Question

If \(a=\lim_{n\rightarrow\infty}\cos^{2n}x\), \((x=n\pi)\) and \(b=\lim_{n\rightarrow\infty}\cos^{2n}x\), \((x\ne m\pi)\), then numerical value of the area of the triangle whose vertices are \((a,b)\), \((-2,1)\) and \((2,1)\) is:

• A. 2
• B. 4
• C. 1
• D. $\frac{1}{2}$

Hint:

Notice that the vertices $V_2(-2, 1)$ and $V_3(2, 1)$ share the same $y$-coordinate, which means they form a flat horizontal base line along $y = 1$. The length of this base is $2 - (-2) = 4$ units. The third vertex is at $(1, 0)$, so the vertical height of the triangle is $1 - 0 = 1$ unit. Using $\frac{1}{2} \times \text{base} \times \text{height}$ gives $\frac{1}{2} \times 4 \times 1 = 2$ instantly!

Answer 1

The Correct Option is A

Concept: The behavior of the infinite power limit $\lim_{n \rightarrow \infty} y^n$ depends entirely on the size of the base value $y$: $$\lim_{n \rightarrow \infty} y^n = \begin{cases} 1, & \text{if } y = 1
0, & \text{if } 0 \le y < 1 \end{cases}$$ Once the coordinates for the vertices are found using these limits, the geometric area of the triangle can be computed using a standard coordinate matrix determinant. Step 1: Evaluate the limit constant $a$.
The first condition states that $x$ is an integer multiple of $\pi$ ($x = n\pi$). For these values, the cosine function outputs alternating peak values: $\cos(n\pi) = (-1)^n$. Squaring this expression removes the negative sign: $\cos^2(n\pi) = 1$. Now evaluate the limit as the exponent goes to infinity: $$a = \lim_{n \rightarrow \infty} (1)^n = 1$$

Step 2: Evaluate the limit constant $b$.
The second condition states that $x$ is not an integer multiple of $\pi$ ($x \neq m\pi$). For all these intermediate angles, the output value of the cosine function sits strictly between $-1$ and $1$, which means its square is always a fractional value less than 1: $$0 \le \cos^2 x < 1$$ Raising a fraction less than 1 to an infinite power causes it to shrink down to 0: $$b = \lim_{n \rightarrow \infty} (\cos^2 x)^n = 0$$ Thus, the first vertex of the triangle is located at $(a, b) = (1, 0)$.

Step 3: Calculate the area of the triangle using a determinant.
The three vertices of the triangle are $V_1(1, 0)$, $V_2(-2, 1)$, and $V_3(2, 1)$. Substitute these coordinates into the standard triangle area formula: $$\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1
x_2 & y_2 & 1
x_3 & y_3 & 1 \end{pmatrix} \right| = \frac{1}{2} \left| \det \begin{pmatrix} 1 & 0 & 1
-2 & 1 & 1
2 & 1 & 1 \end{pmatrix} \right|$$ Expand the determinant along the first row: $$\text{Area} = \frac{1}{2} \left| 1 \cdot (1 \cdot 1 - 1 \cdot 1) - 0 + 1 \cdot ((-2) \cdot 1 - 2 \cdot 1) \right|$$ $$\text{Area} = \frac{1}{2} \left| 1 \cdot (0) - 0 + 1 \cdot (-2 - 2) \right| = \frac{1}{2} \left| -4 \right| = \frac{1}{2} \times 4 = 2$$ This matches option (A).