The function defined by $f(x)=\left\{\begin{matrix}|x-3|&;&x≥1\\\frac{1}{4}x^2-\frac{3}{2}x+\frac{13}{4}&;&x<1\end{matrix}\right.$ is
Answer & explanation
Correct answer: option 3
Since |x - 3| = x - 3 , if x ≥ 3 = -x + 3, if x < 3
$∴f(x)=\left\{\begin{matrix}\frac{1}{4}x^2-\frac{3}{2}x+\frac{13}{4}&;&x<1\\3-x&;&1≤x<3\\x-3&;&x≥3\end{matrix}\right.$
Now proceed to check the continuity and differentiability at x = 1
$f'(x)=\left\{\begin{matrix}\frac{x}{2}-\frac{3}{2}&;&x<1\\-1&;&1≤x<3\\1&;&x>3\end{matrix}\right.$
but at only = x =1
LHD = RHD = -1
for x = 3 (function non differentiable as LHD ≠ RHD)