Let a be a positive integer such that the limit _ x arrow 1 ((1)/(x-1)-(1)/(x^a-2 x+1)) exists and is equal to b:( where b ≠ 0)

Correct answer: B. ^ -1 ( b) is equal to 3-π . b= _ x arrow 1 (x^a-2 x+1-x+1)/((x-1)(x^a-2 x+1))= _ x arrow 1 (x^a-3 x+2)/((x-1)(x^a+1-2 x)) ; x=1+hb= _ h arrow 0 ((1+h)^a-3(1+h)+2)/(h[(1+h)^a+1-2(1+h)])= _ h arrow 0 ((1+a h+ .)-3 h-1)/(h[1+a h .1-2 h])= _ h arrow 0 ((a-3) h+(a(a-1))/(2) h^2)/((a-2) h^2+ .)a=3 and b=3 a+b=6

JEE Advanced

Exam: JEE Advanced
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Advanced · Mathematics · Inverse Trigonometric Functions

Let a be a positive integer such that the limit _ x arrow 1...

Let

a be a positive integer such that the limit

\lim _{x \rightarrow 1} \left(\frac{1}{x-1}-\frac{1}{x^a-2 x+1}\right) exists and is equal to

b:(

where

b \neq 0)

Options

A $\tan ^{-1}(\tan a)$ is equal to $3-\pi$ .
B $\tan ^{-1}(\tan b)$ is equal to $3-\pi$ .
C $\tan ^{-1}(\tan (a+b))$ is equal to $5-2 \pi$ .
D $\tan ^{-1}(\tan (a-b))$ is equal to 1.

Correct answer

B. $\tan ^{-1}(\tan b)$ is equal to $3-\pi$ .

Step-by-step solution

b=\lim _{x \rightarrow 1} \frac{x^a-2 x+1-x+1}{(x-1)\left(x^a-2 x+1\right)}=\lim _{x \rightarrow 1} \frac{x^a-3 x+2}{(x-1)\left(x^a+1-2 x\right)} ; \quad x=1+hb=\lim _{h \rightarrow 0} \frac{(1+h)^a-3(1+h)+2}{h\left[(1+h)^a+1-2(1+h)\right]}=\lim _{h \rightarrow 0} \frac{(1+a h+\ldots \ldots .)-3 h-1}{h[1+a h \ldots \ldots .1-2 h]}=\lim _{h \rightarrow 0} \frac{(a-3) h+\frac{a(a-1)}{2} h^2}{(a-2) h^2+\ldots \ldots .}a=3 \quad

and

\quad b=3 \quad \Rightarrow \quad a+b=6

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