G Tewani Solutions for Chapter: Applications of Derivatives, Exercise 1: DPP

If a curve with the equation of the form $y=a{x}^4+b{x}^3+cx+d$ has zero gradients at the point $\left(0, 1\right)$ and also touches the $x$-axis at the point $(-1, 0)$ then the values of $x$ for which the curve has a negative gradient are:

From the point $\left(1,1\right)$ tangents are drawn to the curve represented parametrically as $x=2t-t^2$ and $y=t+t^2.$ The distance between the points of contact is equal to

The value of parameter $t$ so that the line $(4-t)x+t y+\left(a^3-1\right)=0$ is normal to the curve $xy=1$ may lie in the interval:

The portion of the tangent at any point on the curve $x=a{t}^3$, $y=a{t}^4$ between the axes is divided by the abscissa of the point of contact externally in the ratio:

Equation of a line which is tangent to both the curves $y=x^2+1$ and $y=-x^2$ is:

For the functions defined parametrically by the equations

$f\left(t\right)=x=\left\{\begin{array}{cc}2t+t^2\sin \frac{1}{t}& ; t\ne 0\\ 0& ; t=0\end{array}\right.$     and

$g\left(t\right)=y=\left\{\begin{array}{cc}\frac{1}{t}\sin t^2& ; t\ne 0\\ 0& ; t=0\end{array}\right.$.

For the curve $x=2a\sin t+a\sin t{\cos }^{2}t$ ; $y=-a{\cos }^{3}t$

The curve $y=a{x}^3+b{x}^2+cx$ is inclined at $45°$ to $x$-axis at (0,0) but it touches $x$-axis at $(1, 0)$, then: