Let $C$ be the curve $y = x^3$ (where $x$ takes all real values). The tangent at $A(t, t^3)$ meets the curve again at $B(T, T^3)$. If the gradient at $B$ is $K$ times the gradient at $A$,then $K$ is equal to

  • A
    $4$
  • B
    $2$
  • C
    $- 2$
  • D
    $\frac{1}{4}$

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If $f(x) = 4x^3 - x^2 - 2x + 1$ and $g(x) = \begin{cases} \min \{f(t) : 0 \le t \le x\} & ; 0 \le x \le 1 \\ 3 - x & ; 1 < x \le 2 \end{cases}$,then the value of $g\left( \frac{1}{4} \right) + g\left( \frac{3}{4} \right) + g\left( \frac{5}{4} \right)$ is:

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 $f : (0, \pi) \rightarrow \mathbb{R}$ be a twice differentiable function such that $\lim _{t \rightarrow x} \frac{f(x) \sin t - f(t) \sin x}{t-x} = \sin^2 x$ for all $x \in (0, \pi)$. If $f \left(\frac{\pi}{6}\right) = -\frac{\pi}{12}$,then which of the following statement$(s)$ is (are) $TRUE$?
$(A) f \left(\frac{\pi}{4}\right) = \frac{\pi}{4 \sqrt{2}}$
$(B) f(x) < \frac{x^4}{6} - x^2$ for all $x \in (0, \pi)$
$(C)$ There exists $\alpha \in (0, \pi)$ such that $f^{\prime}(\alpha) = 0$
$(D) f^{\prime \prime}\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = 0$

If $y=(x+1)(x^2+1)(x^4+1)(x^8+1)$,then $\lim _{x \rightarrow-1} \frac{dy}{dx}=$

Let $f: R \to R$ be a differentiable function such that $f(2) = 6$ and $f'(2) = \frac{1}{48}.$ Then $\lim_{x \to 2} \int_{6}^{f(x)} \frac{4t^3}{x - 2} dt$ equals

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