CUET Preparation Today
Let a function $y=y(x)$ be defined parametrically by $x(t)=2t-|t|, y(t)=t^2+t|t|, t \in R .$ Then $\frac{dy}{dx}$ at x= 5 isequal to |
20 10 40 -20 |
20 |
The correct answer is Option (1) → 20 At $x=5$ $x(t)=2t-t$, $y(t)=t^2+t^2=2t^2$ $\frac{dx}{dt}=1$, $\frac{dy}{dt}=4t$ $⇒\frac{dy}{dx}=4t$ $⇒\left.\frac{dy}{dx}\right|_{t=5}=4×5=20$ |
