Here x = 2t – |t-1| and y = 2\\(t^2\\) + t|t|.<\/p>\n
Now when t < 0;<\/p>\n
x = 2t – {-(t-1)} = 3t – 1 and y = 2\\(t^2\\) – \\(t^2\\) = \\(t^2\\) \\(\\implies\\) = y = \\({1\\over 9}{(x+1)}^2\\)<\/p>\n
= when 0 \\(\\le\\) t < 1<\/p>\n
x = 2t – {-(t-1)} = 3t – 1 and y = 2\\(t^2\\) – \\(t^2\\) = 3\\(t^2\\) \\(\\implies\\) = y = \\({1\\over 3}{(x+1)}^2\\)<\/p>\n
when t \\(\\ge\\) 1;<\/p>\n
x = 2t – (t-1) = t + 1 and y = 2\\(t^2\\) + \\(t^2\\) = 3\\(t^2\\) \\(\\implies\\) = y = 3\\({(x+1)}^2\\)<\/p>\n
Thus, y = f(x) = \\({1\\over 9}{(x+1)}^2\\), x < -1<\/p>\n
y = f(x) = \\({1\\over 3}{(x+1)}^2\\), -1\\(\\le\\)x < 2<\/p>\n
y = f(x) = 3\\({(x+1)}^2\\), x\\(\\ge\\) 2<\/p>\n
We have to check differentiability at x = -1 and 2.<\/p>\n
Differentiabilty at x = -1;<\/p>\n
LHD = f'(\\(-1)^-\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\)\\(f(-1-h) – f(-1)\\over -h\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\({1\\over 9}(-1-h+1)^2 – 0\\over -h\\) = 0<\/p>\n
RHD = f'(\\(-1)^+\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\)\\(f(-1+h) – f(-1)\\over h\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\({1\\over 3}(-1+h+1)^2 – 0\\over h\\) = 0<\/p>\n
Hence f(x) is differentiable at x = -1<\/p>\n
\\(\\implies\\) continuous at x = -1.<\/p>\n
To check differentiability at x = 2;<\/p>\n
LHD = f'(\\(2)^-\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\)\\({1\\over 3}(2-h+1)^2 – 3\\over -h\\) = 2<\/p>\n
RHD = f'(\\(2)^+\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\)\\(3(2+h-1)^2 – 3\\over h\\) = 6<\/p>\n
Hence f(x) is not differentiable at x = 2.<\/p>\n
But continuous at x = 2, because LHD and RHD both are finite.<\/p>\n
f(x) is continuous for all x and differentiable for all x, except x = 2.<\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : Here x = 2t – |t-1| and y = 2\\(t^2\\) + t|t|. Now when t < 0; x = 2t – {-(t-1)} = 3t – 1 and y = 2\\(t^2\\) – \\(t^2\\) = \\(t^2\\) \\(\\implies\\) = y = \\({1\\over 9}{(x+1)}^2\\) = when 0 \\(\\le\\) t < 1 x = 2t – {-(t-1)} = …<\/p>\n