Question

Let $f(x) = x^2 + bx + c \forall x \in R, (b, c \in R)$ attains it's least value at x = -1 and the graph of f(x) cuts y-axis at y = 2. If f(x) = a has two distinct real roots, then complete set of values of a is

• A. (1,∞)
• B. (-2,-1)
• C. (0, 1)
• D. (1, 2)

Answer 1

Answer: (A)

(1,∞)

Answer 2

The given quadratic function is $f(x) = x^2 + bx + c$. Since the coefficient of $x^2$ is positive, the parabola opens upwards and attains a minimum value at its vertex. The x-coordinate of the vertex is given by $-b/(2 \times 1) = -b/2$. We are given that the function attains its least value at $x = -1$. Therefore, $-b/2 = -1$, so $b = 2$.

The function becomes $f(x) = x^2 + 2x + c$. The graph of $f(x)$ cuts the y-axis at $y = 2$, meaning $f(0) = 2$. Substituting $x = 0$ into the function gives $f(0) = 0^2 + 2(0) + c = c$, so $c = 2$.

Thus, the quadratic function is $f(x) = x^2 + 2x + 2$.

Now, we need to find the values of $a$ such that the equation $f(x) = a$ has two distinct real roots. This means $x^2 + 2x + 2 = a$, which can be rearranged to $x^2 + 2x + (2 - a) = 0$.

For a quadratic equation to have two distinct real roots, its discriminant $\Delta = B^2 - 4AC$ must be positive. In this case, $\Delta = (2)^2 - 4(1)(2 - a) = 4 - 8 + 4a = 4a - 4$.

For two distinct real roots, we require $\Delta > 0$, so $4a - 4 > 0$, which simplifies to $4a > 4$, or $a > 1$.

Therefore, the complete set of values of $a$ for which $f(x) = a$ has two distinct real roots is the interval $(1, \infty)$.