Q111 / 155 · WBJEE 2026 · Mathematics
If $f(x) = x(1331x^2 - 3630x + 3300)$, then for $a = \cos^2(\tan^{-1}(\sin(\cot^{-1} 3)))$
- Correct answerA$f(a + 1) = 2331$
- B$f'(a) = 11$
- Correct answerC$\lim_{x \to a}f(x) = 1000$
- D$\displaystyle\int_{0}^{a}(f(x) - 1000)\,dx = \dfrac{2500}{11}$
Solution
Given $f(x) = 1331x^3 - 3630x^2 + 3300x$
We can rewrite the function by completing the cube:
$f(x) = (11x)^3 - 3(11x)^2(10) + 3(11x)(10)^2 - 10^3 + 10^3$
$f(x) = (11x - 10)^3 + 1000$
Now, evaluating the expression for $a$:
Let $\theta = \cot^{-1} 3 \Rightarrow \cot \theta = 3 \Rightarrow \sin \theta = \dfrac{1}{\sqrt{1 + 3^2}} = \dfrac{1}{\sqrt{10}}$
Let $\phi = \tan^{-1}(\sin \theta) = \tan^{-1}\left(\dfrac{1}{\sqrt{10}}\right) \Rightarrow \tan \phi = \dfrac{1}{\sqrt{10}}$
$a = \cos^2 \phi = \dfrac{1}{\sec^2 \phi} = \dfrac{1}{1 + \tan^2 \phi} = \dfrac{1}{1 + \dfrac{1}{10}} = \dfrac{10}{11}$
Evaluating the given options with $a = \dfrac{10}{11}$:
For option (A):
$f(a + 1) = f\left(\dfrac{21}{11}\right) = \left(11 \times \dfrac{21}{11} - 10\right)^3 + 1000 = 11^3 + 1000 = 1331 + 1000 = 2331$
For option (B):
$f'(x) = 33(11x - 10)^2 \Rightarrow f'(a) = 33\left(11 \times \dfrac{10}{11} - 10\right)^2 = 0$
For option (C):
$\lim_{x \to a} f(x) = f(a) = \left(11 \times \dfrac{10}{11} - 10\right)^3 + 1000 = 1000$
For option (D):
$\displaystyle\int_{0}^{a} (f(x) - 1000) \, dx = \displaystyle\int_{0}^{10/11} (11x - 10)^3 \, dx = \left[ \dfrac{(11x - 10)^4}{44} \right]_{0}^{10/11} = 0 - \dfrac{(-10)^4}{44} = -\dfrac{10000}{44} = -\dfrac{2500}{11}$
Answer: $f(a + 1) = 2331$; $\lim_{x \to a}f(x) = 1000$
We can rewrite the function by completing the cube:
$f(x) = (11x)^3 - 3(11x)^2(10) + 3(11x)(10)^2 - 10^3 + 10^3$
$f(x) = (11x - 10)^3 + 1000$
Now, evaluating the expression for $a$:
Let $\theta = \cot^{-1} 3 \Rightarrow \cot \theta = 3 \Rightarrow \sin \theta = \dfrac{1}{\sqrt{1 + 3^2}} = \dfrac{1}{\sqrt{10}}$
Let $\phi = \tan^{-1}(\sin \theta) = \tan^{-1}\left(\dfrac{1}{\sqrt{10}}\right) \Rightarrow \tan \phi = \dfrac{1}{\sqrt{10}}$
$a = \cos^2 \phi = \dfrac{1}{\sec^2 \phi} = \dfrac{1}{1 + \tan^2 \phi} = \dfrac{1}{1 + \dfrac{1}{10}} = \dfrac{10}{11}$
Evaluating the given options with $a = \dfrac{10}{11}$:
For option (A):
$f(a + 1) = f\left(\dfrac{21}{11}\right) = \left(11 \times \dfrac{21}{11} - 10\right)^3 + 1000 = 11^3 + 1000 = 1331 + 1000 = 2331$
For option (B):
$f'(x) = 33(11x - 10)^2 \Rightarrow f'(a) = 33\left(11 \times \dfrac{10}{11} - 10\right)^2 = 0$
For option (C):
$\lim_{x \to a} f(x) = f(a) = \left(11 \times \dfrac{10}{11} - 10\right)^3 + 1000 = 1000$
For option (D):
$\displaystyle\int_{0}^{a} (f(x) - 1000) \, dx = \displaystyle\int_{0}^{10/11} (11x - 10)^3 \, dx = \left[ \dfrac{(11x - 10)^4}{44} \right]_{0}^{10/11} = 0 - \dfrac{(-10)^4}{44} = -\dfrac{10000}{44} = -\dfrac{2500}{11}$
Answer: $f(a + 1) = 2331$; $\lim_{x \to a}f(x) = 1000$
Hence, the correct answer is options: 'A', 'C'.