If $f(x) = \begin{cases} ax+b, & \text{if } x \leq 1 \\ ax^2+c, & \text{if } 1 < x \leq 2 \\ \frac{dx^2+1}{x}, & \text{if } x > 2 \end{cases}$ is differentiable on $\mathbb{R}$, then $ad-bc = $

  • A
    $0$
  • B
    $1$
  • C
    $-1$
  • D
    $2$

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Similar Questions

Let $f$ be a real-valued function defined on the interval $(0, \infty)$ by $f(x)=\ln x+\int_0^x \sqrt{1+\sin t} \, dt$. Then which of the following statement$(s)$ is (are) true?
$(A)$ $f^{\prime \prime}(x)$ exists for all $x \in(0, \infty)$
$(B)$ $f^{\prime}(x)$ exists for all $x \in(0, \infty)$ and $f^{\prime}$ is continuous on $(0, \infty)$,but not differentiable on $(0, \infty)$
$(C)$ there exists $\alpha>1$ such that $|f^{\prime}(x)|<|f(x)|$ for all $x \in(\alpha, \infty)$
$(D)$ there exists $\beta>0$ such that $|f(x)|+|f^{\prime}(x)| \leq \beta$ for all $x \in(0, \infty)$

If $y=|\cos x-\sin x|+|\tan x-\cot x|$,then $\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{3}}+\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{6}}=$

If $f(x) = \begin{cases} \int_{0}^{x} (5 + |1-t|) \, dt, & x > 2 \\ 5x + 1, & x \leq 2 \end{cases}$,then:

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{\int_2^{\sec ^2 x} f(t) d t}{x^2-\frac{\pi^2}{16}}$ equals

Let $f:\left[-\frac{1}{2}, 2\right] \rightarrow R$ and $g:\left[-\frac{1}{2}, 2\right] \rightarrow R$ be functions defined by $f(x)=\left[x^2-3\right]$ and $g(x)=|x| f(x)+|4 x-7| f(x)$,where $[y]$ denotes the greatest integer less than or equal to $y$ for $y \in R$. Then
$(A)$ $f$ is discontinuous exactly at three points in $\left[-\frac{1}{2}, 2\right]$
$(B)$ $f$ is discontinuous exactly at four points in $\left[-\frac{1}{2}, 2\right]$
$(C)$ $g$ is $NOT$ differentiable exactly at four points in $\left(-\frac{1}{2}, 2\right)$
$(D)$ $g$ is $NOT$ differentiable exactly at five points in $\left(-\frac{1}{2}, 2\right)$

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