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

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

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Similar Questions

Let a function $f: R \rightarrow R$ be defined as :
$f(x)=\begin{cases} \int_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x, & x \leq 4 \end{cases}$
where $b \in R$. If $f$ is continuous at $x=4$,then which of the following statements is $NOT$ true?

Give the correct order of initials $T$ or $F$ for the following statements. Use $T$ if the statement is true and $F$ if it is false.
Statement-$1$: If $f: R \rightarrow R$ and $c \in R$ is such that $f$ is increasing in $(c - \delta, c)$ and $f$ is decreasing in $(c, c + \delta)$,then $f$ has a local maximum at $c$. Where $\delta$ is a sufficiently small positive quantity.
Statement-$2$: Let $f: (a, b) \rightarrow R, c \in (a, b)$. Then $f$ cannot have both a local maximum and a point of inflection at $x = c$.
Statement-$3$: The function $f(x) = x^2 |x|$ is twice differentiable at $x = 0$.
Statement-$4$: Let $f: [c - 1, c + 1] \rightarrow [a, b]$ be a bijective map such that $f$ is differentiable at $c$ and $f'(c) \neq 0$,then $f^{-1}$ is also differentiable at $f(c)$.

Let $f: R \to R$ be a twice differentiable function such that the quadratic equation $f(x)m^{2}-2f^{\prime}(x)m+f^{\prime\prime}(x)=0$ in $m$ has two equal roots for every $x \in R$. If $f(0)=1$, $f^{\prime}(0)=2$ and $(\alpha, \beta)$ is the largest interval in which the function $g(x) = f(\log_{e}x-x)$ is increasing, then $\alpha+\beta$ is equal to:

Match each function in List-$I$ to its derivative given in List-$II$.
List-$I$List-$II$
$(A) \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$$(I) \cos x-\sin x$
$(B) \tan ^{-1}\left(\frac{1-x}{1+x}\right)$$(II) \frac{-1}{1+x^2}$
$(C) e^{\log (\sin x+\cos x)}$$(III) \frac{2}{1+x^2}$
$(D) \sqrt{1-\sin 2 x} \text{ for } (0 < x < \frac{\pi}{4})$$(IV) \cos x+\sin x$
$(V) -\sin x-\cos x$

The correct match is:

$\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\frac{{{e^{{e^{{x^2}}}}} - e}}{x}} \right)$ is

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