Differential Equations

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