BITSAT 2024 Paper 2
Exam BITSAT
Year 2024
Paper Paper 2
Subject Mathematics
Chapter Continuity and Differentiability BITSAT 2024 · Paper 2 · Mathematics · Continuity and Differentiability
If f(x)=\ array cc (x^2 ( x))/( (1+x)), & x ≠ 0 0, & x=0 array . ,...
If
f(x)=\left\{ x 2 log ( cos x ) log ( 1 + x ) , x ≠ 0 0 , x = 0 \begin{array}{cc}\frac{x^2 \log (\cos x)}{\log (1+x)}, & x \neq 0 \\ 0, & x=0\end{array} l o g ( 1 + x ) x 2 l o g ( c o s x ) , 0 , x = 0 x = 0 \right. , then at
x = 0 , f ( x ) x=0, f(x) x = 0 , f ( x ) is
Options A not continuous B continuous but not differentiable C differentiable D not continuous, but differentiable Correct answer C. differentiable
Step-by-step solution
lim x → 0 f ( x ) = lim x → 0 x 2 log ( cos x ) log ( 1 + x ) = lim x → 0 − x ⋅ log ( cos x ) log ( 1 + x ) x = lim x → 0 x ⋅ log ( cos x ) = 0 ⋅ log 1 = 0 ∵ lim x → 0 f ( x ) = f ( 0 ) ∴ f ( x ) \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{x^2 \log (\cos x)}{\log (1+x)}=\lim _{x \rightarrow 0}-\frac{x \cdot \log (\cos x)}{\frac{\log (1+x)}{x}}=\lim _{x \rightarrow 0} x \cdot \log (\cos x)=0 \cdot \log 1=0\because \lim _{x \rightarrow 0} f(x)=f(0)\therefore f(x) lim x → 0 f ( x ) = lim x → 0 l o g ( 1 + x ) x 2 l o g ( c o s x ) = lim x → 0 − x l o g ( 1 + x ) x ⋅ l o g ( c o s x ) = lim x → 0 x ⋅ log ( cos x ) = 0 ⋅ log 1 = 0 ∵ lim x → 0 f ( x ) = f ( 0 ) ∴ f ( x ) is continuous at
x = 0 x=0 x = 0
Now,
lim x → 0 f ( x + h ) − f ( x ) h lim h → 0 f ( h ) − f ( 0 ) h = lim h → 0 h 2 log cos h − 0 log ( 1 + h ) lim h → 0 h 2 log ( cos h ) 1 = 0 ∴ f ( x ) \lim _{x \rightarrow 0} \frac{f(x+h)-f(x)}{h}\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{h^2 \log \cos h-0}{\log (1+h)}\lim _{h \rightarrow 0} \frac{h^2 \log (\cos h)}{1}=0\therefore f(x) lim x → 0 h f ( x + h ) − f ( x ) lim h → 0 h f ( h ) − f ( 0 ) = lim h → 0 l o g ( 1 + h ) h 2 l o g c o s h − 0 lim h → 0 1 h 2 l o g ( c o s h ) = 0 ∴ f ( x ) is differentiable at
x = 0 x=0 x = 0
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