81.

The derivative of ${\tan }^{-1}\left(\frac{\sqrt{1 + {\mathrm{x}}^2} - 1}{\mathrm{x}}\right)$ with respect to ${\tan }^{-1}\left(\frac{2\mathrm{x}\sqrt{1 - {\mathrm{x}}^2}}{1 - 2{\mathrm{x}}^2}\right)$ at x = 0 is

• $\frac{1}{8}$

• $\frac{1}{4}$

• $\frac{1}{2}$

• 1

Let f(x + y) = f(x) f(y) and f(x) = 1 + sin(2x) g(x) where g(x) is continuous. Then, f'(x) equals

• f(x) g(0)

• 2f(x) g(0)

• 2g(0)

• None of the above

If $$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{{\mathrm{ax}}^2 - \mathrm{b}, \mathrm{a} \le \mathrm{x} < 1 \\ 2, \mathrm{x} = 1 \\ \mathrm{x} + 1, 1 < \mathrm{x} \le 2\right\} \end{gathered}$$ Then, the value of the pair (a, b) for which f(x) cannot be continuous at x = 1, is

• (2, 0)

• (1, - 1)

• (4, 2)

• (1, 1)

Which of the following function is not differentiable at x = 1 ?

• f(x) = $\tan \left(\left|\mathrm{x} - 1\right|\right) + \left|\mathrm{x} - 1\right|$

• f(x) = $\sin \left(\left|\mathrm{x} - 1\right|\right) - \left|\mathrm{x} - 1\right|$

• f(x) = $\left|\left({\mathrm{x}}^2 - 1\right)\left(\mathrm{x} - 1\right)\left(\mathrm{x} - 2\right)\right|$

• None of the above

Using Rolle's theorem, the equation a₀xⁿ + a₁x^{n - 1} + ... + a_n = 0 has atleast one root between 0 and 1, if 

• $\frac{{\mathrm{a}}_0}{\mathrm{n}} + \frac{{\mathrm{a}}_1}{\mathrm{n} - 1} + ... + {\mathrm{a}}_{\mathrm{n} - 1} = 0$

• $\frac{{\mathrm{a}}_0}{\mathrm{n} - 1} + \frac{{\mathrm{a}}_1}{\mathrm{n} - 2} + ... + {\mathrm{a}}_{\mathrm{n} - 2} = 0$

• na₀ + (n - 1)a₁ + ... + a_{n - 1} = 0

• $\frac{{\mathrm{a}}_0}{\mathrm{n} + 1} + \frac{{\mathrm{a}}_1}{\mathrm{n}} + ... + {\mathrm{a}}_{\mathrm{n}} = 0$

The value of cfrom the Lagrange's mean value theorem for which f(x) = $\sqrt{25 - {\mathrm{x}}^2}$ in [1, 5], is

• 5

• 1

• $\sqrt{15}$

• None of these

Let f'(x), be differentiable $\forall$a. If f(1) = - 2 and f'(x) $\ge$ 2 $\forall$ x $\in$ [1, 6], then

• f(6) < 8

• f(6) $\ge$ 8

• f(6) $\ge$ 5

• f(6) $\le$ 5

$$\begin{gathered} \mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = \left\{{\left(1 + \left|\sin \left(\mathrm{x}\right)\right|\right)}^{\raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$\left|\sin \left(\mathrm{x}\right)\right|$}\right.}, - \frac{\mathrm{\pi }}{6} < \mathrm{x} < 0 \\ \mathrm{b}, \mathrm{x} = 0 \\ {\mathrm{e}}^{\raisebox{1ex}{$\tan \left(2\mathrm{x}\right)$}\!\left/ \!\raisebox{-1ex}{$\tan \left(3\mathrm{x}\right)$}\right.}, 0 < \mathrm{x} < - \frac{\mathrm{\pi }}{6}\right\} \end{gathered}$$

then the value of a and b, if f is continuous at x = 0, are respectively

• $\frac{2}{3}, \frac{3}{2}$

• $\frac{2}{3}, {\mathrm{e}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

• $\frac{3}{2}, {\mathrm{e}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• None of these

$$\begin{gathered} \mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = \left\{\mathrm{mx} + 1, \mathrm{x} \le \frac{\mathrm{\pi }}{2} \\ \sin \left(\mathrm{x} + \mathrm{n}\right), \mathrm{x} > \frac{\mathrm{\pi }}{2}\right\} \mathrm{is} \mathrm{continuous} \mathrm{at} \mathrm{x} = \frac{\mathrm{\pi }}{2}, \mathrm{then} \end{gathered}$$

• m = 1, n = 0

• $\mathrm{m} = \frac{\mathrm{n\pi }}{2} + 1$

• $\mathrm{n} = \mathrm{m}\frac{\mathrm{\pi }}{2}$

• $\mathrm{m} = \mathrm{n} = \frac{\mathrm{\pi }}{2}$

$$\begin{gathered} \mathrm{If} \mathrm{f}\left(\mathrm{x}\right) = \left\{{\mathrm{x}}^2, \mathrm{x} \le 0 \\ 2 \sin \left(\mathrm{x}\right), \mathrm{x} > 0\right\}, \mathrm{then} \mathrm{x} = 0 \mathrm{is} \end{gathered}$$

• point of minima

• point of maxima

• point of discontinuity

• None of the above