Madhya Pradesh Board Class 12 Mathematics — NCERT Solutions

Given: $A = \{1, 2, 3, \dots, 14\}$, $R = \{(x, y): 3x - y = 0\}$, i.e., $y = 3x$.

Listing R: $R = \{(1,3),(2,6),(3,9),(4,12)\}$

Reflexive: For reflexivity, $(a,a) \in R$ for all $a \in A$, i.e., $3a = a \Rightarrow a = 0$, which is not in $A$. For example, $(1,1) \notin R$. Hence R is not reflexive.

Symmetric: $(1,3) \in R$ but $3(3) = 9 \neq 1$, so $(3,1) \notin R$. Hence R is not symmetric.

Transitive: We need: if $(x,y) \in R$ and $(y,z) \in R$, then $(x,z) \in R$. Check: $(1,3) \in R$ and $(3,9) \in R$. Is $(1,9) \in R$? $3(1) = 3 \neq 9$. So $(1,9) \notin R$. Hence R is not transitive.

Conclusion: R is neither reflexive, nor symmetric, nor transitive.

Given: $R = \{(x,y): y = x+5,\ x < 4\}$

Listing R: $R = \{(1,6),(2,7),(3,8)\}$

Reflexive: $(1,1) \notin R$ since $1 \neq 1+5$. Hence R is not reflexive.

Symmetric: $(1,6) \in R$ but $(6,1) \notin R$ since $6 \not< 4$. Hence R is not symmetric.

Transitive: We need $(x,y) \in R$ and $(y,z) \in R \Rightarrow (x,z) \in R$. The pairs in R have second elements $6,7,8$, none of which is less than 4, so no pair $(y,z)$ exists in R for any $(x,y) \in R$. The condition is vacuously satisfied. Hence R is transitive.

Conclusion: R is neither reflexive nor symmetric, but it is transitive.

Differentiate $\sin 2x$:
$$\frac{d}{dx}(\sin 2x)=2\cos 2x.$$
So an antiderivative of $\cos 2x$ is $\frac12\sin 2x$.

Since the chapter asks for the antiderivative of $\sin 2x$, use the standard rule
$$\int \sin ax\,dx=-\frac{1}{a}\cos ax + C.$$
With $a=2$,
$$\int \sin 2x\,dx=-\frac12\cos 2x + C.$$

Using the standard result from the chapter,
$$\int \cos ax\,dx=\frac{1}{a}\sin ax + C.$$
Here $a=3$, so
$$\int \cos 3x\,dx=\frac13\sin 3x + C.$$

Given: $\sin^{-1}\left(-\dfrac{1}{2}\right)$

Concept: The principal value branch of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

Working:

Let $\sin^{-1}\left(-\dfrac{1}{2}\right) = y$. Then $\sin y = -\dfrac{1}{2}$.

We know that $\sin\left(-\dfrac{\pi}{6}\right) = -\dfrac{1}{2}$ and $-\dfrac{\pi}{6} \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.

Answer: The principal value of $\sin^{-1}\left(-\dfrac{1}{2}\right) = \boxed{-\dfrac{\pi}{6}}$.

Given: $\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$

Concept: The principal value branch of $\cos^{-1}$ is $[0, \pi]$.

Working:

Let $\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = y$. Then $\cos y = \dfrac{\sqrt{3}}{2}$.

We know that $\cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}$ and $\dfrac{\pi}{6} \in [0, \pi]$.

Answer: The principal value of $\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right) = \boxed{\dfrac{\pi}{6}}$.

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 $= \dfrac{5}{2}$

Given: $\dfrac{d^4y}{dx^4} + \sin(y''') = 0$

Order: The highest order derivative present is $\dfrac{d^4y}{dx^4}$ (i.e., the 4th derivative), so the order is 4.

Degree: The equation contains $\sin(y''')$, which is a transcendental (non-polynomial) function of the derivative $y'''$. Therefore, the equation is not a polynomial in its derivatives.

Degree is not defined.

Given: $\Delta = \left|\begin{array}{cc}2 & 4\\-5 & -1\end{array}\right|$

Formula: For a $2\times2$ determinant, $\left|\begin{array}{cc}a & b\\c & d\end{array}\right| = ad - bc$

Working:
$$\Delta = (2)(-1) - (4)(-5) = -2 + 20 = 18$$

Answer: $\Delta = 18$

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.

Given: $f(x) = 5x - 3$

Concept: A function $f$ is continuous at $x = c$ if $\lim_{x \to c} f(x) = f(c)$.

At $x = 0$:
$$f(0) = 5(0) - 3 = -3$$
$$\lim_{x \to 0} f(x) = \lim_{x \to 0}(5x - 3) = 5(0) - 3 = -3$$
Since $\lim_{x \to 0} f(x) = f(0) = -3$, $f$ is continuous at $x = 0$.

At $x = -3$:
$$f(-3) = 5(-3) - 3 = -15 - 3 = -18$$
$$\lim_{x \to -3} f(x) = 5(-3) - 3 = -18$$
Since $\lim_{x \to -3} f(x) = f(-3) = -18$, $f$ is continuous at $x = -3$.

At $x = 5$:
$$f(5) = 5(5) - 3 = 25 - 3 = 22$$
$$\lim_{x \to 5} f(x) = 5(5) - 3 = 22$$
Since $\lim_{x \to 5} f(x) = f(5) = 22$, $f$ is continuous at $x = 5$.

Hence, $f(x) = 5x - 3$ is continuous at $x = 0$, $x = -3$, and $x = 5$. $\blacksquare$

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 $0,\ -\dfrac{1}{\sqrt{2}},\ \dfrac{1}{\sqrt{2}}$.

Given: Area of a circle $A = \pi r^2$.

Formula used: Rate of change of area with respect to radius $= \dfrac{dA}{dr}$.

$$\frac{dA}{dr} = 2\pi r$$

(a) When $r = 3$ cm:
$$\frac{dA}{dr}\bigg|_{r=3} = 2\pi(3) = 6\pi \text{ cm}^2/\text{cm}$$

(b) When $r = 4$ cm:
$$\frac{dA}{dr}\bigg|_{r=4} = 2\pi(4) = 8\pi \text{ cm}^2/\text{cm}$$

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 $P(F|E) = \dfrac{1}{3}$.