The slope of the tangent to the curve ${\left(y-x^5\right)}^2=x{\left(1+x^2\right)}^2$at the point $\left(1,3\right)$is.

If $a,b,c\in R\text{and}a+b+c=0,$ then the quadratic equation $3a{x}^2+2bx+c=0$ has (a) at least one root in $\left[0,1\right]$ (b) at least one root in $\left[1,2\right]$ (c) at least one root in $\left[\frac{3}{2},2\right]$ (d) none of these

if $\left|f\left(x_1\right)-f\left(x_2\right)\right|\le {\left(x_1-x_2\right)}^2$Find the equation of gent to the curve $y=f\left(x\right)$ at the point $\left(1,2\right).$

The triangle formed by the tangent to the curve $f\left(x\right)=x^2+bx-b$ at the point $\left(1,1\right)$ and the coordinate axes, lies in the first quadrant. If its area is 2, then the value of $b$ is (a) $-1$ (b) $3$ (c) $-3$ (d) $1$

The point on the curve $4x^2+a^2{y}^2=4a^2$ ; $4<a^2<8$ that is farthest from the point (0,-2) is