Let $a$ be a positive integer such that the limit $\displaystyle\lim_{x \to 1}\left(\frac{1}{x-1} - \frac{1}{x^a - 2x + 1}\right)$ exists and is equal to $b$ (where $b \neq 0$). Then:
$\tan^{-1}(\tan a)$ is equal to $3 - \pi$
$\tan^{-1}(\tan b)$ is equal to $3 - \pi$
$\tan^{-1}(\tan(a+b))$ is equal to $5 - 2\pi$
$\tan^{-1}(\tan(a-b))$ is equal to 1
Step-by-Step Solution
Key Concept: To evaluate this $\infty - \infty$ limit, first combine the fractions to obtain a $\frac{0}{0}$ indeterminate form. For the limit to exist as a finite, non-zero value, the multiplicity of the $(x-1)$ factor in the numerator and denominator must be equal, which helps determine the parameter 'a'. Subsequently, carefully apply the property of inverse trigonometric functions, $\tan^{-1}(\tan x) = x - n\pi$, considering the principal value range.
Step 1: Combine the fractions and analyze the indeterminate form.
The given limit is
$$ \lim_{x \to 1}\left(\frac{1}{x-1} - \frac{1}{x^a - 2x + 1}\right) $$
Combine the fractions:
$$ \lim_{x \to 1}\frac{x^a - 2x + 1 - (x-1)}{(x-1)(x^a - 2x + 1)} = \lim_{x \to 1}\frac{x^a - 3x + 2}{(x-1)(x^a - 2x + 1)} $$
Let $N(x) = x^a - 3x + 2$ and $D(x) = (x-1)(x^a - 2x + 1)$.
At $x=1$, $N(1) = 1^a - 3(1) + 2 = 1 - 3 + 2 = 0$.
At $x=1$, $x^a - 2x + 1 = 1^a - 2(1) + 1 = 1 - 2 + 1 = 0$.
Thus, $D(1) = (1-1)(1^a - 2(1) + 1) = 0 \cdot 0 = 0$.
The limit is of the indeterminate form $\frac{0}{0}$.
Step 2: Determine the value of $a$.
For the limit to exist and be a non-zero finite value $b$, the factor $(x-1)$ in the denominator must be cancelled out by a corresponding factor in the numerator. Since $x^a - 2x + 1$ has a root at $x=1$, it can be written as $(x-1)Q(x)$ for some polynomial $Q(x)$.
The denominator is $(x-1)^2 Q(x)$.
For the limit to be finite and non-zero, the numerator $N(x) = x^a - 3x + 2$ must have $(x-1)^2$ as a factor.
If $(x-1)^2$ is a factor of $N(x)$, then $N(1)=0$ and $N'(1)=0$.
We already established $N(1)=0$.
Now, compute $N'(x)$:
$$ N'(x) = \frac{d}{dx}(x^a - 3x + 2) = ax^{a-1} - 3 $$
Evaluate $N'(1)$:
$$ N'(1) = a(1)^{a-1} - 3 = a - 3 $$
For $N'(1)=0$, we must have $a-3=0$, which implies $a=3$.
Step 3: Factor the numerator and denominator.
With $a=3$, the numerator is $x^3 - 3x + 2$.
Since $(x-1)^2$ is a factor, we can perform polynomial division:
$$ x^3 - 3x + 2 = (x-1)(x^2+x-2) = (x-1)(x-1)(x+2) = (x-1)^2(x+2) $$
The denominator factor $x^a - 2x + 1$ becomes $x^3 - 2x + 1$.
Since $(x-1)$ is a factor:
$$ x^3 - 2x + 1 = (x-1)(x^2+x-1) $$
The full denominator is $(x-1)(x^3 - 2x + 1) = (x-1)(x-1)(x^2+x-1) = (x-1)^2(x^2+x-1)$.
Step 4: Evaluate the limit to find $b$.
Substitute the factored forms into the limit expression:
$$ \lim_{x \to 1}\frac{(x-1)^2(x+2)}{(x-1)^2(x^2+x-1)} = \lim_{x \to 1}\frac{x+2}{x^2+x-1} $$
Now, substitute $x=1$:
$$ b = \frac{1+2}{1^2+1-1} = \frac{3}{1} = 3 $$
Thus, $a=3$ and $b=3$.
Step 5: Calculate the required expressions involving $a$ and $b$.
Given $a=3$ and $b=3$:
1. $\tan^{-1}(\tan a)$:
$\tan^{-1}(\tan 3)$. Since $\frac{\pi}{2} < 3 < \pi$, we have $\tan^{-1}(\tan 3) = 3 - \pi$.
2. $\tan^{-1}(\tan b)$:
$\tan^{-1}(\tan 3)$. Since $\frac{\pi}{2} < 3 < \pi$, we have $\tan^{-1}(\tan 3) = 3 - \pi$.
3. $\tan^{-1}(\tan(a+b))$:
$a+b = 3+3 = 6$.
$\tan^{-1}(\tan 6)$. Since $2\pi - \frac{\pi}{2} < 6 < 2\pi$ (approximately $4.71 < 6 < 6.28$), we have $\tan^{-1}(\tan 6) = 6 - 2\pi$.
4. $\tan^{-1}(\tan(a-b))$:
$a-b = 3-3 = 0$.
$\tan^{-1}(\tan 0) = 0$.
Correct Answer: 1, 2, 3, 4