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प्रश्न
Find whether the following function is differentiable at x = 1 and x = 2 or not : \[f\left( x \right) = \begin{cases}x, & & x < 1 \\ 2 - x, & & 1 \leq x \leq 2 \\ - 2 + 3x - x^2 , & & x > 2\end{cases}\] .
उत्तर
We have \[f\left( x \right) = \begin{cases}x, & x < 1 \\ 2 - x, & 1 \leq x \leq 2 \\ - 2 + 3x - x^2 , & x > 2\end{cases}\]
Clearly, f(x), being a polynomial function, is continuous and differentiable for all x < 1, 1 < x < 2 and also for all x > 2.
Thus, the possible points of non-differentiability of f(x) are x = 1 and x = 2.
Now,
f(1) = 2 – 1 = 1
and
f(2) = 2 – 2 = 0
At x = 1,
\[\begin{array}{cl}LHD & = & \lim_{x \to 1^-} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1} \\ = & \lim_{x \to 1} \frac{x - 1}{x - 1} \\ = & 1 \\ RHD & = & \lim_{x \to 1^+} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1} \\ = & \lim_{x \to 1} \frac{\left( 2 - x \right) - 1}{x - 1} \\ = & \lim_{x \to 1} \frac{- \left( x - 1 \right)}{x - 1} \\ = & - 1\end{array}\]
∴ LHD ≠ RHD
So, f(x) is not differentiable at x = 1.
At x = 2,
LHD =`lim_(x →2^-) (f(x) - f (2))/(x-2)`
= `lim_(x →2) (2- x- 0)/(x-2)`
= `lim_(x →2) (-(x-2))/(x-2)`
= -1
RHD = `lim_(x →2^+) (f(x) - f (2))/(x-2)`
= `lim_(x →2) ((-2+3x -x^2)-0)/(x-2)`
= `lim_(x →2) (-(x^2- 3x +2))/(x-2)`
= `lim_(x →2) (-(x-1)(x-2))/(x-2)`
=` lim_(x →2) -(x-1)`
= -1
LHD = RHD
So, f(x) is differentiable at x = 2.
Thus, the given function is differentiable at x = 2, but not at x = 1.
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