CBSE Class 12 Mathematics — NCERT Solutions

(i) For $x=1$, we need $y=3$. But $(1,3)$ is not in the set since $3\cdot 1-3=0$ is true only for the pair $(1,3)$, while reflexive pairs would require $(1,1)$. So the relation is not reflexive. Also, if $(x,y)\in R$, then $y=3x$. Then for symmetry we would need $(y,x)\in R$, i.e. $x=3y$, which is not true in general. For transitivity, $(1,3)$ and $(3,9)$ would be in the relation only if the numbers were in the set, but $9\notin A$; in any case the relation is not transitive because from $y=3x$ and $z=3y$ we get $z=9x$, not $z=3x$. So it is not symmetric and not transitive.

(ii) Here $(x,y)\in R$ means $y=x+5$ and $x<4$. This relation is not reflexive because $(x,x)$ would require $x=x+5$, impossible. It is not symmetric because if $(1,6)\in R$, then $(6,1)$ is not in $R$. It is not transitive because if $(1,6)$ and $(6,11)$ were to be in $R$, the second one is impossible since $6\not<4$; more generally the condition $x<4$ prevents chaining in the required way.

(iii) $(x,y)\in R$ means $y$ is divisible by $x$.

• Reflexive: For every $x\in \{1,2,3,4,5,6\}$, $x$ divides $x$, so $(x,x)\in R$.
• Symmetric: Not always. For example, $(2,4)\in R$, but $(4,2)\notin R$.
• Transitive: If $x$ divides $y$ and $y$ divides $z$, then $x$ divides $z$. So it is transitive.

(iv) $(x,y)\in R$ means $x-y$ is an integer. Since $x,y\in \mathbb Z$, $x-y$ is always an integer.

• Reflexive: $x-x=0$ is an integer.
• Symmetric: If $x-y$ is an integer, then $y-x=-(x-y)$ is also an integer.
• Transitive: If $x-y$ and $y-z$ are integers, then $x-z=(x-y)+(y-z)$ is an integer.

So it is reflexive, symmetric and transitive.

(v) For human beings in a town:
(a) Work at the same place: reflexive, symmetric and transitive? It is reflexive if a person works at the same place as himself, symmetric and transitive in the usual school-level sense, so it is an equivalence relation.
(b) Live in the same locality: also reflexive, symmetric and transitive, hence an equivalence relation.
(c) Exactly 7 cm taller than: not reflexive, not symmetric, and not transitive.
(d) Is wife of: not reflexive, not symmetric, not transitive.
(e) Is father of: not reflexive, not symmetric, not transitive.

For $\mathbb R=\{(a,b):a\le b^2\}$ on the set of real numbers:

• Not reflexive: We need $a\le a^2$ for every real $a$. Take $a=\tfrac12$. Then $\tfrac12\nleq \tfrac14$, so $(\tfrac12,\tfrac12)\notin R$.
• Not symmetric: $(0,1)\in R$ because $0\le 1^2=1$, but $(1,0)\notin R$ because $1\nleq 0^2=0$.
• Not transitive: Take $(1,1)\in R$ and $(1,2)\in R$ since $1\le1$ and $1\le4$. But $(1,2)$ does not help produce $(1,2)$ from transitivity in the given form; a standard counterexample is $a=4,b=2,c=1$: $4\le 2^2$ and $2\le1^2$ is false, so the relation does not satisfy transitivity in general. Hence the relation is neither reflexive nor symmetric nor transitive.

We know that

$$\frac{d}{dx}(\cos 2x)=-2\sin 2x$$

so

$$\frac{d}{dx}\left(-\frac{1}{2}\cos 2x\right)=\sin 2x.$$

Hence an anti derivative is $-\frac{1}{2}\cos 2x + C$.

We know that

$$\frac{d}{dx}(\sin 3x)=3\cos 3x.$$

So to get $\cos 3x$, we divide by $3$:

$$\frac{d}{dx}\left(\frac{1}{3}\sin 3x\right)=\cos 3x.$$

Hence the anti derivative is $\frac{1}{3}\sin 3x + C$.

The principal value branch of $\sin^{-1}x$ is in the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. Since

$$\sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2},$$

we get

$$\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}.$$

The principal value branch of $\cos^{-1}x$ is in the interval $[0,\pi]$. Since

$$\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2},$$

we get

$$\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}.$$

Given: Ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$, so $a^2 = 16$, $b^2 = 9$, i.e., $a = 4$, $b = 3$.

Formula used: Area of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ is $\pi ab$.

Working:

From the ellipse equation: $y = \dfrac{3}{4}\sqrt{16 - x^2}$ (taking positive square root for upper half).

By symmetry about both axes:
$$\text{Area} = 4\int_{0}^{4} y\, dx = 4\int_{0}^{4} \frac{3}{4}\sqrt{16 - x^2}\, dx = 3\int_{0}^{4}\sqrt{16 - x^2}\, dx$$

Using the standard result $\displaystyle\int_{0}^{a}\sqrt{a^2 - x^2}\, dx = \dfrac{\pi a^2}{4}$, with $a = 4$:
$$\text{Area} = 3 \times \frac{\pi (4)^2}{4} = 3 \times \frac{16\pi}{4} = 3 \times 4\pi = 12\pi$$

Answer: Area $= 12\pi$ square units.

Given: Matrix $A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}$

(i) Order of the matrix:
The matrix has 3 rows and 4 columns.
$$\therefore \text{Order} = 3 \times 4$$

(ii) Number of elements:
Number of elements $= m \times n = 3 \times 4 = \mathbf{12}$

(iii) Elements:

• $a_{13}$ = element in row 1, column 3 $= 19$
• $a_{21}$ = element in row 2, column 1 $= 35$
• $a_{33}$ = element in row 3, column 3 $= -5$
• $a_{24}$ = element in row 2, column 4 $= 12$
• $a_{23}$ = element in row 2, column 3 $= \dfrac{5}{2}$

The highest derivative is $\dfrac{d^4y}{dx^4}$, so the order is 4.

The equation is polynomial in derivatives, and the highest power of the highest order derivative $\dfrac{d^4y}{dx^4}$ is $1$, so the degree is 1.

For a $2\times 2$ determinant,

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc$$

So,

$$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}=2(-1)-4(-5)=-2+20=18$$

Given: Displacement = 40 km, direction = 30° east of north.

To represent this graphically:

• Draw the north direction (positive y-axis) as reference.
• From the initial point O, draw an arrow (vector) of length representing 40 km (using a suitable scale, e.g., 1 cm = 10 km, so length = 4 cm).
• The arrow makes an angle of 30° towards the east (right) from the north direction.

The directed line segment $\overrightarrow{OA}$ with $|\overrightarrow{OA}|$ = 40 km, inclined at 30° east of north, represents the required displacement graphically.

For any real number $c$,

$$\lim_{x\to c}(5x-3)=5c-3=f(c).$$

So $f(x)=5x-3$ is continuous at every real number. Hence it is continuous at $x=0$, $x=-3$ and $x=5$.

Given: A line makes angles $\alpha = 90^\circ$, $\beta = 135^\circ$, $\gamma = 45^\circ$ with the $x$-, $y$- and $z$-axes respectively.

Formula: Direction cosines are $l = \cos\alpha,\ m = \cos\beta,\ n = \cos\gamma$.

Working:
$$l = \cos 90^\circ = 0$$
$$m = \cos 135^\circ = -\frac{1}{\sqrt{2}}$$
$$n = \cos 45^\circ = \frac{1}{\sqrt{2}}$$

Verification: $l^2 + m^2 + n^2 = 0 + \dfrac{1}{2} + \dfrac{1}{2} = 1$ ✓

Answer: The direction cosines are $0,\ -\dfrac{1}{\sqrt{2}},\ \dfrac{1}{\sqrt{2}}$.

For a circle, the area is

$A=\pi r^2$.

So the rate of change of area with respect to radius is

$\dfrac{dA}{dr}=\dfrac{d}{dr}(\pi r^2)=2\pi r$.

• When $r=3$ cm: $\dfrac{dA}{dr}=2\pi(3)=6\pi$.
• When $r=4$ cm: $\dfrac{dA}{dr}=2\pi(4)=8\pi$.

So the area changes at rates $6\pi$ cm$^2$/cm and $8\pi$ cm$^2$/cm respectively.

The feasible region is the triangle with corner points $(0,0)$, $(4,0)$ and $(0,4)$.

Evaluate $Z=3x+4y$ at each corner point:

• At $(0,0)$: $Z=3(0)+4(0)=0$
• At $(4,0)$: $Z=3(4)+4(0)=12$
• At $(0,4)$: $Z=3(0)+4(4)=16$

The maximum value is $16$ at $(0,4)$.

Given: $P(E) = 0.6$, $P(F) = 0.3$, $P(E \cap F) = 0.2$

Formula used: $P(E|F) = \dfrac{P(E \cap F)}{P(F)}$ and $P(F|E) = \dfrac{P(E \cap F)}{P(E)}$

Finding $P(E|F)$:
$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0.2}{0.3} = \frac{2}{3}$$

Finding $P(F|E)$:
$$P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{0.2}{0.6} = \frac{1}{3}$$

Answer: $P(E|F) = \dfrac{2}{3}$ and $P(F|E) = \dfrac{1}{3}$.