If the inequality `sin^2x+acosx+a^2>1+cosx` holds for any `x in R ,` then the largest negative integral value of a is `-4` (b) 3 (c) `-2` (d) `-1`

Options

A. -4
B. -3
C. -2
D. -1

Explanation

To solve the inequality \( \sin^2 x + a \cos x + a^2 > 1 + \cos x \) for any \( x \in \mathbb{R} \), we will follow these steps:

### Step 1: Rewrite the Inequality
Start with the given inequality:
\[ \sin^2 x + a \cos x + a^2 > 1 + \cos x \]
We can rearrange this to:
\[ \sin^2 x + a \cos x + a^2 - \cos x > 1 \]
This simplifies to:
\[ \sin^2 x + (a - 1) \cos x + a^2 - 1 > 0 \]

### Step 2: Use the Pythagorean Identity
Recall that \( \sin^2 x = 1 - \cos^2 x \). Substitute this into the inequality:
\[ 1 - \cos^2 x + (a - 1) \cos x + a^2 - 1 > 0 \]
This simplifies to:
\[ -\cos^2 x + (a - 1) \cos x + a^2 - 1 > 0 \]
Multiplying through by -1 (which reverses the inequality):
\[ \cos^2 x - (1 - a) \cos x - (a^2 - 1) < 0 \]

### Step 3: Analyze the Quadratic
Let \( y = \cos x \). The inequality now reads:
\[ y^2 - (1 - a)y - (a^2 - 1) < 0 \]
This is a quadratic inequality in \( y \). For this quadratic to be negative for all \( y \) (which ranges from -1 to 1), the discriminant must be less than or equal to zero.

### Step 4: Calculate the Discriminant
The discriminant \( D \) of the quadratic \( y^2 - (1 - a)y - (a^2 - 1) \) is given by:
\[ D = (1 - a)^2 + 4(a^2 - 1) \]
Expanding this:
\[ D = (1 - 2a + a^2) + 4a^2 - 4 = 5a^2 - 2a - 3 \]
For the quadratic to be negative for all \( y \), we need:
\[ D \leq 0 \]

### Step 5: Solve the Discriminant Inequality
Set the discriminant less than or equal to zero:
\[ 5a^2 - 2a - 3 \leq 0 \]
We can solve this quadratic inequality by finding its roots using the quadratic formula:
\[ a = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 5 \cdot (-3)}}{2 \cdot 5} \]
Calculating the discriminant:
\[ D = 4 + 60 = 64 \]
Thus, the roots are:
\[ a = \frac{2 \pm 8}{10} = 1 \text{ and } -\frac{3}{5} \]
The quadratic \( 5a^2 - 2a - 3 \) opens upwards (since the coefficient of \( a^2 \) is positive), so it is negative between its roots:
\[ -\frac{3}{5} \leq a \leq 1 \]

### Step 6: Find the Largest Negative Integral Value
The largest negative integral value of \( a \) that satisfies this inequality is:
\[ -1 \]

### Conclusion
Thus, the largest negative integral value of \( a \) is:
\[ \boxed{-1} \]