Question

Find the equation of the normal to the curve $2y=x^2$, which passes through the point (2, 1).

OR Separate the interval $\left[\begin{array}{c}0,\frac{\pi }{2}\end{array}\right]$ into subintervals in which $f\left(x\right)=si{n}^{4}x+co{s}^{4}x$ is strictly increasing or strictly decreasing.

Answer 1

Let the normal be at $(x_1,y_1)$ to the curve $2y=x^2\therefore 2y_1=x_1^2...\left(a\right)$

Also $\frac{dy}{dx}=x$ $\therefore$ slope of normal at $(x_1,y_1)=-\frac{1}{x_1}.$

Equation of normal : $y-y_1=\frac{-1}{x_1}(x-x_1)...\left(i\right)$

As (i) passes through (2, 1) so, $1-y_1=\frac{-1}{x_1}(2-x_1)\Rightarrow x_1{y}_1=\mathrm{2...}\left(ii\right)$

Solving (a) and (ii), we get $x_1=2^{2/3},y_1=2^{1/3}$

Hence the required equation is $y-2^{1/3}=\frac{-1}{2^{2/3}}(x-2^{2/3})$ i.e., $x+2^{2/3}y=2+2^{2/3}.$

OR Given $f\left(x\right)=si{n}^{4}x+co{s}^{4}x,x\in [0,\frac{\pi }{2}]\Rightarrow f^{\prime }\left(x\right)=4si{n}^{3}xcosx-4co{s}^{3}xsinx$

$\Rightarrow f^{\prime }\left(x\right)=4si{n}^{3}xcosx-4co{s}^{3}xsinx\Rightarrow f^{\prime }\left(x\right)=-4sinxcosx(co{s}^{2}x-si{n}^{2}x)$

$\Rightarrow f^{\prime }\left(x\right)=-2sin2xcos2x=-sin4x$

For critical points, $f^{\prime }\left(x\right)=-sin4x=0\Rightarrow sin4x=0\Rightarrow 4x=0,\pm \pi ,\pm 2\pi ,\pm 3\pi ,\pm 4\pi ,....$

$\therefore x=0,\frac{\pi }{4},\frac{\pi }{2}\in [0,\frac{\pi }{2}]$

$\begin{array}{ccc}\text{Interval}& \text{Sign of f'(x)}& \text{f(x) is strictly}\\ (0,\frac{\pi }{4})& \text{Negative}& \text{Decreasing}\\ (\frac{\pi }{4},\frac{\pi }{2})& \text{Positive}& \text{Increasing}\end{array}$