Question

If \[ (4^{\sec^{2}\alpha})x^{2}+2x+\left(\beta^{2}-\beta+\frac{1}{2}\right)=0 \] has real roots, then the value/value(s) of \[ (\cos\alpha+\cos^{-1}\beta) \] is/are:

• A. $1+\frac{\pi}{3}$ if $\alpha = 2n\pi$
• B. $-1-\frac{\pi}{3}$ if $\alpha = (2n+1)\pi$
• C. $-1+\frac{\pi}{3}$ if $\alpha = (2n+1)\pi$
• D. $-1+\frac{\pi}{3}$ if $\alpha = 2n\pi$

Hint:

Whenever a discriminant inequality forces a condition like $\text{Expression} \le 1$, and your component factors have minimum values that multiply to exactly 1, the inequalities lock up completely! The system collapses from an infinite range down to a single fixed intersection point.

Answer 1

The Correct Option is A

Concept: For a quadratic equation $Ax^2 + Bx + C = 0$ to have real roots, its mathematical discriminant must be non-negative ($D = B^2 - 4AC \ge 0$). We can use this condition to establish absolute numerical boundaries for the individual parameter components. Step 1: Set up the discriminant inequality for real roots.
For the given quadratic equation, the coefficients are $A = 4^{\sec^2\alpha}$, $B = 2$, and $C = \beta^2 - \beta + \frac{1}{2}$. Calculate the discriminant and set it $\ge 0$: \[ D = B^2 - 4AC \ge 0 \quad \Rightarrow \quad (2)^2 - 4\left(4^{\sec^2\alpha}\right)\left(\beta^2 - \beta + \frac{1}{2}\right) \ge 0 \] Divide the entire inequality expression by 4 to simplify the terms: \[ 1 - \left(4^{\sec^2\alpha}\right)\left(\beta^2 - \beta + \frac{1}{2}\right) \ge 0 \quad \Rightarrow \quad \left(4^{\sec^2\alpha}\right)\left(\beta^2 - \beta + \frac{1}{2}\right) \le 1 \quad \cdots (1) \]

Step 2: Analyze the lower bound bounds of each factor component.
Let us evaluate the minimum possible values that each independent factor in inequality (1) can reach:
• For the first factor: We know that $\sec^2\alpha \ge 1$ for any real angle $\alpha$. Therefore, the exponential base expression satisfies: \[ 4^{\sec^2\alpha} \ge 4^1 = 4 \]
• For the second factor: Complete the square for the quadratic expression in terms of $\beta$: \[ \beta^2 - \beta + \frac{1}{2} = \left(\beta - \frac{1}{2}\right)^2 + \frac{1}{4} \] Since the squared term is always non-negative, the minimum value of this factor is exactly $\frac{1}{4}$ (occurring when $\beta = \frac{1}{2}$).

Step 3: Isolate the exact parameter values from the boundary condition.
Let us multiply the minimum bounds of our two factors together: \[ \text{Minimum Product} = 4 \times \frac{1}{4} = 1 \] Notice that our discriminant constraint from equation (1) requires this product to be less than or equal to 1. The only way a product of terms can be $\le 1$ when their absolute minimum combined value is exactly 1 is if both factors sit simultaneously at their absolute minimum values: \[ 4^{\sec^2\alpha} = 4 \quad \Rightarrow \quad \sec^2\alpha = 1 \quad \Rightarrow \quad \cos^2\alpha = 1 \quad \Rightarrow \quad \cos\alpha = \pm 1 \] \[ \beta^2 - \beta + \frac{1}{2} = \frac{1}{4} \quad \Rightarrow \quad \beta = \frac{1}{2} \]

Step 4: Calculate the target angle expression values.
Now substitute our isolated parameter values into the target expression $\cos\alpha + \cos^{-1}\beta$. First evaluate the constant inverse trigonometric term: $\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}$. This gives two target value branches based on the sign of $\cos\alpha$:
• Branch 1: If $\cos\alpha = 1$ ($\alpha = 2n\pi$): \[ \text{Value} = 1 + \frac{\pi}{3} \quad \text{(Matches option A for even integer periodic alignments)} \]
• Branch 2: If $\cos\alpha = -1$ ($\alpha = (2n+1)\pi$): \[ \text{Value} = -1 + \frac{\pi}{3} \quad \text{(Matches option C for odd integer periodic alignments)} \] Therefore, the valid solution branches correspond to options (A) and (C).