Question

If \[ f(x)=x(1331x^{2}-3630x+3300), \] then for \[ a=\cos^{2}\left(\tan^{-1}\left(\sin(\cot^{-1}3)\right)\right): \]

• A. $f(a+1)=2331$
• B. $f^{\prime}(a)=11$
• C. $\lim_{x\rightarrow a}f(x)=1000$
• D. $\int_{0}^{a}(f(x)-1000)dx=\frac{2500}{11}$

Hint:

Whenever a high-degree polynomial contains large, unusual coefficients like 1331, 3630, and 3300, it is almost always a hidden power expansion of a prime number ($1331 = 11^3$). Identifying this base pattern allows you to collapse the entire polynomial into a simple shifted form immediately.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires calculating a nested trigonometric value and then evaluating a cubic polynomial at that value. Since polynomials are continuous, the limit as \( x \to a \) is simply \( f(a) \).
Step 2: Key Formula or Approach:
1. Triangle method for inverse trig: \( \cot^{-1} 3 = \theta \).
2. Expand \( f(x) \): \( 1331x^3 - 3630x^2 + 3300x \).
Step 3: Detailed Explanation:
Calculating \( a \):
Let \( \cot^{-1} 3 = \theta \). Then \( \cot \theta = 3 \). In a right triangle, Base = 3, Perpendicular = 1, Hypotenuse = \( \sqrt{10} \).
\( \sin \theta = 1/\sqrt{10} \).
Next, \( \tan^{-1}(1/\sqrt{10}) = \phi \). Then \( \tan \phi = 1/\sqrt{10} \). In a triangle, P = 1, B = \( \sqrt{10} \), H = \( \sqrt{11} \).
\( \cos \phi = \sqrt{10}/\sqrt{11} \).
Finally, \( a = \cos^2 \phi = 10/11 \).
Evaluating \( f(a) \):
\( f(10/11) = (10/11) [ 1331(100/121) - 3630(10/11) + 3300 ] \).
\( f(10/11) = (10/11) [ 11(100) - 330(10) + 3300 ] \).
\( f(10/11) = (10/11) [ 1100 - 3300 + 3300 ] = (10/11)(1100) = 1000 \).
Since \( f(a) = 1000 \), the limit \( \lim_{x \to a} f(x) \) is indeed 1000.
Step 4: Final Answer:
The limit is 1000, which corresponds to option (C).