Given a real valued function f(x), defined as f(x)=tan2{x}(x−[x])2,x>0 1 , x=0{x}.cot{x},x<0 where [x] and {x} represents greatest integer function and fractional point of x, respectively, then which is incorrect?

limx→0+f(x)=limx→0+tan2xx2=1limx→0−f(x)=limx→0−{x}cot{x}=cot1⇒cot−1(limx→0−f(x))2=cot−1cot1=1tan−1(limx→0+f(x))=tan−1(1)=π4

Given a real valued function f(x), defined as f(x)=tan2{x}(x−[x])2,x>0 1 , x=0{x}.cot{x},x<0 where [x] and {x} represents greatest integer function and fractional point of x, respectively, then which is incorrect?

limx→0+f(x)=limx→0+tan2xx2=1limx→0−f(x)=limx→0−{x}cot{x}=cot1⇒cot−1(limx→0−f(x))2=cot−1cot1=1tan−1(limx→0+f(x))=tan−1(1)=π4

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Given a real valued function f(x), defined as $f (x) = \{\begin{cases} \frac{\tan^{2} {x}}{{(x - [x])}^{2}}, & x > 0 \\ 1, & x = 0 \\ \sqrt{{x} . \cot {x}}, & x < 0 \end{cases}$ where [x] and {x} represents greatest integer function and fractional point of x, respectively, then which is incorrect?

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$\lim_{x \to 0} f (x) e x i s t s$

$\tan^{- 1} (\lim_{x \to 0^{+}} f (x)) = \frac{π}{4}$

$\lim_{x \to 0^{+}} f (x) = 1$

$\cot^{- 1} {(\lim_{x \to 0^{-}} f (x))}^{2} = 1$

answer is D.

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Detailed Solution

$\lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{+}} \frac{\tan^{2} x}{x^{2}} = 1$

$\lim_{x \to 0^{-}} f (x) = \lim_{x \to 0^{-}} \sqrt{{x} \cot {x}} = \sqrt{\cot 1}$

$\Rightarrow \cot^{- 1} {(\lim_{x \to 0^{-}} f (x))}^{2} = \cot^{- 1} \cot 1 = 1$

$\tan^{- 1} (\lim_{x \to 0^{+}} f (x)) = \tan^{- 1} (1) = \frac{π}{4}$

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Given a real valued function f(x), defined as f(x)=tan2{x}(x−[x])2,x>0 1 , x=0{x}.cot{x},x<0 where [x] and {x} represents greatest integer function and fractional point of x, respectively, then which is incorrect?

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Given a real valued function f(x), defined as $f (x) = \{\begin{cases} \frac{\tan^{2} {x}}{{(x - [x])}^{2}}, & x > 0 \\ 1, & x = 0 \\ \sqrt{{x} . \cot {x}}, & x < 0 \end{cases}$ where [x] and {x} represents greatest integer function and fractional point of x, respectively, then which is incorrect?