sin t$.
Expand: $t\left(6 + \lambda - \frac{\lambda t^2}{2}\right) - \mu\left(t - \frac{t^3}{6}\right) + O(t^5) = (6 + \lambda - \mu)t + \left(-\frac{\lambda}{2} + \frac{\mu}{6}\right)t^3 + \ldots$
Divided by $t^3$, a finite limit needs $6 + \lambda - \mu = 0$, and then the limit is $-\frac{\lambda}{2} + \frac{\mu}{6} = -1$.
With $\mu = 6 + \lambda$: $-\frac{\lambda}{2} + 1 + \frac{\lambda}{6} = -1 \Rightarrow -\frac{\lambda}{3} = -2 \Rightarrow \lambda = 6$, $\mu = 12$.
$\lambda + \mu = 18$.
Answer: (B) $18$. | ||||
| 18 | Limits, Continuity and DifferentiabilityLimits of functions | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | Single correctMedium | |
A medium single correct question on limits of functions, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Replace $x$ by $\frac{x}{2^k}$ in $f(2x) - f(x) = x$: $f\left(\frac{x}{2^{k-1}}\right) - f\left(\frac{x}{2^k}\right) = \frac{x}{2^k}$.
Adding for $k = 1, \ldots, n$ (the sum telescopes): $f(x) - f\left(\frac{x}{2^n}\right) = x\left(\frac12 + \frac14 + \ldots + \frac1{2^n}\right) = x\left(1 - \frac1{2^n}\right)$.
As $n \to \infty$ this tends to $x$, so $G(x) = x$. (By continuity $f\left(\frac{x}{2^n}\right) \to f(0) = 1$, which also gives $f(x) = 1 + x$.)
$\sum_{r=1}^{10}G(r^2) = \sum_{r=1}^{10} r^2 = \frac{10\cdot11\cdot21}{6} = 385$.
Answer: (B) $385$. | ||||
| 19 | Integral CalculusProperties of definite integrals | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on properties of definite integrals, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Split the numerator: $\left(1 + \sqrt{|x|-x}\right)e^x + \sqrt{|x|-x}\,e^{-x} = e^x + \sqrt{|x|-x}\,(e^x + e^{-x})$.
So the integrand is $\frac{e^x}{e^x + e^{-x}} + \sqrt{|x| - x}$.
First part: with $I_1 = \int_{-1}^{1}\frac{e^x}{e^x+e^{-x}}dx$, replacing $x$ by $-x$ gives $I_1 = \int_{-1}^{1}\frac{e^{-x}}{e^x+e^{-x}}dx$; adding, $2I_1 = \int_{-1}^1 1\,dx = 2$, so $I_1 = 1$.
Second part: for $x \ge 0$, $|x| - x = 0$; for $x < 0$, $|x| - x = -2x$. So $\int_{-1}^{0}\sqrt{-2x}\,dx = \sqrt2\int_0^1\sqrt u\,du = \frac{2\sqrt2}{3}$.
Value $= 1 + \frac{2\sqrt2}{3}$.
Answer: (D). | ||||
| 20 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Write $f(x) = 1 - 2x + e^x\int_0^x e^{-t}f(t)\,dt$. At $x = 0$, $f(0) = 1$.
Differentiate: $f'(x) = -2 + e^x\int_0^x e^{-t}f(t)\,dt + f(x) = -2 + (f(x) - 1 + 2x) + f(x)$.
So $f' - 2f = 2x - 3$, a linear equation with integrating factor $e^{-2x}$:
$\left(fe^{-2x}\right)' = (2x - 3)e^{-2x}$, and $\int(2x-3)e^{-2x}dx = -\frac{(2x-3)e^{-2x}}{2} - \frac{e^{-2x}}{2} = (1 - x)e^{-2x}$.
$f(x) = 1 - x + Ce^{2x}$; $f(0) = 1$ gives $C = 0$. So $f(x) = 1 - x$.
The line $y = 1 - x$ meets the axes at $(1, 0)$ and $(0, 1)$: area $= \frac12\cdot1\cdot1 = \frac12$.
Answer: (C) $\frac12$. | ||||
| 21 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | NumericalMedium | |
A medium numerical question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations With $z = x + iy$: $A$ is the circle $(x - 2)^2 + (y - 1)^2 = 9$.
$z - iz = x + iy - ix + y = (x + y) + i(y - x)$, so $B$ is the line $x + y = 2$.
Put $y = 2 - x$: $(x-2)^2 + (1-x)^2 = 9 \Rightarrow 2x^2 - 6x - 4 = 0 \Rightarrow x^2 - 3x - 2 = 0$ (two real roots, discriminant 17). $x_1 + x_2 = 3$, $x_1x_2 = -2$, $x_1^2 + x_2^2 = 9 + 4 = 13$.
$|z|^2 = x^2 + (2 - x)^2 = 2x^2 - 4x + 4$.
Sum $= 2(13) - 4(3) + 8 = 22$.
Answer: 22. | ||||
| 22 | Matrices and DeterminantsAlgebra of matrices | Class 12: Matrices; Class 12: Determinants | NumericalMedium | |
A medium numerical question on algebra of matrices, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants $A$ is a rotation of the $xz$-plane by $\theta$ (the middle row and column untouched). Multiplying out, $A^2$ is the same matrix with $2\theta$, and $A^T$ is the same matrix with $-\theta$.
$A^2 = A^T$ needs $\cos2\theta = \cos\theta$ and $\sin2\theta = -\sin\theta$, i.e. $3\theta$ a multiple of $2\pi$; in $(0, \pi)$, $\theta = \frac{2\pi}{3}$. Then $A^3 = A\cdot A^2 = AA^T = I$ (rotation by $2\pi$).
Since $A$ and $I$ commute, $(A+I)^3 + (A-I)^3 = 2A^3 + 6A$. So the matrix is $2A^3 + 6A - 6A = 2A^3 = 2I$.
Sum of diagonal elements $= 2 \times 3 = 6$.
Answer: 6. | ||||
| 23 | Conic SectionsEllipse | Class 11: Conic Sections | NumericalHard | |
A hard numerical question on ellipse, from Conic Sections. NCERT: Class 11: Conic Sections Point $P$: the line touches $C$ (centre $(0, 1)$, radius $\sqrt2$) at the foot of the perpendicular from the centre. Move from $(0,1)$ along the normal $(1,1)$: $(t, 1 + t)$ on $x + y = 3$ gives $t = 1$, so $P = (1, 2)$.
Points of the line at distance $\frac{2\sqrt2}{3}$ from $P$: $(1 \pm s, 2 \mp s)$ with $s\sqrt2 = \frac{2\sqrt2}{3}$, $s = \frac23$. These are $\left(\frac53, \frac43\right)$ and $\left(\frac13, \frac83\right)$, one on each side of $P$, so $P$ is the midpoint of $QR$.
Which is which: the tangent to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_0, y_0)$ is $\frac{xx_0}{a^2} + \frac{yy_0}{b^2} = 1$; matching $x + y = 3$ gives $x_0 = \frac{a^2}{3}$, $y_0 = \frac{b^2}{3}$. For $E_1$ ($a > b$) $x_0 > y_0$: $Q = \left(\frac53, \frac43\right)$ ($a^2 = 5$, $b^2 = 4$), and $R = \left(\frac13, \frac83\right)$ for $E_2$ ($a^2 = 1$, $b^2 = 8$). Both satisfy $a^2 + b^2 = 9$, the tangency condition.
$x_1y_1 + x_2y_2 + x_3y_3 = 2 + \frac{20}{9} + \frac{8}{9} = \frac{46}{9}$.
$9\times\frac{46}{9} = 46$.
Answer: 46. | ||||
| 24 | Limits, Continuity and DifferentiabilityDifferentiability | Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability | NumericalMedium | |
A medium numerical question on differentiability, from Limits, Continuity and Differentiability. NCERT: Class 11: Limits and Derivatives; Class 12: Continuity and Differentiability Compare the odd powers on each interval.
$x \le -1$: $x^{21} \le \ldots \le x^3 \le x$, so $f(x) = x$.
$-1 \le x \le 0$: the higher the power the closer to 0, so $f(x) = x^{21}$.
$0 \le x \le 1$: $x \ge x^3 \ge \ldots$, so $f(x) = x$.
$x \ge 1$: $f(x) = x^{21}$.
The pieces agree at $-1$, $0$ and $1$, so $f$ is continuous everywhere: $n = 0$.
Slopes at the joins: at $x = -1$, left 1 and right $21x^{20} = 21$; at $x = 0$, left $21\cdot0 = 0$ and right 1; at $x = 1$, left 1 and right 21. All three differ, so $f$ is not differentiable at $-1, 0, 1$: $m = 3$.
$m + n = 3$.
Answer: 3. | ||||
| 25 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | NumericalMedium | |
A medium numerical question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Where $y = 4\sqrt x$ meets $y = |x - 5|$:
for $x \ge 5$: $x - 5 = 4\sqrt x \Rightarrow (\sqrt x - 5)(\sqrt x + 1) = 0 \Rightarrow x = 25$;
for $x < 5$: $5 - x = 4\sqrt x \Rightarrow (\sqrt x + 5)(\sqrt x - 1) = 0 \Rightarrow x = 1$.
The region lies over $1 \le x \le 25$:
$A = \int_1^{25}4\sqrt x\,dx - \int_1^{25}|x - 5|\,dx = \frac83\left(125 - 1\right) - \left(\int_1^5(5-x)dx + \int_5^{25}(x-5)dx\right)$
$= \frac{992}{3} - (8 + 200) = \frac{992 - 624}{3} = \frac{368}{3}$.
$3A = 368$.
Answer: 368. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Oscillations and WavesStanding waves and beats | Class 11: Oscillations; Class 11: Waves | Single correctEasy | |
A easy single correct question on standing waves and beats, from Oscillations and Waves. NCERT: Class 11: Oscillations; Class 11: Waves A pipe closed at one end, with air column $L$, has fundamental frequency $f = \frac{v}{4L}$.
Filling $\frac15$ of its volume with water leaves an air column $L' = \frac45L$ (still closed at one end, now by the water surface).
$f' = \frac{v}{4L'} = \frac{v}{4\cdot\frac45L} = \frac54f$.
Change $= \frac{f' - f}{f}\times100\% = +25\%$.
Answer: (B) $25\%$. | ||||
| 27 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctMedium | |
A medium single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Electric flux $\Phi_E = EA$. $[E] = \frac{[F]}{[q]} = \frac{\mathrm{MLT^{-2}}}{\mathrm{AT}} = \mathrm{MLT^{-3}A^{-1}}$, so $[\Phi_E] = \mathrm{ML^3T^{-3}A^{-1}}$.
Magnetic flux $\Phi_B = BA$. From $F = qvB$, $[B] = \frac{\mathrm{MLT^{-2}}}{\mathrm{AT\cdot LT^{-1}}} = \mathrm{MT^{-2}A^{-1}}$, so $[\Phi_B] = \mathrm{ML^2T^{-2}A^{-1}}$.
$\frac{[\Phi_E]}{[\Phi_B]} = \mathrm{M^0L^1T^{-1}A^0}$ (a speed; indeed $E = cB$ for a light wave).
$Q = 1$, $R = -1$.
Answer: (B) $(1, -1)$. | ||||
| 28 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | StatementsEasy | |
A easy statements question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion A. $\vec r\times\vec L$ has dimensions of (length × angular momentum), not torque. Wrong.
B. $\vec r\times\vec p = \vec L$, and $\vec\tau = \frac{d\vec L}{dt}$. Correct.
C. $\frac{d\vec p}{dt} = \vec F$, so $\vec r\times\frac{d\vec p}{dt} = \vec r\times\vec F$. Correct.
D. For rotation about a fixed axis, $\tau = I\alpha$. Correct.
E. The definition $\vec\tau = \vec r\times\vec F$. Correct.
So B, C, D and E.
Answer: (C) B, C, D and E Only. | ||||
| 29 | Laws of MotionEquilibrium of concurrent forces | Class 11: Laws of Motion | Single correctEasy | |
A easy single correct question on equilibrium of concurrent forces, from Laws of Motion. NCERT: Class 11: Laws of Motion With both angles measured from the horizontal ceiling:
Horizontal balance: $T_1\cos\theta_1 = T_2\cos\theta_2 \Rightarrow \sqrt3\cos\theta_1 = \cos\theta_2$.
$\theta_1 = 30^\circ$ would need $\cos\theta_2 = \frac32 > 1$, impossible; $\theta_1 = 45^\circ$ gives $\cos\theta_2 = \sqrt{\frac32} > 1$, impossible. $\theta_1 = 60^\circ$ gives $\cos\theta_2 = \frac{\sqrt3}{2}$, $\theta_2 = 30^\circ$.
Vertical balance: $T_1\sin60^\circ + T_2\sin30^\circ = mg \Rightarrow \sqrt3T_2\cdot\frac{\sqrt3}{2} + \frac{T_2}{2} = 2T_2 = mg$, so $T_2 = \frac{mg}{2}$.
Answer: (B) $\theta_1 = 60^\circ$, $\theta_2 = 30^\circ$ with $T_2 = \frac{mg}{2}$. | ||||
| 30 | GravitationEscape velocity | Class 11: Gravitation | Assertion and reasonEasy | |
A easy assertion and reason question on escape velocity, from Gravitation. NCERT: Class 11: Gravitation Assertion A: on the surface the potential energy is $-\frac{GMm}{R} = -mgR$ (using $GM = gR^2$); at infinity it is 0. To just reach infinity the kinetic energy must be $\frac{GMm}{R} = mgR$ (equivalently $\frac12mv_e^2$ with $v_e = \sqrt{2gR}$). It is $mgR$, not $\frac12mgR$, so A is false.
Reason R: gravitational potential energy $-\frac{GMm}{r}$ is negative at every finite distance and rises to 0 at infinity, so its maximum value, 0, is reached at infinity. R is true.
Answer: (D) A is false but R is true. | ||||
| 31 | KinematicsUniform circular motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on uniform circular motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane $\vec r = a(\hat i\cos\omega t + \hat j\sin\omega t)$ is uniform circular motion of radius $a$.
$\vec v = a\omega(-\hat i\sin\omega t + \hat j\cos\omega t)$, $\vec a = -a\omega^2(\hat i\cos\omega t + \hat j\sin\omega t) = -\omega^2\vec r$.
$\vec F = m\vec a = -m\omega^2\vec r$: directed opposite to $\vec r$ (towards the centre).
Checking the others: $\vec F \perp \vec P$ (the velocity is tangential), $\vec F \perp \vec L$ (since $\vec L = ma^2\omega\hat k$), and $\vec L\times\vec P$ points along $\hat k\times(\text{tangent}) = -\hat r$, i.e. along $\vec F$, not opposite to it.
Answer: (A) Opposite to the direction of $\vec r$. | ||||
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