CBSE Class 11 Maths Exemplar Chapter 1 Sets NCERT Solutions

To write the sets in roster form, we need to find the elements that satisfy the given conditions.

(i) For set A, we need to solve the equation 2x + 11 = 15 for x \in R.

Subtracting 11 from both sides gives 2x = 15 - 11, which simplifies to 2x = 4.

Dividing by 2, we get x = 2.

Therefore, the set A in roster form is A = \{2\}.

(ii) For set B, we need to solve the equation x^2 = x for x \in R.

Rearranging the equation, we get x^2 - x = 0.

Factoring out x, we have x(x - 1) = 0.

This gives two possible values for x: x = 0 or x = 1.

Therefore, the set B in roster form is B = \{0, 1\}.

(iii) For set C, we need to find the positive factors of a prime number p.

By definition, a prime number has exactly two distinct positive factors: 1 and the number itself.

Therefore, the set C in roster form is C = \{1, p\}.

We will solve the given equations to find the elements for each set.

(i) For set D, we solve t^3 = t where t \in R.

Rearranging, we get t^3 - t = 0.

Factoring out t, we have t(t^2 - 1) = 0.

Further factoring the difference of squares, t(t - 1)(t + 1) = 0.

The solutions are t = 0, t = 1, and t = -1.

Thus, the set D in roster form is D = \{-1, 0, 1\}.

(ii) For set E, we solve \frac{w-2}{w+3} = 3 where w \in R.

Assuming w \neq -3, we multiply both sides by w+3:

w - 2 = 3(w + 3)

w - 2 = 3w + 9

Subtracting w from both sides: -2 = 2w + 9.

Subtracting 9 from both sides: -11 = 2w.

Dividing by 2, we get w = -\frac{11}{2}.

Since w = -\frac{11}{2} is not equal to -3, this is a valid solution.

Thus, the set E in roster form is E = \left\{-\frac{11}{2}\right\}.

(iii) For set F, we solve x^4 - 5x^2 + 6 = 0 where x \in R.

This is a quadratic equation in x^2. Let y = x^2. Then the equation becomes y^2 - 5y + 6 = 0.

Factoring the quadratic, we get (y - 2)(y - 3) = 0.

So, y = 2 or y = 3.

Substituting back x^2 for y:

x^2 = 2 \implies x = \pm \sqrt{2}

x^2 = 3 \implies x = \pm \sqrt{3}.

Thus, the set F in roster form is F = \{-\sqrt{3}, -\sqrt{2}, \sqrt{2}, \sqrt{3}\}. The order of elements does not matter in roster form.

The number is given by $N\; =\; 2^\{p-1\}(2^p\; -\; 1)$, where $2^p\; -\; 1$ is a prime number. This form is characteristic of even perfect numbers. The positive factors of $2^\{p-1\}$ are $1,\; 2,\; 2^2,\; 2^3,\; \backslash dots,\; 2^\{p-1\}$. Since $2^p\; -\; 1$ is a prime number, its positive factors are 1 and $2^p\; -\; 1$. The factors of $N$ are formed by multiplying the factors of $2^\{p-1\}$ with the factors of $2^p\; -\; 1$. The factors of $N$ are: $1\; \backslash times\; 1,\; 2\; \backslash times\; 1,\; 2^2\; \backslash times\; 1,\; \backslash dots,\; 2^\{p-1\}\; \backslash times\; 1$ (which are $1,\; 2,\; 2^2,\; \backslash dots,\; 2^\{p-1\}$) and $1\; \backslash times\; (2^p\; -\; 1),\; 2\; \backslash times\; (2^p\; -\; 1),\; 2^2\; \backslash times\; (2^p\; -\; 1),\; \backslash dots,\; 2^\{p-1\}\; \backslash times\; (2^p\; -\; 1)$. However, the problem statement implies that $2^p-1$ is a prime factor. The set Y consists of the positive factors of the number $2^\{p-1\}(2^p-1)$. The factors of $2^\{p-1\}$ are $\backslash \{1,\; 2,\; 2^2,\; \backslash dots,\; 2^\{p-1\}\backslash \}$. The factors of the prime number $2^p-1$ are $\backslash \{1,\; 2^p-1\backslash \}$. The set Y contains all unique products of one factor from the first set and one factor from the second set. The factors are $1\; \backslash times\; 1,\; 2\; \backslash times\; 1,\; \backslash dots,\; 2^\{p-1\}\; \backslash times\; 1$ and $1\; \backslash times\; (2^p-1),\; 2\; \backslash times\; (2^p-1),\; \backslash dots,\; 2^\{p-1\}\; \backslash times\; (2^p-1)$. This simplifies to the set of factors of $2^\{p-1\}$ and the set of multiples of $2^p-1$ up to $2^\{p-1\}(2^p-1)$. The set Y is given by the factors of $2^\{p-1\}$ and the prime factor $2^p-1$ itself, assuming $2^p-1$ is not already one of the powers of 2. Thus, the set Y in roster form is $Y\; =\; \backslash \{1,\; 2,\; 2^2,\; 2^3,\; \backslash dots,\; 2^\{p-1\},\; 2^p\; -\; 1\backslash \}$.

We will analyze each statement.

(i) The statement is 35 \in \{x \mid x \text{ has exactly four positive factors}\}.

Let's find the positive factors of 35. The factors are 1, 5, 7, and 35. There are exactly four positive factors.

Therefore, the statement is True.

(ii) The statement is 128 \in \{y \mid \text{the sum of all the positive factors of } y \text{ is } 2y\}.

The positive factors of 128 (= 2^7) are 1, 2, 4, 8, 16, 32, 64, 128.

The sum of these factors is 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255.

We need to check if this sum is equal to 2y, which is 2 \times 128 = 256.

Since 255 \neq 256, the sum of the factors is not 2y.

Therefore, the statement is False.

(iii) The statement is 3 \notin \{x \mid x^4 - 5x^3 + 2x^2 - 112x + 6 = 0\}.

To check if 3 is a root of the polynomial P(x) = x^4 - 5x^3 + 2x^2 - 112x + 6, we substitute x = 3 into the polynomial:

P(3) = (3)^4 - 5(3)^3 + 2(3)^2 - 112(3) + 6

P(3) = 81 - 5(27) + 2(9) - 336 + 6

P(3) = 81 - 135 + 18 - 336 + 6

P(3) = (81 + 18 + 6) - (135 + 336)

P(3) = 105 - 471

P(3) = -366.

Since P(3) \neq 0, 3 is not a root of the equation.

Therefore, the statement 3 \notin \{x \mid x^4 - 5x^3 + 2x^2 - 112x + 6 = 0\} is True.

(iv) The statement is 496 \notin \{y \mid \text{the sum of all the positive factors of } y \text{ is } 2y\}.

A number y for which the sum of its positive factors is 2y is called a perfect number.

Let's find the factors of 496.

496 = 2^4 \times 31^1.

The sum of the factors is given by the formula \sigma(y) = \prod_{i=1}^{k} \frac{p_i^{a_i+1}-1}{p_i-1} for y = p_1^{a_1} \dots p_k^{a_k}.

For 496, the sum of factors is:

\sigma(496) = \left(\frac{2^{4+1}-1}{2-1}\right) \times \left(\frac{31^{1+1}-1}{31-1}\right)

\sigma(496) = \left(\frac{2^5-1}{1}\right) \times \left(\frac{31^2-1}{30}\right)

\sigma(496) = (32-1) \times \left(\frac{961-1}{30}\right)

\sigma(496) = 31 \times \left(\frac{960}{30}\right)

\sigma(496) = 31 \times 32 = 992.

Now, we check if \sigma(496) = 2 \times 496.

2 \times 496 = 992.

Since \sigma(496) = 992 = 2 \times 496, 496 is a perfect number.

The statement claims that 496 is NOT in the set of perfect numbers, which is false.

Therefore, the statement is False.