1. **State the problem:** Given sets and functions: - $A = \{x \in \mathbb{Z} : x^2 < 11\}$ - $B = \{x \in \mathbb{N} : \text{$x$ মৌলিক সংখ্যা এবং } 1 < x \leq 5\}$ - $f(x) = \frac{4x + 1}{4x - 1}$ - (ক) $g(t) = \frac{t^4 + t^2 + 1}{t^2}$, find $g(-3^{-1})$ - (খ) Show $A \cup B = (A - B) \cup (B - A) \cup (A \cap B)$ - (গ) If $\frac{f(x+2) - 1}{f(x - 2) - 1} + 1^r = 0$, find $x$. 2. **(ক) Finding $g(-3^{-1})$: - First, compute $-3^{-1} = -\frac{1}{3}$. - Replace $t$ by $-\frac{1}{3}$ in $g(t)$: $$g\left(-\frac{1}{3}\right) = \frac{\left(-\frac{1}{3}\right)^4 + \left(-\frac{1}{3}\right)^2 + 1}{\left(-\frac{1}{3}\right)^2}$$ - Calculate numerator terms: $$\left(-\frac{1}{3}\right)^4 = \frac{1}{81}, \quad \left(-\frac{1}{3}\right)^2 = \frac{1}{9}$$ - Sum numerator: $$\frac{1}{81} + \frac{1}{9} + 1 = \frac{1}{81} + \frac{9}{81} + \frac{81}{81} = \frac{91}{81}$$ - Denominator: $$\left(-\frac{1}{3}\right)^2 = \frac{1}{9}$$ - Divide numerator by denominator: $$g\left(-\frac{1}{3}\right) = \frac{91/81}{1/9} = \frac{91}{81} \times 9 = \frac{91 \times 9}{81} = \frac{91}{9}$$ 3. **(খ) Verify set equality:** - Identity to show: $$A \cup B = (A - B) \cup (B - A) \cup (A \cap B)$$ - Recall: $$A - B = \{x | x \in A, x \notin B\}$$ $$B - A = \{x | x \in B, x \notin A\}$$ - The right side union is exactly the symmetric difference plus the intersection, which combine to form $A \cup B$. - This is a standard set identity for union as the union of symmetric difference and intersection. - Therefore, the statement holds. 4. **(গ) Solve for $x$ given:** $$\frac{f(x+2) - 1}{f(x - 2) - 1} + 1^r = 0$$ Since $1^r = 1$ for any real $r$, Rewrite: $$\frac{f(x+2) - 1}{f(x - 2) - 1} + 1 = 0 \implies \frac{f(x+2) - 1}{f(x - 2) - 1} = -1$$ Multiply both sides: $$f(x+2) - 1 = -\left(f(x - 2) - 1\right) = -f(x - 2) + 1$$ Rearranged: $$f(x+2) + f(x - 2) = 2$$ Recall: $$f(t) = \frac{4t + 1}{4t - 1}$$ Calculate: $$f(x+2) = \frac{4(x+2) + 1}{4(x+2) - 1} = \frac{4x + 8 + 1}{4x + 8 - 1} = \frac{4x + 9}{4x + 7}$$ $$f(x - 2) = \frac{4(x-2) + 1}{4(x-2) - 1} = \frac{4x - 8 + 1}{4x - 8 -1} = \frac{4x -7}{4x -9}$$ Sum: $$\frac{4x+9}{4x+7} + \frac{4x -7}{4x - 9} = 2$$ Multiply both sides by $(4x+7)(4x - 9)$: $$(4x + 9)(4x - 9) + (4x -7)(4x + 7) = 2(4x +7)(4x -9)$$ Calculate each term: - $(4x+9)(4x-9) = (4x)^2 - 9^2 = 16x^2 - 81$ - $(4x -7)(4x +7) = (4x)^2 - 7^2 = 16x^2 - 49$ - Right side: $$2(4x +7)(4x -9) = 2(16x^2 - 36x + 28x - 63) = 2(16x^2 - 8x - 63) = 32x^2 - 16x - 126$$ Sum left side: $$16x^2 - 81 + 16x^2 - 49 = 32x^2 - 130$$ Equation: $$32x^2 - 130 = 32x^2 - 16x - 126$$ Subtract $32x^2$ both sides: $$-130 = -16x - 126$$ Add $126$ to both sides: $$-4 = -16x$$ Divide both sides by $-16$: $$x = \frac{1}{4}$$ **Final answers:** - (ক) $g(-3^{-1}) = \frac{91}{9}$ - (খ) Set equality holds true. - (গ) $x = \frac{1}{4}$