Question:

Step 1: For $f(x) = x^2 + \lambda x + a^2 + a + 1 = 0$ to have real roots, the discriminant must be non-negative: $\Delta \geq 0$, i.e., $\lambda^2 - 4(a^2 + a + 1) \geq 0$, so $\lambda^2 \geq 4(a^2 + a + 1)$. Step 2: Find the minimum value of $a^2 + a + 1$. Completing the square: $a^2 + a + 1 = \left(a + \frac{1}{2}\right)^2 + \frac{3}{4} \geq \frac{3}{4}$. So $\lambda^2 \geq 4 \cdot \frac{3}{4} = 3$, giving $|\lambda| \geq \sqrt{3} \approx 1.732$. The least positive integral value of $\lambda$ is $2$. So option (a) is correct. Step 3: For $\lambda = 2$: $\Delta \geq 0 \Rightarrow 4 - 4(a^2 + a + 1) \geq 0 \Rightarrow a^2 + a \leq 0 \Rightarrow a(a+1) \leq 0 \Rightarrow a \in [-1, 0]$. So option (b) is correct. Step 4: For option (c), if both roots of $f(x) = 0$ and $2x^2 - x + 6 = 0$ are identical, then $\frac{1}{2} = \frac{\lambda}{-1} = \frac{a^2+a+1}{6}$. From $\frac{1}{2} = \frac{\lambda}{-1}$, $\lambda = -\frac{1}{2}$. From $\frac{a^2+a+1}{6} = \frac{1}{2}$, $a^2 + a + 1 = 3$, $a^2 + a - 2 = 0$, $(a+2)(a-1) = 0$, so $a = -2$ or $a = 1$. Sum of all possible values of $a$ is $-2 + 1 = -1$. So option (c) is correct. Step 5: For option (d), $f(1+x) = f(1-x)$ means $f$ is symmetric about $x = 1$. The axis of symmetry of $f(x) = x^2 + \lambda x + (a^2+a+1)$ is $x = -\frac{\lambda}{2}$. So $-\frac{\lambda}{2} = 1 \Rightarrow \lambda = -2$, not $\lambda = 2$. So option (d) is incorrect.