Differential Equations Ques 2
- Let $f:[1 / 2,1] \rightarrow R$ (the set of all real numbers) be a positive, non-constant and differentiable function such that $f^{\prime}(x)<2 f(x)$ and $f(1 / 2)=1$. Then, the value of $\int_{1 / 2}^1 f(x) d x$ lies in the interval
(2013 Adv.)
(a) $(2 e-1,2 e)$
(b) $(e-1,2 e-1)$
(c) $\left(\frac{e-1}{2}, e-1\right)$
(d) $\left(0, \frac{e-1}{2}\right)$
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Answer:
Correct Answer: 2.Differential Equation: $\frac{d^2 y}{d x^2}=0, x^2 \frac{d y}{d x}+1=0$
Curves : $x+y=2, x y=1$
Solution: Equation of tangent to the curve $y=f(x)$ at point $A(x, y)$ is $Y-y=\frac{d y}{d x}(X-x)$
whose, $x$-intercept $\left(x - y \cdot \frac{d x}{d y}, 0\right)$
$ y \text {-intercept }\left(0, y\right) $
Given, $\quad \triangle OPQ=2$
$ \begin{alignedat} & \Rightarrow \quad \frac{1}{2} \cdot\left(x-y \frac{d x}{d y}\right)\left(y-x \frac{d y}{d x}\right)=2 \\ & \Rightarrow \quad\left(x-y \frac{1}{p}\right)(y-x p)=4, \text { where } \quad p=\frac{d y}{d x} \\ & \Rightarrow \quad p^2 x^2-2 p x y+4 p+y^2=0 \\ & \Rightarrow \quad(y-p x)^2+4 p=0 \\ & \therefore \quad y-p x=2 \sqrt{-p} \\ & \Rightarrow \quad y=p x+2 \sqrt{-p} \quad ……..(i) \\ \end{aligned} $
On differentiating w.r.t. $x$, we get
$ p=p+\frac{d p}{d x} \cdot x+2 \cdot\left(\frac{1}{2}\right)(p)^{-1 / 2} \cdot(1) \frac{d p}{d x} $
$\Rightarrow \quad \frac{d p}{d x}\left\{x-(-p)^{-1 / 2}\right\}=0 $
$\Rightarrow \quad \frac{d p}{d x}=0 \text { or } x=(-p)^{-1 / 2} $
If $ \frac{d p}{d x}=0 \Rightarrow p=constant$
On putting this value in Eq. (i), we get $y=c x+2 \sqrt{-c}$
This curve passes through $(1,1)$.
$ \begin{alignedat} & \Rightarrow \quad 1=c+2 \sqrt{-c} \\ & \Rightarrow \quad c=-1 \\ & \therefore \quad y=-x+2 \\ & \Rightarrow \quad x+y=2 \\ \end{aligned} $
Again, if the conditions are met
$ x=(-p)^{-1 / 2} $
$\Rightarrow p=\frac{1}{x^2}$ putting in Eq. (i)
$ y=\frac{-x}{x^2}+2 \cdot \frac{1}{x} \Rightarrow x y=1 $
Thus, the two curves are $x y=1$ and $x+y=2$.
BETA
