Class 12 Maths NCERT Solutions

Each chapter below opens with the key idea and then walks through representative NCERT exercise questions from start to finish — the step where the marks are won or lost. Follow each line of working with a pencil before checking your own attempt.

• Chapter 1 — Relations and Functions: bijective functions and their inverses, equivalence-style relations.
• Chapter 2 — Inverse Trigonometric Functions: principal values and the standard inverse identities.
• Chapter 3 — Matrices: operations, transpose identities, inverses via the adjoint.
• Chapter 4 — Determinants: evaluation, properties and Cramer's rule for linear systems.
• Chapter 5 — Continuity and Differentiability: continuity tests, non-differentiability at |x| = 0, log differentiation.
• Chapter 6 — Application of Derivatives: tangent slopes, monotonicity, maxima and minima, rates of change.
• Chapter 7 — Integrals: substitution, integration by parts, definite integrals.
• Chapter 8 — Application of Integrals: area under and between curves.
• Chapter 9 — Differential Equations: order and degree, separable form, first-order linear equations.
• Chapter 10 — Vector Algebra: dot and cross products, projections, unit vectors.
• Chapter 11 — Three Dimensional Geometry: direction cosines, distance of a point from a plane.
• Chapter 12 — Linear Programming: the corner-point method for maximisation.
• Chapter 13 — Probability: Bayes' theorem, random variables and the binomial distribution.

Q1. Show that the function $f:\mathbb{R}\to \mathbb{R}$ defined by $f(x)=4x+3$ is bijective, and find $f^{-1}$.

Solution. One-one: if $f(x_1)=f(x_2)$ then $4x_1+3=4x_2+3$ so $x_1=x_2$. Onto: for any $y\in \mathbb{R}$, take $x=\frac{y-3}{4}$; then $f(x)=y$. Being both one-one and onto, f is bijective, and $f^{-1}(x)=\frac{x-3}{4}$.

Q2. On $\mathbb{N}$, define R by $xRy⟺2x+y=10$. Is R reflexive, symmetric or transitive?

Solution. $R=\{(1,8),(2,6),(3,4),(4,2)\}$. Not reflexive: 1R1 would need 2 + 1 = 10, false. Not symmetric: 1R8 holds but 8R1 would need 16 + 1 = 10, false. Not transitive: the only first-coordinate successions give nothing chainable (8 has no partner in R). So R has none of the three properties.

Q3. Let $f(x)=x+2$ and $g(x)=2x+1$. Find $g\circ f$ and $f\circ g$, and show they differ.

Solution. $g(f(x))=2(x+2)+1=2x+5$ while $f(g(x))=(2x+1)+2=2x+3$. Since $2x+5\ne 2x+3$, composition is not commutative: $g\circ f\ne f\circ g$.

Q1. Find the principal value of ${\sin }^{-1}\left(-\frac{1}{2}\right)$.

Solution. The principal-value interval for ${\sin }^{-1}$ is $[-\pi /2,\pi /2]$. Since $\sin (-\pi /6)=-1/2$, the principal value is $-\pi /6$.

Q2. Prove that ${\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}=\frac{\pi }{4}$.

Solution. ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$ (valid since $xy<1$). With $x=\frac{1}{2},y=\frac{1}{3}$:$\frac{x+y}{1-xy}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1$, so the sum is ${\tan }^{-1}1=\frac{\pi }{4}$.

Q3. Find ${\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)$.

Solution. $\cos \frac{7\pi }{6}=-\frac{\sqrt{3}}{2}$. The principal-value interval of ${\cos }^{-1}$ is $[0,\pi ]$, and $\cos \frac{5\pi }{6}=-\frac{\sqrt{3}}{2}$, therefore ${\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)=\frac{5\pi }{6}$.

Q1. For $A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)$ and $B=\left(\begin{array}{cc}5& 6\\ 7& 8\end{array}\right)$, verify that $AB\ne BA$.

Solution. $AB=\left(\begin{array}{cc}1\cdot 5+2\cdot 7& 1\cdot 6+2\cdot 8\\ 3\cdot 5+4\cdot 7& 3\cdot 6+4\cdot 8\end{array}\right)=\left(\begin{array}{cc}19& 22\\ 43& 50\end{array}\right)$. But $BA=\left(\begin{array}{cc}5\cdot 1+6\cdot 3& 5\cdot 2+6\cdot 4\\ 7\cdot 1+8\cdot 3& 7\cdot 2+8\cdot 4\end{array}\right)=\left(\begin{array}{cc}23& 34\\ 31& 46\end{array}\right)$. As the two products differ, matrix multiplication is not commutative.

Q2. For the same A and B, verify $(AB{)}^T=B^T{A}^T$.

Solution. From Q1, $AB=\left(\begin{array}{cc}19& 22\\ 43& 50\end{array}\right)$ so $(AB{)}^T=\left(\begin{array}{cc}19& 43\\ 22& 50\end{array}\right)$. On the other side, $B^T=\left(\begin{array}{cc}5& 7\\ 6& 8\end{array}\right)$ and $A^T=\left(\begin{array}{cc}1& 3\\ 2& 4\end{array}\right)$, giving $B^T{A}^T=\left(\begin{array}{cc}5+14& 15+28\\ 6+16& 18+32\end{array}\right)=\left(\begin{array}{cc}19& 43\\ 22& 50\end{array}\right)$. Both sides match.

Q3. Using the adjoint, find $A^{-1}$ for $A=\left(\begin{array}{cc}2& 5\\ 1& 3\end{array}\right)$.

Solution. $|A|=2\cdot 3-5\cdot 1=1\ne 0$, so A is invertible. $\text{adj}(A)=\left(\begin{array}{cc}3& -5\\ -1& 2\end{array}\right)$, hence $A^{-1}=\frac{1}{|A|}\text{adj}(A)=\left(\begin{array}{cc}3& -5\\ -1& 2\end{array}\right)$.

Q1. Evaluate $\left|\begin{array}{cc}3& 4\\ 5& 6\end{array}\right|$.

Solution. $\left|\begin{array}{cc}3& 4\\ 5& 6\end{array}\right|=3\cdot 6-4\cdot 5=18-20=-2$.

Q2. Solve by Cramer's rule: $x+y=3,2x+3y=7$.

Solution. $D=\left|\begin{array}{cc}1& 1\\ 2& 3\end{array}\right|=3-2=1$, $D_x=\left|\begin{array}{cc}3& 1\\ 7& 3\end{array}\right|=9-7=2$, $D_y=\left|\begin{array}{cc}1& 3\\ 2& 7\end{array}\right|=7-6=1$. So $x=D_x/D=2$ and $y=D_y/D=1$. Check: 2 + 1 = 3 and 4 + 3 = 7.

Q3. Show that $\left|\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right|=0$ without expanding.

Solution. Row 3 = 2(Row 2) − Row 1 because $2(4,5,6)-(1,2,3)=(7,8,9)$. Rows are linearly dependent, so the determinant is zero — no expansion needed.

Q1. Test the function $f(x)=|x|$ for continuity and differentiability at x = 0.

Solution. Continuity: ${\lim }_{x\to 0}|x|=0=f(0)$, so f is continuous at 0. Derivative: LHD ${\lim }_{h\to 0^{-}}\frac{|0+h|-0}{h}={\lim }_{h\to 0^{-}}\frac{-h}{h}=-1$, RHD ${\lim }_{h\to 0^{+}}\frac{h}{h}=1$. Since left and right derivatives differ, f is not differentiable at 0 — continuous yet not differentiable.

Q2. Differentiate $y=x^x$, x > 0.

Solution. Take logarithms: $\ln y=x\ln x$. Differentiate both sides: $\frac{1}{y}\frac{dy}{dx}=\ln x+1$, hence $\frac{dy}{dx}=y(\ln x+1)=x^x(\ln x+1)$.

Q1. Find the equation of the tangent to $y=x^3-x+1$ at x = 1.

Solution. $\frac{dy}{dx}=3x^2-1$, so the slope at x = 1 is $3-1=2$. Since $y(1)=1-1+1=1$, the tangent through (1, 1) is $y-1=2(x-1)$, i.e. $y=2x-1$.

Q2. Find the minimum value of $f(x)=x^2-4x+9$ on $\mathbb{R}$.

Solution. $f^{\prime }(x)=2x-4=0$ gives $x=2$. $f^{\prime \prime }(x)=2>0$, so x = 2 is a point of minima and $f(2)=4-8+9=5$ is the minimum value.

Q3. The radius of a circle grows at 1 cm/s. How fast does the area increase when r = 3 cm?

Solution. $A=\pi r^2$, so $\frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi (3)(1)=6\pi$ cm²/s.

Q1. Evaluate $\int x{e}^{x}dx$.

Solution. Integrate by parts with $u=x,dv=e^{x}dx$, so $du=dx,v=e^x$:$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C=e^x(x-1)+C$.

Q2. Evaluate $\int \frac{2x}{x^2+1}dx$.

Solution. Substitute $t=x^2+1$, $dt=2xdx$: $\int \frac{2x}{x^2+1}dx=\int \frac{dt}{t}=\ln |t|+C=\ln (x^2+1)+C$.

Q3. Evaluate ${\int }_0^1{x}^{2}dx$.

Solution. ${\left[\frac{x^3}{3}\right]}_0^1=\frac{1}{3}-0=\frac{1}{3}$.

Q1. Find the area enclosed between $y=x^2$ and $y=x$.

Solution. Intersection: $x^2=x\Rightarrow x(x-1)=0$, so x = 0 and x = 1. The line lies above the parabola, giving$A={\int }_0^1(x-x^2)dx={\left[\frac{x^2}{2}-\frac{x^3}{3}\right]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$.

Q2. Find the area under $y=x^3$ from x = 0 to x = 2.

Solution. $A={\int }_0^2{x}^{3}dx={\left[\frac{x^4}{4}\right]}_0^2=\frac{16}{4}=4$ square units.

Q1. Write the order and degree of $y^{\prime \prime }+(y^{\prime }{)}^3-2y=0$.

Solution. Order is the highest derivative, 2 (from $y^{\prime \prime }$). Degree is the power of the highest derivative after clearing radicals/fractions of derivatives — 1.

Q2. Solve $\frac{dy}{dx}=\frac{x}{y}$ given $y(0)=1$.

Solution. Variable-separable: $ydy=xdx$. Integrate: $\frac{y^2}{2}=\frac{x^2}{2}+C$. From $y(0)=1$, $C=\frac{1}{2}$, hence $y^2=x^2+1$.

Q3. Solve $\frac{dy}{dx}+y=e^x$.

Solution. Integrating factor $I.F.=e^{\int 1dx}=e^x$. Then $y{e}^x=\int e^x\cdot e^{x}dx=\int e^{2x}dx=\frac{e^{2x}}{2}+C$, so $y=\frac{e^x}{2}+C{e}^{-x}$.

Q1. Given $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+2\hat{k}$, find the projection of $\vec{a}$ on $\vec{b}$.

Solution. $\vec{a}\cdot \vec{b}=2+6+8=16$ and $|\vec{b}|=\sqrt{1+4+4}=3$. Projection $=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\frac{16}{3}$.

Q2. Find a unit vector along $\vec{c}=2\hat{i}-2\hat{j}+\hat{k}$.

Solution. $|\vec{c}|=\sqrt{4+4+1}=3$, so the unit vector is $\hat{c}=\frac{1}{3}(2\hat{i}-2\hat{j}+\hat{k})$.

Q1. Find the direction cosines of the line through (1, 2, 3) and (2, 3, 5).

Solution. Direction ratios $(1,1,2)$, magnitude $\sqrt{1+1+4}=\sqrt{6}$. Direction cosines $\left(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}}\right)$.

Q2. Find the distance of the point (1, 1, 2) from the plane $x+2y+3z=10$.

Solution. $d=\frac{|1+2+6-10|}{\sqrt{1^2+2^2+3^2}}=\frac{|-1|}{\sqrt{14}}=\frac{1}{\sqrt{14}}$ units.

Q1. Maximise $z=3x+5y$ subject to $x+y\le 4$, $x+2y\le 6$, $x,y\ge 0$.

Solution. Corner points of the feasible region: $(0,0),(4,0),(0,3)$ and the intersection $(x+y=4)\cap (x+2y=6)$ → subtract to get $y=2$, $x=2$, i.e. (2, 2). Values: $z(0,0)=0,z(4,0)=12,z(0,3)=15,z(2,2)=16$. Maximum is $z=16$ at (2, 2).

Q1. If $P(A)=\frac{1}{2}$, $P(B|A)=\frac{9}{10}$ and $P(B|A^c)=\frac{1}{5}$, find $P(A|B)$.

Solution. Bayes' theorem: $P(A|B)=\frac{P(A)P(B|A)}{P(A)P(B|A)+P(A^c)P(B|A^c)}$ $=\frac{\frac{1}{2}\cdot \frac{9}{10}}{\frac{1}{2}\cdot \frac{9}{10}+\frac{1}{2}\cdot \frac{1}{5}}=\frac{9/20}{9/20+1/10}=\frac{9/20}{11/20}=\frac{9}{11}$.

Q2. A fair die is thrown 6 times. Find the probability of getting exactly three sixes.

Solution. Binomial with $n=6$, $p=\frac{1}{6}$. $P(\text{3 sixes})={}^6{C}_3{\left(\frac{1}{6}\right)}^3{\left(\frac{5}{6}\right)}^3=20\cdot \frac{125}{6^6}=\frac{2500}{46656}=\frac{625}{11664}$.