Let \(f:[-1,2]\to R\) be given by \(f(x)=2{x}^{2}+x+\left[{x}^{2}\right]-[x]\) , where \([t]\) denotes the greatest inte…

Doubtful points: \(-1,0,1,\sqrt{2},\sqrt{3},2\)

At \(x=\sqrt{2},\sqrt{3}\),

\(f(x)=(2x^2+x-[x])+[x^2]\)

Here \(2x^2+x-[x]\) is continuous, but \([x^2]\) is discontinuous.

So, \(f(x)\) is discontinuous at \(x=\sqrt{2},\sqrt{3}\).

At \(x=-1\):

R.H.L. \(\Rightarrow f(-1^+)=2-1-(-1)+0=2\)

\(f(-1)=2-1-(-1)+1=3\)

So, \(f(x)\) is discontinuous at \(x=-1\).

At \(x=2\):

L.H.L. \(\Rightarrow f(2^-)=8+2-1+3=12\)

\(f(2)=8+2-2+4=12\)

R.H.L. \(\Rightarrow f(2^+)=8+2-2+4=12\)

So, \(f(x)\) is continuous at \(x=2\).

At \(x=0\):

L.H.L. \(\Rightarrow f(0^-)=0+0-(-1)+0=1\)

\(f(0)=0\)

So, \(f(x)\) is discontinuous at \(x=0\).

At \(x=1\):

L.H.L. \(\Rightarrow f(1^-)=2+1-0+0=3\)

\(f(1)=2+1-1+1=3\)

R.H.L. \(\Rightarrow f(1^+)=2+1-1+1=3\)

So, \(f(x)\) is continuous at \(x=1\).