The function $\text{f(x)}=\text{e}^{|\text{x}|}$ is:
- Continuous everywhere but not differentiable at x = 0.
- Continuous and differentiable everywhere.
- Not continuous at x = 0.
- None of these.
Solution:
Let $\text{u(x)}=|\text{x}|$ and $\text{v(x)}=\text{e}^\text{x}$
$\therefore\ \text{f(x)}=\text{vou(x)}=\text{v}[\text{u(x)]}$
$=\text{v}|\text{x}|=\text{e}^{|\text{x}|}$
Since, u(x) and v(x) are both continuous functions.
So, f(x) is also continuous function but u(x) = |x| is not differentiable at x = 0, whereas v(x) = ex is differentiable at everywhere.
Hence, f(x) is continuous everywhere but not differentiable at x = 0.
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$1.$ The correct statement$(s)$ is(are)
$(A)$ $f^{\prime}(1) < 0$
$(B)$ $f(2) < 0$
$(C)$ $f^{\prime}(x) \neq 0$ for any $x \in(1,3)$
$(D)$ $f^{\prime}(x)=0$ for some $x \in(1,3)$
$2.$ If $\int_1^3 x^2 F^{\prime}(x) d x=-12$ and $\int_1^3 x^3 F^{\prime \prime}(x) d x=40$, then the correct expression$(s)$ is(are)
$(A)$ $9 f^{\prime}(3)+f^{\prime}(1)-32=0$
$(B)$ $\int_1^3 f(x) d x=12$
$(C)$ $9 f^{\prime}(3)-f^{\prime}(1)+32=0$
$(D)$ $\int_1^3 f(x) d x=-12$
Give the answer question $1$ and $2.$