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Class 11 | CBSE | MATHEMATICS
TEXT BOOK EXAMPLES
Question
Write the solution set of the equation
\[ x² + x − 2 = 0 \]Solution
The given equation can be written as:
\[ x² + x − 2 = 0 \] \[ (x − 1)(x + 2) = 0 \]Therefore,
\[ x − 1 = 0 \quad \text{or} \quad x + 2 = 0 \] \[ x = 1 \quad \text{or} \quad x = −2 \]Hence, the solution set in roster form is:
\[ \{1,\;−2\} \]Question
Write the set
\[ \{x : x \text{ is a positive integer and } x² < 40\} \]in the roster form.
Solution
We find the values of x such that:
\[ x² < 40 \] \[ 1² = 1 < 40 \] \[ 2² = 4 < 40 \] \[ 3² = 9 < 40 \] \[ 4² = 16 < 40 \] \[ 5² = 25 < 40 \] \[ 6² = 36 < 40 \] \[ 7² = 49 > 40 \quad (\text{not allowed}) \]Therefore, the required set in roster form is:
\[ \{1,\;2,\;3,\;4,\;5,\;6\} \]Question
Write the set
\[ A = \{1,\;4,\;9,\;16,\;25,\;\ldots\} \]in the set-builder form.
Solution
The given elements are squares of natural numbers.
Hence,
A = { x : x is the square of a natural number }Alternatively,
\[ A = \{x : x = n²,\; n \in N\} \]Question
Write the set
\[ \left\{\frac{1}{2},\;\frac{2}{3},\;\frac{3}{4},\;\frac{4}{5},\;\frac{5}{6},\;\frac{6}{7}\right\} \]in the set-builder form.
Solution
In each element, the numerator is one less than the denominator.
The numerator starts from 1 and does not exceed 6.
Hence, the set-builder form is:
\[ \{x : x = \frac{n}{n+1},\; n \in N\} \] \[ 1 \le n \le 6 \]