CBSE Class 11 Mathematics — NCERT Solutions

A set is a well-defined collection of objects, meaning there is no ambiguity about whether an object belongs to the collection or not.

(i) Yes, it is a set. The months of a year beginning with 'J' are January, June, and July — clearly and unambiguously defined. So this is a set: {January, June, July}.

(ii) No, it is not a set. The term 'most talented' is subjective and varies from person to person. There is no definite criterion, so the collection is not well-defined.

(iii) No, it is not a set. The term 'best-cricket batsmen' is subjective. Different selectors may choose different players, so the collection is not well-defined.

(iv) Yes, it is a set. The collection of all boys in your class is well-defined — for any boy, it can be determined whether he belongs to your class or not.

(v) Yes, it is a set. The natural numbers less than 100 are precisely 1, 2, 3, …, 99. This is a well-defined collection.

(vi) Yes, it is a set. The novels written by Munshi Prem Chand are well-defined — one can verify whether a given novel was written by him or not.

(vii) Yes, it is a set. Even integers are well-defined: …, −4, −2, 0, 2, 4, … There is no ambiguity.

(viii) Yes, it is a set. The questions in this chapter are fixed and well-defined.

(ix) No, it is not a set. The term 'most dangerous' is subjective and not well-defined. Different people may have different opinions.

Given: $A = \{1, 2, 3, 4, 5, 6\}$

We check whether each number belongs to A or not:

(i) $5 \in A$ — since 5 is an element of A.

(ii) $8 \notin A$ — since 8 is not an element of A.

(iii) $0 \notin A$ — since 0 is not an element of A.

(iv) $4 \in A$ — since 4 is an element of A.

(v) $2 \in A$ — since 2 is an element of A.

(vi) $10 \notin A$ — since 10 is not an element of A.

Given: Two ordered pairs are equal: $\left(\dfrac{x}{3} + 1,\ y - \dfrac{2}{3}\right) = \left(\dfrac{5}{3},\ \dfrac{1}{3}\right)$.

Concept: Two ordered pairs $(a, b)$ and $(c, d)$ are equal if and only if $a = c$ and $b = d$.

Equating first elements:
$$\frac{x}{3} + 1 = \frac{5}{3}$$
$$\frac{x}{3} = \frac{5}{3} - 1 = \frac{5-3}{3} = \frac{2}{3}$$
$$x = 2$$

Equating second elements:
$$y - \frac{2}{3} = \frac{1}{3}$$
$$y = \frac{1}{3} + \frac{2}{3} = \frac{3}{3} = 1$$

Answer: $x = 2$ and $y = 1$.

Given: $n(\mathrm{A}) = 3$ and $\mathrm{B} = \{3, 4, 5\}$, so $n(\mathrm{B}) = 3$.

Concept: If $n(\mathrm{A}) = p$ and $n(\mathrm{B}) = q$, then $n(\mathrm{A} \times \mathrm{B}) = p \times q$.

Calculation:
$$n(\mathrm{A} \times \mathrm{B}) = n(\mathrm{A}) \times n(\mathrm{B}) = 3 \times 3 = 9$$

Answer: The number of elements in $\mathrm{A} \times \mathrm{B}$ is $\mathbf{9}$.

We use the conversion formula: Radian measure = $\dfrac{\pi}{180} \times$ Degree measure.

(i) 25°
$$25^\circ = \frac{\pi}{180} \times 25 = \frac{25\pi}{180} = \frac{5\pi}{36} \text{ radian}$$

(ii) −47°30′
First convert minutes to degrees: $30' = \dfrac{30}{60}^\circ = \dfrac{1}{2}^\circ$

So $-47^\circ 30' = -47\dfrac{1}{2}^\circ = -\dfrac{95}{2}^\circ$
$$-\frac{95}{2}^\circ = \frac{\pi}{180} \times \left(-\frac{95}{2}\right) = -\frac{95\pi}{360} = -\frac{19\pi}{72} \text{ radian}$$

(iii) 240°
$$240^\circ = \frac{\pi}{180} \times 240 = \frac{240\pi}{180} = \frac{4\pi}{3} \text{ radian}$$

(iv) 520°
$$520^\circ = \frac{\pi}{180} \times 520 = \frac{520\pi}{180} = \frac{26\pi}{9} \text{ radian}$$

We use the conversion formula: Degree measure = $\dfrac{180}{\pi} \times$ Radian measure.

(i) $\dfrac{11}{16}$ radian
$$\frac{11}{16} \times \frac{180}{\pi} = \frac{11}{16} \times \frac{180 \times 7}{22} = \frac{11 \times 180 \times 7}{16 \times 22} = \frac{11 \times 1260}{352} = \frac{13860}{352} = 39\frac{3}{8}^\circ$$
$$= 39^\circ + \frac{3}{8} \times 60' = 39^\circ 22' 30''$$

(ii) $-4$ radian
$$-4 \times \frac{180}{\pi} = -4 \times \frac{180 \times 7}{22} = -\frac{5040}{22} = -229\frac{1}{11}^\circ$$
$$= -\left(229^\circ + \frac{1}{11} \times 60'\right) = -\left(229^\circ 5' \frac{5}{11}''\right) \approx -229^\circ 5' 27''$$

(iii) $\dfrac{5\pi}{3}$ radian
$$\frac{5\pi}{3} \times \frac{180}{\pi} = \frac{5 \times 180}{3} = 300^\circ$$

(iv) $\dfrac{7\pi}{6}$ radian
$$\frac{7\pi}{6} \times \frac{180}{\pi} = \frac{7 \times 180}{6} = 210^\circ$$

Given: $(5i)\left(-\dfrac{3}{5}i\right)$

Working:
$$(5i)\left(-\frac{3}{5}i\right) = 5 \times \left(-\frac{3}{5}\right) \times i \times i = -3 \times i^2$$
Since $i^2 = -1$:
$$= -3 \times (-1) = 3$$

Answer: $3 + 0i$, i.e., $a = 3,\ b = 0$.

$24x<100$

Divide both sides by $24$:

$x<\frac{100}{24}=\frac{25}{6}$.

So the solutions depend on the set of numbers used in the subparts below.

Given: Digits available: 1, 2, 3, 4, 5 (total 5 digits). We need to form 3-digit numbers.

(i) Repetition allowed:

Each of the 3 places (hundreds, tens, units) can be filled by any of the 5 digits.

By the Fundamental Principle of Counting:
$$\text{Number of 3-digit numbers} = 5 \times 5 \times 5 = 125$$

(ii) Repetition not allowed:

• Hundreds place: 5 choices
• Tens place: 4 choices (one digit used)
• Units place: 3 choices (two digits used)

$$\text{Number of 3-digit numbers} = 5 \times 4 \times 3 = 60$$

Using the binomial theorem,
$$(1-2x)^5=\sum_{r=0}^{5} {^5C_r}(1)^{5-r}(-2x)^r.$$
Now compute term by term:
$$=1-5(2x)+10(2x)^2-10(2x)^3+5(2x)^4-(2x)^5.$$
So,
$$(1-2x)^5=1-10x+40x^2-80x^3+80x^4-32x^5.$$

Given: $a_n = n(n+2)$

Substituting $n = 1, 2, 3, 4, 5$:

$$a_1 = 1(1+2) = 1 \times 3 = 3$$

$$a_2 = 2(2+2) = 2 \times 4 = 8$$

$$a_3 = 3(3+2) = 3 \times 5 = 15$$

$$a_4 = 4(4+2) = 4 \times 6 = 24$$

$$a_5 = 5(5+2) = 5 \times 7 = 35$$

The first five terms are: $3, 8, 15, 24, 35$.

Given: Vertices of the quadrilateral are $A(-4, 5)$, $B(0, 7)$, $C(5, -5)$ and $D(-4, -2)$.

Area of quadrilateral ABCD can be found by dividing it into two triangles: $\triangle ABC$ and $\triangle ACD$.

Area of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ is:
$$\text{Area} = \frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$$

Area of $\triangle ABC$ with $A(-4,5)$, $B(0,7)$, $C(5,-5)$:
$$= \frac{1}{2}|(-4)(7-(-5))+0((-5)-5)+5(5-7)|$$
$$= \frac{1}{2}|(-4)(12)+0+5(-2)|$$
$$= \frac{1}{2}|-48-10| = \frac{1}{2}(58) = 29 \text{ sq. units}$$

Area of $\triangle ACD$ with $A(-4,5)$, $C(5,-5)$, $D(-4,-2)$:
$$= \frac{1}{2}|(-4)((-5)-(-2))+5((-2)-5)+(-4)(5-(-5))|$$
$$= \frac{1}{2}|(-4)(-3)+5(-7)+(-4)(10)|$$
$$= \frac{1}{2}|12-35-40| = \frac{1}{2}(63) = \frac{63}{2} \text{ sq. units}$$

Total Area of quadrilateral ABCD:
$$= 29 + \frac{63}{2} = \frac{58+63}{2} = \frac{121}{2} = 60.5 \text{ sq. units}$$

Given: Centre $(h,k) = (0,2)$, radius $r = 2$.

Formula: $(x-h)^2 + (y-k)^2 = r^2$

Solution:
$$(x-0)^2 + (y-2)^2 = 2^2$$
$$x^2 + (y-2)^2 = 4$$

Expanding:
$$x^2 + y^2 - 4y + 4 = 4$$
$$\boxed{x^2 + y^2 - 4y = 0}$$

Given: A point lies on the x-axis.

Any point on the x-axis is of the form $(x, 0, 0)$.

Therefore, the y-coordinate and z-coordinate of the point are both $\mathbf{0}$.

Given: $\lim_{x\to 3}(x + 3)$

Concept: For a polynomial function, the limit is found by direct substitution.

Working:
$$\lim_{x\to 3}(x + 3) = 3 + 3 = 6$$

Answer: $6$

Given: Data: 4, 7, 8, 9, 10, 12, 13, 17, $n = 8$

Step 1: Find the Mean
$$\bar{x} = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10$$

Step 2: Find $|x_i - \bar{x}|$ for each observation

Step 3: Calculate Mean Deviation
$$\text{M.D.}(\bar{x}) = \frac{\sum|x_i - \bar{x}|}{n} = \frac{24}{8} = 3$$

Answer: Mean Deviation about the mean $= 3$

Given: A die is rolled. Sample space S = {1, 2, 3, 4, 5, 6}.

Event E = {4} (die shows 4)

Event F = {2, 4, 6} (die shows even number)

Check for mutual exclusivity: Two events are mutually exclusive if their intersection is empty, i.e., E ∩ F = φ.

$$E \cap F = \{4\} \cap \{2, 4, 6\} = \{4\} \neq \phi$$

Since E ∩ F ≠ φ, E and F are NOT mutually exclusive.