If x²y⁵ = (x + y)⁷, then $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ is equal to

• y/x²

• x/y

• 1

• 0

If x = $\mathrm{x} = \sec \left(\mathrm{\theta }\right), \mathrm{y} = \tan \left(\mathrm{\theta }\right)$, then the value of $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ at $\mathrm{\theta } = \frac{\mathrm{\pi }}{4}$ is

• 0

• 1

• - 1

• 2

 If x = f(t) and y =g(t), then the value of $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}$ is

• $\frac{\mathrm{f}\text{'}\left(\mathrm{t}\right)\mathrm{g}\text{'}\text{'}\left(\mathrm{t}\right) - \mathrm{g}\text{'}\left(\mathrm{t}\right)\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)}{{\left\{\mathrm{f}\text{'}\left(\mathrm{t}\right)\right\}}^3}$

• $\frac{\mathrm{f}\text{'}\left(\mathrm{t}\right)\mathrm{g}\text{'}\text{'}\left(\mathrm{t}\right) - \mathrm{g}\text{'}\left(\mathrm{t}\right)\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right)}{{\left\{\mathrm{f}\text{'}\left(\mathrm{t}\right)\right\}}^2}$

• $\frac{\text{'}\left(\mathrm{t}\right)\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right) - \mathrm{g}\text{'}\text{'}\left(\mathrm{t}\right)\mathrm{f}\text{'}\left(\mathrm{t}\right)}{{\left\{\mathrm{f}\text{'}\left(\mathrm{t}\right)\right\}}^2}$

• $\frac{\mathrm{g}\text{'}\left(\mathrm{t}\right)\mathrm{f}\text{'}\text{'}\left(\mathrm{t}\right) - \mathrm{g}\text{'}\text{'}\left(\mathrm{t}\right)\mathrm{f}\text{'}\left(\mathrm{t}\right)}{{\left\{\mathrm{f}\text{'}\left(\mathrm{t}\right)\right\}}^3}$

The value of a and b such that the function

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{- 2\sin \left(\mathrm{x}\right), - \mathrm{\pi } \le \mathrm{x} \le - \frac{\mathrm{\pi }}{2} \\ \mathrm{asin}\left(\mathrm{x}\right) + \mathrm{b}, - \frac{\mathrm{\pi }}{2} < \mathrm{x} < \frac{\mathrm{\pi }}{2} \\ \cos \left(\mathrm{x}\right), \frac{\mathrm{\pi }}{2} \le \mathrm{x} \le \mathrm{\pi }\right\} \end{gathered}$$

is continuous in $\left[- \mathrm{\pi }, \mathrm{\pi }\right]$, are

• - 1, 0

• 1, 0

• 1, 1

• - 1, 1

If g is the inverse of f and f'(x) = $\frac{1}{1 + {\mathrm{x}}^2}$, then g'(x) is equal to

• 1 + [g(x)]²

• $\frac{- 1}{1 + \left[\mathrm{g}\right(\mathrm{x}){]}^2}$

• $\frac{1}{2\left(1 + {\mathrm{x}}^2\right)}$

• None of these

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

• 3/2, e^3/2

• - 2/3, e^{- 3/2}

• 2/3, e^2/3

• None of these

827.

If f(x) = e^xg(x), g(0) = 2, g'(0) = 1 then f'(0) is

• 1

• 3

• 2

• 0

At the point x = 1, the function

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right) = \left\{{\mathrm{x}}^3 - 1, 1 < \mathrm{x} < \infty \\ \mathrm{x} - 1, - \infty < \mathrm{x} \le 1\right\} \end{gathered}$$

• continuous and differentiable

• continuous and not differentiable

• discontinuous and differentiable

• discontinuous and not differentiable

If x = 2cos(t) - cos(2t), y = 2sin(t) - sin(2t), then the value of ${\overline{)\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}}}_{\mathrm{t} = \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

• 3/2

• 5/2

• 5/3

• - 3/2

If f(x) = $$\begin{gathered} \left\{\frac{\log \left(1 +\hspace{0.17em}2\mathrm{ax}\right) - \log \left(1 - \mathrm{bx}\right)}{\mathrm{x}}, \mathrm{x} \ne 0 \\ \mathrm{k}, \mathrm{x} = 0\right\} \end{gathered}$$ is contonuous at x = 0, then value of k is

• b + a

• b - 2a

• 2a - b

• 2a + b