Differential Equations

$ \begin{aligned} & \lim _{1 \rightarrow x} \frac{t^{10} f(x)-x^{10} f(t)}{t^9-x^9}=1 \\\\ & \Rightarrow \lim _{1 \rightarrow x} \frac{10 t^9 f(x)-x^{10} f^{\prime}(t)}{9 t^8}=1 \\\\ & \Rightarrow 10 x f(x)-x^2 f^{\prime}(x)=9 \\\\ & \Rightarrow x^2 f^{\prime}(x)=10 x f(x)-9 \\\\ & \Rightarrow f^{\prime}(x)=\frac{10 f(x)}{x}-\frac{9}{x^2} \\\\ & \Rightarrow \frac{d y}{d x}-\frac{10}{x} y=-\frac{9}{x^2} \\\\ & \Rightarrow y \cdot \frac{1}{x^{10}}=\int-\frac{9}{x^2} \cdot \frac{1}{x^{10}} d x \\\\ & \Rightarrow \frac{y}{x^{10}}=\frac{9}{11 x^{11}}+c \\\\ & \because f(1)=2 \Rightarrow \frac{2}{1}=\frac{9}{11}+c \Rightarrow c=\frac{13}{11} \\\\ & \therefore f(x)=\frac{9}{11 x}+\frac{13}{11} x^{10} \end{aligned} $

$\Rightarrow$ Option (B) is correct.

1. We are given a relationship between the integral of the function $f(t)$ and the function $f(x)$ :

$3 \int\limits_1^x f(t) d t=x f(x)-\frac{x^3}{3}, x \in[1, \infty)$

2. Differentiate both sides of the above equation with respect to $x$ :

$ 3(f(x) \cdot 1-0)=1 . f(x)+x f^{\prime}(x)-\frac{3 x^2}{3} $

$ \Rightarrow $ $3f(x) = f(x) + xf'(x) - x^2$

$ \Rightarrow $ $2f(x) = xf'(x) - x^2$

3. We can rewrite this as a first order linear differential equation:

$xf'(x) - 2f(x) = x^2$

4. The general form of a first order linear differential equation is:

$y'(x) + p(x)y = q(x)$

Here $y = f(x)$, $p(x) = -\frac{2}{x}$, and $q(x) = x$.

5. The integrating factor is obtained by exponentiating the integral of $p(x)$ with respect to $x$:

$e^{\int p(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2\ln|x|} = \frac{1}{x^2}$

6. $ \therefore $ Solution of D.E :

$ \begin{gathered} y \cdot\left(\frac{1}{x^2}\right)=\int(x) \frac{1}{x^2} d x \\\\ \frac{y}{x^2}=\ln x+c \\\\ y=x^2 ln x+c x^2 \end{gathered} $

$ \therefore $ The general solution is :

$y(x) = x^2(\ln x + c)$

Here, $y(x)$ is the same as $f(x)$, i.e. the function we're trying to find.

7. Given that $f(1) = \frac{1}{3}$, let's use this condition to find the constant $c$ :

$\frac{1}{3} = 1^2(\ln 1 + c) \Rightarrow c = \frac{1}{3}$

Therefore, the function $f(x)$ we are looking for is:

$f(x) = x^2(\ln x + \frac{1}{3})$

We want to find $f(e)$, so substituting $x=e$ into the function, we get :

$f(e) = e^2(\ln e + \frac{1}{3}) = e^2(1 + \frac{1}{3}) = \frac{4e^2}{3}$

So, the correct answer is

Option C : $\frac{4e^2}{3}$

${{dy} \over {dx}} = {1 \over {8\sqrt x + \sqrt {9 + \sqrt x } \sqrt {4 + \sqrt {9 + \sqrt x } } }}$

$ \Rightarrow y = \sqrt {4 + \sqrt { + \sqrt x } } + c$

Now, $y(0) = \sqrt 7 + c$

$ \Rightarrow c = 0$

$y(256) = \sqrt {4 + \sqrt {9 + \sqrt {16} } } = \sqrt {4 + 5} = 3$

Given, $\frac{d y}{d x}+\left(\frac{x}{x^2-1}\right) y=\frac{x^4+2 x}{\sqrt{1-x^2}}$

The above differential equation is a linear differential equation.

Integrating factor I.F. $=e^{\int \frac{x}{x^2-1} d x}$

$\begin{aligned} & \Rightarrow \text { I.F. }=e^{\frac{1}{2} \int \frac{-2 x d x}{1-x^2}} \\ & \Rightarrow \text { I.F. }=e^{\frac{1}{2} \ln \left(1-x^2\right)} \quad\left(\int \frac{f^{\prime}(x) d x}{f(x)}=\ln f(x)+c\right) \\ & \Rightarrow \text { I.F. }=e^{\ln \sqrt{1-x^2}} \\ & \Rightarrow \text { I.F. }=\sqrt{1-x^2} \end{aligned}$

Now, the solution of the given differential equation is

$\begin{aligned} & \Rightarrow y \cdot \text { (I.F. })=\int\left(\frac{x^4+2 x}{\sqrt{1-x^2}}\right)(\text { L.F. }) d x+\text { C } \\ & \Rightarrow y \sqrt{1-x^2}=\int\left(\frac{x^4+2 x}{\sqrt{1-x^2}}\right) \sqrt{1-x^2} d x+\text { C } \\ & \Rightarrow y \sqrt{1-x^2}=\frac{x^5}{5}+2 \times \frac{x^2}{2}+\text { C } \\ & \Rightarrow y=f(x)=\frac{\frac{x^5}{5}+x^2+\mathrm{C}}{\sqrt{1-x^2}} \end{aligned}$

Given, $f(0)=0$

$\begin{array}{ll} \Rightarrow & 0=\frac{0+0+C}{\sqrt{1-0}} \\ \Rightarrow & C=0 \\ \therefore & f(x)=\frac{\frac{x^5}{5}+x^2}{\sqrt{1-x^2}}=\frac{x^5}{5 \sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}} \end{array}$

Let $\mathrm{I}=\int_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x$

$\Rightarrow \mathrm{I}=\int_{\frac{-\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}}\left(\frac{x^5}{5 \sqrt{1-x^2}}+\frac{x^2}{\sqrt{1-x^2}}\right) d x$

Here, $\frac{x^5}{5 \sqrt{1-x^2}}$ is an odd function and $\frac{x^2}{\sqrt{1-x^2}}$ is an even function.

$\begin{aligned} & \therefore \mathrm{I}=0+2 \int_0^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} d x \\ & \Rightarrow \mathrm{I}=2 \int_0^{\frac{\sqrt{3}}{2}}\left(\frac{x^2-1+1}{\sqrt{1-x^2}}\right) d x \\ & \Rightarrow \mathrm{I}=2 \int_0^{\frac{\sqrt{3}}{2}}\left(-\sqrt{1-x^2}+\frac{1}{\sqrt{1-x^2}}\right) d x \\ & \Rightarrow \mathrm{I}=2\left[-\left(\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right)+\sin ^{-1} x\right]_0^{\frac{\sqrt{3}}{2}} \\ & \Rightarrow \mathrm{I}=2\left[-\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right]_0^{\frac{\sqrt{3}}{2}} \\ & \Rightarrow \mathrm{I}=2\left[-\frac{\sqrt{3}}{4} \times \frac{1}{2}+\frac{1}{2} \times \frac{\pi}{3}\right] \\ & \Rightarrow \mathrm{I}=\frac{\pi}{3}-\frac{\sqrt{3}}{4} \end{aligned}$

Hint:

(i) The solution of linear differential equation $\frac{d y}{d x}+\mathrm{P}(x) \cdot y=\mathrm{Q}(x)$ is

$y \text { (I.F.) } =\int \mathrm{Q}(x) \cdot(\text { I.F. }) d x+\text { C }$

Where, $\quad \text { L.F. } =e^{\int p(x) d x}$

(ii) Recall the property

$\int_{-a}^a f(x) d x=\left\{\begin{array}{l} 2 \int_0^a f(x) d x, \text { When } f(x) \text { is an even } \\ \text { function. } \\ 0, \text { When } f(x) \text { is an odd function. } \end{array}\right.$

Given, the slope of curve at $(x,y)$ is

$\begin{aligned} & \frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0 . \\ \therefore \quad & \frac{d y}{d x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right) \quad \text{... (i)} \end{aligned}$

Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$ in the equation (i)

$\begin{aligned} & \Rightarrow \quad v+x \frac{d v}{d x}=\frac{v x}{x}+\sec \left(\frac{v x}{x}\right) \\ & \Rightarrow \quad v+x \frac{d v}{d x}=v+\sec (v) \\ & \Rightarrow \quad x \frac{d v}{d x}=\sec v \\ & \Rightarrow \quad \cos v d v=\frac{d x}{x} \\ & \Rightarrow \quad \int \cos v d v=\int \frac{d x}{x} \\ & \Rightarrow \quad \sin v=\ln x+c \\ & \Rightarrow \quad \sin \left(\frac{y}{x}\right)=\ln x+c \quad \text{... (i)} \end{aligned}$

Put $x=1$ and $y=\frac{\pi}{6}$ in the equation (ii)

$\begin{aligned} \Rightarrow & & \sin \frac{\pi}{6} & =\log 1+c \\ \Rightarrow & & \frac{1}{2} & =0+c \\ \Rightarrow & & c & =\frac{1}{2} \end{aligned}$

Put $c=\frac{1}{2}$ in the equation (ii)

$\Rightarrow \quad \sin \frac{y}{x}=\log x+\frac{1}{2}$

Hints:

(i) the slope of a curve at $(x, y)$ is equal to $\frac{d y}{d x}$

(ii) For the solution of homogeneous differential equation $\frac{d y}{d x}=f\left(\frac{y}{x}\right)$ put $y=v x$ i.e. $\frac{d y}{d x}=v+x \frac{d v}{d x}$.

(A) We have $f'(x) > 0,\forall x \in (0,\pi /2)$. Therefore,

$f(0) < 0$ and $f(\pi /2) > 0$

Hence, there is no one solution.

(B) Let us consider that $(a,b,c)$ is direction ratio of the intersected line. Therefore,

$ak + 4b + c = 0$

$4a + kb + 2c = 0$

${a \over {8 - k}} = {b \over {4 - 2k}} = {c \over {{k^2} - 16}}$

We need to have

$2(8 - k) + 2(4 - 2k) + ({k^2} - 16) = 0$

$ \Rightarrow k = 2,4$.

(C) Let us consider

$f(x) = |x + 2| + |x + 1| + |x - 1| + |x - 2|$

Therefore, $k$ can take values: 2, 3, 4, 5.

(D) $\int {{{dy} \over {y + 1}} = \int {dx} } $

$ \Rightarrow f(x) = 2{e^x} - 1$

$ \Rightarrow f(\ln 2) = 3$

(A) We have ${(x - 3)^2}{{dy} \over {dx}} + y = 0$

$\int {{{dx} \over {{{(x - 3)}^2}}} = - \int {{{dy} \over y}} } $

$ \Rightarrow {1 \over {x - 3}} = \ln |y| + c$

So the domain is $R \to \{ 3\} $.

(B) On substituting $x = t + 3$, we get

$\int\limits_{ - 2}^2 {(t + 2)(t + 1)t(t - 1)(t - 2)dt = \int\limits_{ - 2}^2 {t({t^2} - 1)({t^2} - 4)dt = 0} } $ (being odd function)

(C) $f(x) = {5 \over 4} - {\left( {\sin x - {1 \over 2}} \right)^2}$

The maximum value occurs when $\sin x = 1/2$.

(D) $f'(x) > 0$ if $\cos x > \sin x$.

The given differential equation is

$x\sqrt {{x^2} - 1} dy - y\sqrt {{y^2} - 1} dx = 0$

$ = \int {{{dy} \over {y\sqrt {{y^2} - 1} }} = \int {{{dx} \over {x\sqrt {{x^2} - 1} }}} } $

$ \Rightarrow {\sec ^{ - 1}}y = {\sec ^{ - 1}}x + c$

$y = \sec [{\sec ^{ - 1}}x + C]$ [$\because$ $y(2) = {2 \over {\sqrt 3 }}$]

${2 \over {\sqrt 3 }} = \sec ({\sec ^{ - 1}}2 + C)$

$ = {\sec ^{ - 1}}{2 \over {\sqrt 3 }} - {\sec ^{ - 1}}2 = C$

$C = {\pi \over 6} - {\pi \over 3} = {{ - \pi } \over 6}$ [$\therefore$ $y = \sec \left[ {{{\sec }^{ - 1}}x - {\pi \over 6}} \right]$]

Statement 1 is true

Also, ${1 \over y} = \cos \left[ {\cos - {\pi \over x} - {\pi \over 6}} \right]$

$\cos \left( {{{\cos }^{ - 1}}{1 \over x}} \right)\cos {\pi \over 6} + \sin ({\cos ^{ - 1}}{1 \over x})\sin {\pi \over 6}$

$ = {1 \over y} = {{\sqrt 3 } \over {2x}} + {1 \over 2}\sqrt {1 - {1 \over {{x^2}}}} $

$\therefore$ Statement 2 is false.

Given, $ \frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{y}$

$\Rightarrow \int \frac{y}{\sqrt{1-y^{2}}} d y=\int d x$

$\Rightarrow -\sqrt{1-y^{2}}=x+c \Rightarrow(x+c)^{2}+y^{2}=1$

Here, centre $(-c, 0)$ and radius $=1$

To find $y''(0)$, we need to differentiate the given equation twice with respect to $x$ and then substitute $x=0$.

Differentiating the equation $x\cos \,y + y\,cos\,x = \pi $ with respect to $x$, we get:

$ \cos\,y - x\sin\,y \cdot y' + y'\cos\,x - y\sin\,x = 0$

Now, let's differentiate this equation again with respect to $x$:

$- \sin\,y \cdot y' - \sin\,y \cdot y' - x\cos\,y \cdot (y')^2 - x\sin\,y \cdot y'' + y''\cos\,x - y'\sin\,x - y'\sin\,x - y\cos\,x = 0$

Simplifying the equation, we get:

$-2\sin\,y \cdot y' - x\cos\,y \cdot (y')^2 - x\sin\,y \cdot y'' + y''\cos\,x - 2y'\sin\,x - y\cos\,x = 0$

Now, let's substitute $x = 0$ in the original equation and the first derivative equation. This will help us find the values of $y(0)$ and $y'(0)$ respectively:

From the original equation, $x\cos \,y + y\,cos\,x = \pi$, substituting $x = 0$, we get:

$y(0) \cdot 1 = \pi$

Therefore, $y(0) = \pi$.

Now, substituting $x = 0$ in the first derivative equation, we get:

$ \cos\,y(0) + y'(0) \cdot 1 = 0$

Substituting $y(0) = \pi$, we get:

$ \cos \pi + y'(0) = 0$

Therefore, $y'(0) = 1$.

Finally, let's substitute $x = 0$, $y(0) = \pi$, and $y'(0) = 1$ in the second derivative equation:

$-2\sin\pi \cdot 1 - 0 \cdot \cos \pi \cdot (1)^2 - 0 \cdot \sin \pi \cdot y''(0) + y''(0)\cos 0 - 2 \cdot 1 \cdot \sin 0 - \pi \cdot \cos 0 = 0$

Simplifying the equation, we get:

$y''(0) - \pi = 0$

Therefore, $y''(0) = \pi$.

Given, $ \frac{d y}{d x}+12 y=\cos \left(\frac{\pi}{12} x\right) $

This is a linear differential equation.

Where P = 12 and Q = $\cos \left(\frac{\pi}{12} x\right)$

$ \text { I.F. }=e^{\int P d x}=e^{\int 12 d x}=e^{12 x} $

Solution of differencial equation is

$y \times I.F = \int {Q \times \left( {I.F} \right)} dx$

$y \times \text { I.F. }=\int e^{12 x} \cdot \cos \left(\frac{\pi}{12} x\right) d x $

$ \Rightarrow y e^{12 x}=\frac{e^{12 x}}{12^2+\left(\frac{\pi}{12}\right)^2}\left(12 \cos \frac{\pi}{12} x+\frac{\pi}{12} \sin \frac{\pi}{12} x\right)+C $

[As $\int {{e^{ax}}.\cos \left( {bx} \right)dx} = {{{e^{ax}}} \over {{a^2} + {b^2}}}\left\{ {a\cos \left( {bx} \right) + b\sin \left( {bx} \right)} \right\} + C$]

$ \Rightarrow y=\frac{12}{(12)^4+\pi^2}\left\{(12)^2 \cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right)\right\} +\frac{C}{e^{12 x}} $

$ \begin{aligned} & \text { Given }, y(0)=0 \\\\ & \Rightarrow 0=\frac{12 \cdot\left(12^2\right)}{12^4+\pi^2}+C \\\\ & \Rightarrow C=\frac{-12^3}{12^4+\pi^2} \\\\ & \therefore y=\frac{12}{12^4+\pi^2}\left[12^2 \cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right)-12^2 \cdot e^{-12 x}\right] \end{aligned} $

$y(x)$ is not a periodic function as $e^{-12 x}$ is a non-periodic function.

$ \therefore $ Option (D) is a wrong statement.

$ \begin{aligned} & \text { Now } \frac{d y}{d x}=\frac{12}{12^4+\pi^2}[{-12 \pi \sin \left(\frac{\pi x}{12}\right)+\frac{\pi^2}{12} \cos \left(\frac{\pi x}{12}\right)}+12^3 \mathrm{e}^{-12 x}] \end{aligned} $

Values of ${-12 \pi \sin \left(\frac{\pi x}{12}\right)+\frac{\pi^2}{12} \cos \left(\frac{\pi x}{12}\right)}$ varies between $\left(-\sqrt{144 \pi^2+\frac{\pi^4}{144}},-12 \pi \sqrt{1+\frac{\pi^2}{12^4}}\right)$

So, $\mathrm{f}(\mathrm{x})$ is neither increasing nor decreasing.

$ \therefore $ Option (A) and (B) are wrong statements.

For some $\beta \in R, y=\beta$ intersects $y=f(x)$ at infinitely many points.

So option $\mathrm{C}$ is correct.

Let a point P(h, k) on the curve y = y(x), so equation of tangent to the curve at point P is

$y - k = {\left( {{{dy} \over {dx}}} \right)_{h,k}}(x - h)$ ....(i)

Now, the tangent (i) intersect the Y-axis at Y_p, so coordinates Y_p is $\left( {0,k - h{{dy} \over {dx}}} \right)$,

where ${{{dy} \over {dx}}}$ = ${\left( {{{dy} \over {dx}}} \right)}$_(h,k)

So, PY_p = 1 (given)

$ \Rightarrow \sqrt {{h^2} + {h^2}{{\left( {{{dy} \over {dx}}} \right)}^2}} = 1$

$ \Rightarrow {{dy} \over {dx}} = \pm {{\sqrt {1 - {x^2}} } \over x}$

[on replacing h by x]

$ \Rightarrow dy = \pm {{\sqrt {1 - {x^2}} } \over x}dx$

On puting x = sin$\theta $, dx = cos$\theta $d$\theta $, we get

$dy = \pm {{\sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}\cos \theta d\theta $

$ = \pm {{{{\cos }^2}\theta } \over {\sin \theta }}d\theta $

$ = \pm (\cos ec\theta - \sin \theta )d\theta $

$ \Rightarrow y = \pm [1n(\cos ec\theta - \cot \theta ) + \cos \theta ] + C$

$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right) + \cos \theta } \right] + C$

$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}} \right) + \sqrt {1 - {{\sin }^2}\theta } } \right] + C$

$ \Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}} } \right] + C$

[$ \because $ x = sin$\theta $]

$ = \pm \left[ { - 1n{{1 + \sqrt {1 - {x^2}} } \over x} + \sqrt {1 - {x^2}} } \right] + C$

[on rationalization]

$ \because $ The curve is in the first quadrant so y must be positive, so

$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} + C$

As curve passes through (1, 0), so

$0 = 0 - 0 + c \Rightarrow c = 0$, so required curve is

$y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} $

and required differential eaquation is

${{dy} \over {dx}} = - {{\sqrt {1 - {x^2}} } \over x}$

$ \Rightarrow xy' + \sqrt {1 - {x^2}} = 0$

Hence, options (a) and (c) are correct.

$g(x) = \int_{\sin x}^{\sin (2x)} {{{\sin }^{ - 1}}} (t)\,dt$

g'(x) = 2cos 2x sin^$-$1(sin2 x) $-$ cos x sin^$-$1(sin x)

$g'\left( {{\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$

$g'\left( { - {\pi \over 2}} \right) = - 2{\sin ^{ - 1}}(0) = 0$

No option is matching.

Options (A) and (B): The given differential equation is

$ \left[x^2+4 x+4+y(x+2)\right] \frac{d y}{d x}-y^2=0 \quad(x>0) $

which is further simplified as follows:

$ \left[(x+2)^2+y(x+2)\right] \frac{d y}{d x}-y^2=0 $

Substituting $x+2=t$, we get

$ \frac{d x}{d y}=\frac{d t}{d y} $

Now,

$ (x+2)^2+y(x+2)-y^2 \frac{d x}{d y}=0 $

That is,

$ \begin{aligned} &t^2+y t-y^2 \frac{d t}{d y} =0 \\\\ &y^2 \frac{d t}{d y}-y t-t^2 =0 \\\\ &\frac{1}{t^2} \frac{d t}{d y}-\frac{1}{y t} =\frac{1}{y^2} \end{aligned} $

Let $\frac{1}{t}=z$; therefore,

$ \frac{d t}{d y}\left(-\frac{1}{t^2}\right)=\frac{d z}{d y} $

Now,

$\begin{aligned} \frac{-d z}{d y}-\frac{z}{y} & =\frac{1}{y^2} \Rightarrow \frac{d z}{d y}+\frac{z}{y}=\frac{-1}{y^2} \\\\ \Rightarrow d(z y) & =\int-\frac{1}{y} d y \\\\ \Rightarrow z y & =-\ln |y|+\ln c \\\\ \Rightarrow \frac{y}{t} & =-\ln |y|+\ln c\end{aligned}$

$ \Rightarrow \frac{y}{(x+2)}=-\ln |y|+\ln c ~~~~~...(1) $

which passes through the point $(1,3)$. Therefore, from Eq. (1), we get

$ \begin{aligned} &\frac{z}{3} =-\ln 3+c \Rightarrow c=\ln 3 e \\\\ &\frac{y}{x+2} =-\ln |y|+\ln 3 e=\ln \left(\frac{3 e}{|y|}\right) \\\\ &\frac{3 e}{|y|} =e^{y /(x+2)} \\\\ &3 e =|y| e^{y /(x+2)} \end{aligned} $

Substituting $y=(x+2)$, we get $3 e=|x+2| e^1$

$ |x+2|=3 \Rightarrow x+2=-3,3 \Rightarrow x=-5,1 $

Therefore, $x=1$ (since $x \neq-5)$.


That is, the solution curve intersects $y=(x+2)$ exactly at one point and not at two points. Therefore, option (A) is correct and option (B) is incorrect.

Option (C): We have

$ \frac{3 e}{\left|(x+2)^2\right|}=e^{(x+2)} $

which meets at two points for $x<0$ and for $x>0$, there is no intersection point.

Hence, option $(\mathrm{C})$ is incorrect.

Option (D): We have

$ \frac{3 e}{(x+3)^2}=e^{\frac{(x+3)^2}{(x+2)}}=e^{\frac{(x+2)^2+1+2(x+2)}{(x+2)}}=e^{2+\frac{1}{(x+2)}+(x+2)} $

Therefore, there is no intersection point for $x>0$. Hence option (D) is correct.

$f'(x) = 2 - {{f(x)} \over x}$

$ \Rightarrow f'(x) = {1 \over x}f(x) = 2$ is linear differential equation.

Hence, $I.F. = {e^{\int {{1 \over x}dx} }} = {e^{\ln x}} = x$

Thus the solution is given by

$x\,.\,f(x) = \int {2x\,dx + \lambda } $

i.e., $xf(x) = {x^2} + \lambda $

As f(1) $\ne$ 1, we have $\lambda$ $\ne$ 0

$\therefore$ $f(x) = x + {\lambda \over x}$, $\lambda$ $\ne$ 0

Thus, $f'(x) = 1 - {\lambda \over {{x^2}}}$, $\lambda$ $\ne$ 0

Now, $\mathop {\lim }\limits_{x \to {0^ + }} f'\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 - \lambda {x^2}) = 1$

$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} x\left( {{1 \over x} + \lambda x} \right) = \mathop {\lim }\limits_{x \to {0^ + }} (1 + \lambda {x^2}) = 1$

$\mathop {\lim }\limits_{x \to {0^ + }} {x^2}f'(x) = \mathop {\lim }\limits_{x \to {0^ + }} {x^2}\left( {1 - {\lambda \over {{x^2}}}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} ({x^2} - \lambda ) = - \lambda $

Again, $\mathop {\lim }\limits_{x \to {0^ + }} f(x) \to \infty $

Hence the function is not bounded.

Note that $\lambda$ can be $-$ve or +ve.

Given: $\left(1+e^x\right) y^{\prime}+y e^x=1$

$\left(1+e^x\right) \frac{d y}{d x}+y e^x=1$

$\frac{d y}{d x}+\frac{e^x}{1+e^x} y=\frac{1}{1+e^x}$, which is liner differential equation in $y$.

Comparing above equation with $\frac{d y}{d x}+\mathrm{P} y=Q$, we get $Q$, we get

$\begin{aligned} \quad \mathrm{P} & =\frac{e^x}{1+e^x} \text { and } \mathrm{Q}=\frac{1}{1+e^x} \\ \text { So, I.F. } & =e^{\int \mathrm{P} \cdot d x} \\ \text { I.F. } & =e^{\int \frac{e^x}{1+e^x} d x} \\ \text { I.F. } & =e^{\ln \left(1+e^x\right)} \end{aligned}$

$\left\{\because \int \frac{f^{\prime}(x)}{f(x)} d x=\ln f(x)\right\}$

$\text { I.F }=1+e^x$

So, solution of given differential equation is given by

$y \text { (I.F.) }=\int \text { Q.(I.F.) } d x$

$\begin{aligned} y \cdot\left(1+e^x\right) & =\int \frac{1}{1+e^x} \cdot\left(1+e^x\right) d x \\ y\left(1+e^x\right) & =\int 1 d x \\ y\left(1+e^x\right) & =x+\mathrm{C} \\ \because \quad y(0) & =2 \\ \Rightarrow \quad 2\left(1+e^0\right) & =0+\mathrm{C} \\ c & =4 \end{aligned}$

So, $y(x)=\frac{x+4}{1+e^x}\quad \text{... (i)}$

Put $x=-4$ in the equation, we get

$y(-4)=0$

Put $x=-2$ in the above equation (i), we get

$y(-2)=\frac{2}{1+e^{-2}} \neq 0$

For critical points, $y^{\prime}=0$

From e.q, (i), $y\left(1+e^x\right)=x+4$

Differentiating the above equation w.r.t. $x$, we get

$\begin{aligned} y^{\prime}\left(1+e^x\right)+y\left(e^x\right) & =1 \\ 0\left(1+e^x\right)+y e^x & =1 \quad \left\{\because y^{\prime}=0\right\} \\ y e^x-1 & =0 \end{aligned}$

$\quad \begin{aligned} \text { Now, } \text { let } g(x) & =y e^x-1 \\ g(x) & =\frac{(x+4) e^x}{1+e^x}-1 \\ g(x) & =\frac{(x+3) e^x-1}{1+e^x} \end{aligned}$

$\text { Now, } \quad g(-1)=\frac{2 e^{-1}-1}{1+e^{-1}}=\frac{2-e}{1+e}<0 \quad \{\because e=2 \cdot 7 \cdot 8\}$

$\text { And, } \quad g(0)=\frac{3 e^0-1}{1+e^0}=1>0$

So, there exists one value of $x$ in $(-1,0)$ for which

$\begin{aligned} g(x) & =0 \\ \Rightarrow \quad y^{\prime} & =0 \end{aligned}$

There exist a critical point of $y(x)$ in $(-1,0)$

Hint :

(i) Use solution of liner differential equation

$\begin{aligned} \frac{d y}{d x}+\mathrm{P} y & =\mathrm{Q} \text { is given by } \\ y \text { (I.F.) } & =\int \mathrm{Q} \cdot \text { (I.F) } d x, \text { where I.F }=e^{\int p \cdot d x} \end{aligned}$

(ii) For critical points, $y^{\prime}=0$

Let equation of circle whose centre lie on straight line $y=x$ be

$(x-k)^2+(y-k)^2=r^2~~~~...(i)$

Differentiating the above equation w.r.t. $x$, we get

$ \begin{aligned} & 2(x-k)+2(y-k) y^{\prime}=0 \\\\ \Rightarrow & x+y y^{\prime}=k\left(1+y^{\prime}\right)~~~~~...(ii) \\\\ \Rightarrow & k=\frac{x+y y^{\prime}}{1+y^{\prime}} \end{aligned} $

Differentiating the eq. (ii) w.r.t. $x$, we get

$ \begin{aligned} & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=k y^{\prime \prime} \quad\left\{\because(u v)^{\prime}=u v^{\prime}+v u^{\prime}\right\} \\\\ & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=\left(\frac{x+y y^{\prime}}{1+y^{\prime}}\right) y^{\prime \prime} \\\\ & \Rightarrow 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2+y^{\prime}+y y^{\prime} y^{\prime \prime}+\left(y^{\prime}\right)^3 \\\\ & =x y^{\prime \prime}+y y^{\prime} y^{\prime \prime} \\\\ & \Rightarrow y^{\prime \prime}(y-x)+\left(y^{\prime}\right)^2\left(1+y^{\prime}\right)+1+y^{\prime}=0 \\\\ & \Rightarrow y^{\prime \prime}(y-x)+y^{\prime}\left(y^{\prime}+\left(y^{\prime}\right)^2+1\right)+1=0 \end{aligned} $

Comparing the above equation with $p y^{\prime \prime}+\mathrm{Q} y^{\prime}$ $+1=0$, we get

$ \begin{aligned} & P=y-x \text { and } \mathrm{Q}=y^{\prime}+\left(y^{\prime}\right)^2+1 \\\\ \therefore & P+Q=1-x+y+y^{\prime}+\left(y^{\prime}\right)^2 \end{aligned} $

Given, $\frac{d y}{d x}-y \tan x=2 x \sec x$

Comparing with the liner differential form

$\begin{gathered} \frac{d y}{d x}+\mathrm{P} y=\mathrm{Q} \\ \Rightarrow \mathrm{P}=-\tan x \text { and } \mathrm{Q}=2 x \sec x \end{gathered}$

Now, Integrating factor (I.F) $=e^{\int P d x}$

$\begin{aligned} & =e^{\int-\tan x d x} \\ & =e^{-\log \sec x} \\ & =\cos x \end{aligned}$

$\begin{aligned} & \text { The solution will be } y \text { (I.F) }=\int Q \text { (I.F) } d x \\ & \Rightarrow \quad y \cos x=\int 2 x \sec x \cdot \cos x d x \\ & \Rightarrow \quad y \cos x=2 \frac{x^2}{2}+c \\ & \Rightarrow \quad y=x^2 \sec x+c \sec x \\ \end{aligned}$

Now, $y(0)=0$

$\begin{aligned} & \Rightarrow \quad 0=0^2 \sec 0+\mathrm{c} \mathrm{sec} 0 \\ & \Rightarrow \quad c=0 \\ & \therefore \quad y=x^2 \sec x \\ & y^{\prime}=x^2 \sec x \cdot \tan x+2 x \sec x \quad \text{[using Product Rule in Differentiation]}\\ \end{aligned}$

Now,

$\begin{aligned} y\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}\right)^2 \sec \frac{\pi}{4}=\frac{\pi^2}{16} \sqrt{2}=\frac{\pi^2}{8 \sqrt{2}} \\ y^{\prime}\left(\frac{\pi}{4}\right) & =\left(\frac{\pi}{4}\right)^2 \sec \frac{\pi}{4} \tan \frac{\pi}{4}+2 \cdot \frac{\pi}{4} \sec \frac{\pi}{4} \\ & =\frac{\pi^2}{16} \cdot \sqrt{2} \cdot 1+\frac{\pi}{2} \cdot \sqrt{2} \\ & =\frac{\pi^2}{8 \sqrt{2}}+\frac{\pi}{\sqrt{2}} \end{aligned}$

$\begin{aligned} y\left(\frac{\pi}{3}\right) & =\left(\frac{\pi}{3}\right)^2 \sec \frac{\pi}{3} \\ & =\frac{\pi^2}{9} \cdot 2=\frac{2 \pi^2}{9} \\ y^{\prime}\left(\frac{\pi}{3}\right) & =\left(\frac{\pi}{3}\right)^2 \sec \frac{\pi}{3} \cdot \tan \frac{\pi}{3}+2 \cdot \frac{\pi}{3} \cdot \sec \frac{\pi}{3} \\ & =\frac{\pi^2}{9} \cdot 2 \cdot \sqrt{3}+\frac{2 \pi}{3} \cdot 2 \\ & =\frac{2 \pi^2}{3 \sqrt{3}}+\frac{4 \pi}{3} \end{aligned}$