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TS EAMCET 7 May 2018 Shift 1 Solved Paper

Section: Mathematics

Question : 64 of 160

Marks: +1, -0

The function that is not differentiable at x = 1

$f_1$ (x) = |x| , - ∞ < x < ∞
$f_2$ (x) =
$\begin{cases} 1+\sin(x-1) & -\infty < x \le 1 \\ x & x \ge 1 \end{cases}$
$f_3$ (x) =
$\begin{cases} x^{2}+7x-7 & -\infty < x \le 1 \\ \frac{3x-1}{2} & x > 1 \end{cases}$
$f_4$ (x) =
$\begin{cases} \left|x-1\right| + \left|x-2\right| & -\infty < x \le 1 \\ 1+x-x^{3} & x \ge 1 \end{cases}$

Solution:

f_1(x)=\begin{cases} -x & x<0 \\ x & x\ge 0 \end{cases}

⇒

f'_1(x)=\begin{cases} -1 & x<0 \\ 1 & x\ge 0 \end{cases}

Therefore,

f_{1}'(1)=1

f_{2}(x)=\begin{cases} 1+\sin(x-1) & x\le 1 \\ x & x\ge 1 \end{cases}

f'_{2}(x)=\begin{cases} \cos(x-1) & x\le 1 \\ 1 & x\ge 1 \end{cases}

f'_{2}(x)=1

f'_{3}(x)=\begin{cases} x^{2}+7x-7 & x\le 1 \\ \frac{3x-1}{2} & x\ge 1 \end{cases}

f'_{2}(x)=\begin{cases} 2x+7 & x\le 1 \\ \frac{3}{2} & x\ge 1 \end{cases}

since

\text{LHD}

and

\text{RHD}

are not equal.

f_{3}(x)

is not differentiable at

x=1

f_{4}(x)=\begin{cases} \left|x-1\right|+\left|x-2\right| & x\le 1 \\ 1+x-x^{3} & x\ge 1 \end{cases}

f_{4}'(x)=\begin{cases} -2 & x\le 1 \\ 1-3x^{2} & x\ge 1 \end{cases}

f_{4}'(1)=-2

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