# Q 15) $f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$

G Gautam harsolia

Given function is
$f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$
Given function is satisfies for the all real values of x
case (i)  k < 0
$f(k) = 2k$
$\lim_{k \rightarrow 0^-}f(x)= 2k = f(k)$
Hence, function is continuous for all values of x < 0

case (ii)  x = 0
$f(0 )= 0$
L.H.L at x= 0
$\lim_{x\rightarrow 0^-}f(x)= 2(0)= 0$
R.H.L. at x = 0
$\lim_{x\rightarrow 0^+}f(x)= 0$
L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0

case (iii)  k > 0
$f(k)=0$
$\lim_{k\rightarrow 0^+}f(x)= 0= f(k)$
Hence , function is continuous for all values of x > 0

case (iv) k < 1
$f(k) = 0$
$\lim_{x\rightarrow 1^-}f(x)= 0 = f(k)$
Hence , function is continuous for all values of x < 1

case (v)  k > 1
$f(k) = 4k$
$\lim_{x\rightarrow 1^+}f(x)= 4k = f(k)$
Hence , function is continuous for all values of x > 1

case (vi)  x = 1
$f(1)= 0$
$\lim_{x\rightarrow 1^-}f(1)= 0$
$\lim_{x\rightarrow 1^+}f(1)= 4(1) = 4$
Hence, function is not continuous at x = 1

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