Question:

Step 1: Analyze the limit function by considering different cases based on the value of $x$. We need to evaluate $f(x) = \lim_{n \to \infty} \dfrac{x^{2n-1} + ax^3 + bx^2}{x^{2n} + 1}$ for different ranges of $x$. **Case 1:** When $|x| < 1$, we have $x^{2n} \to 0$ as $n \to \infty$, so: $$f(x) = ax^3 + bx^2$$ **Case 2:** When $|x| > 1$, divide numerator and denominator by $x^{2n}$: $$f(x) = \lim_{n \to \infty} \dfrac{x^{-1} + ax^{3-2n} + bx^{2-2n}}{1 + x^{-2n}} = \dfrac{1}{x}$$ **Case 3:** When $x = 1$: $$f(1) = \dfrac{1 + a + b}{2}$$ **Case 4:** When $x = -1$: $$f(-1) = \dfrac{-1 - a + b}{2}$$ Step 2: Apply the continuity condition at $x = 1$. For $f(x)$ to be continuous at $x = 1$, the left-hand limit must equal the right-hand limit and the function value: $$\lim_{x \to 1^-} f(x) = a + b \quad \text{and} \quad \lim_{x \to 1^+} f(x) = 1$$ Therefore: $a + b = 1$ We can verify this is consistent with $f(1) = \dfrac{1+a+b}{2} = \dfrac{1+1}{2} = 1$ ✓ Step 3: Apply the continuity condition at $x = -1$. For continuity at $x = -1$: $$\lim_{x \to -1^-} f(x) = -1 \quad \text{and} \quad \lim_{x \to -1^+} f(x) = -a + b$$ Therefore: $-a + b = -1$ Step 4: Solve the system of equations for $a$ and $b$. We have two equations: $$a + b = 1$$ $$-a + b = -1$$ Adding these equations: $2b = 0 \Rightarrow b = 0$ Substituting back: $a + 0 = 1 \Rightarrow a = 1$ Step 5: Identify the critical points and construct the cubic polynomial $g(x)$. The points are $A(-a, 3) = A(-1, 3)$ and $B(b+1, -1) = B(1, -1)$, which are relative maximum and minimum points respectively. Since these are critical points of $g(x)$, the derivative must have zeros at $x = -1$ and $x = 1$: $$g'(x) = k(x+1)(x-1) = k(x^2 - 1)$$ for some constant $k$. Integrating: $$g(x) = k\left(\dfrac{x^3}{3} - x\right) + C$$ Step 6: Use the point conditions to find $k$ and $C$. From $g(-1) = 3$: $$k\left(\dfrac{(-1)^3}{3} - (-1)\right) + C = 3$$ $$k\left(-\dfrac{1}{3} + 1\right) + C = 3$$ $$\dfrac{2k}{3} + C = 3 \quad \text{...(i)}$$ From $g(1) = -1$: $$k\left(\dfrac{1^3}{3} - 1\right) + C = -1$$ $$k\left(\dfrac{1}{3} - 1\right) + C = -1$$ $$-\dfrac{2k}{3} + C = -1 \quad \text{...(ii)}$$ Adding equations (i) and (ii): $$2C = 2 \Rightarrow C = 1$$ Subtracting equation (ii) from equation (i): $$\dfrac{4k}{3} = 4 \Rightarrow k = 3$$ Therefore: $$g(x) = 3\left(\dfrac{x^3}{3} - x\right) + 1 = x^3 - 3x + 1$$ Step 7: Calculate $g(2)$. $$g(2) = 2^3 - 3(2) + 1 = 8 - 6 + 1 = 3$$ The value of $g(2)$ is **3**, which corresponds to **Option 3**. <div class="key-concept"><strong>Key Concept:</strong> Limit of sequence to find piecewise function, continuity conditions, and cubic polynomial with given extrema.</div> <div class="trap-box"><strong>Trap:</strong> Carefully handle the cases $|x|<1$, $|x|>1$, $x=\pm 1$ separately when evaluating the limit, and ensure continuity at both $x=1$ and $x=-1$.</div>