MHT CET · Maths · Continuity and Differentiability
The function \((f)(x)=[x]^2-\left[x^2\right]\)
(where, \([x]\) is the greatest integer less than or equal to \(x\) ), is discontinuous at
- A all integers.
- B all integers except 0 .
- C all integers except 0 and 1 .
- D all integers except 1 .
Answer & Solution
Correct Answer
(D) all integers except 1 .
Step-by-step Solution
Detailed explanation
for \(x=0\)
\(\left.\begin{array}{l}\text { L.H.L }=\left[O^{-}\right]^2-\left[\left(O^{-}\right)^2\right]=(-1)^2-(0)=1 \\ \text { R.H.L }=\left[O^{+}\right]^2-\left[O^{+}\right]^2=0^2-(0)=0\end{array}\right\}\) discount at \(x=0\)
for \(x=1\)
\(\left.\begin{array}{l}\text { L.H.L }=\left[1^{-}\right]^2-\left[\left(1^{-}\right)^2\right]=0^2-0=0 \\ \text { R.H.L }=\left[1^{+}\right]^2-\left[\left(1^{+}\right)^2\right]=1^2-1=0 \\ \text { F.V. }=[1]^2-\left[1^2\right]=1-1=0\end{array}\right\}\) continuous at \(x=1\)
the function \(f(x)\) is continuous at \(x=1\) and discontinuous at all other integral point
\(\left.\begin{array}{l}\text { L.H.L }=\left[O^{-}\right]^2-\left[\left(O^{-}\right)^2\right]=(-1)^2-(0)=1 \\ \text { R.H.L }=\left[O^{+}\right]^2-\left[O^{+}\right]^2=0^2-(0)=0\end{array}\right\}\) discount at \(x=0\)
for \(x=1\)
\(\left.\begin{array}{l}\text { L.H.L }=\left[1^{-}\right]^2-\left[\left(1^{-}\right)^2\right]=0^2-0=0 \\ \text { R.H.L }=\left[1^{+}\right]^2-\left[\left(1^{+}\right)^2\right]=1^2-1=0 \\ \text { F.V. }=[1]^2-\left[1^2\right]=1-1=0\end{array}\right\}\) continuous at \(x=1\)
the function \(f(x)\) is continuous at \(x=1\) and discontinuous at all other integral point
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