y = log(x²+1)+c

\({ I }_{ n }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { n-1 }{ n } { I }_{ n-2 },n\ge 2 } \)
\(\therefore { I }_{ 7 }=\int _{ 0 }^{ \pi /2 }{ { cos }^{ 7 }xdx=\frac { 6 }{ 7 } \times \frac { 4 }{ 5 } \times \frac { 2 }{ 3 } \times 1 } =\frac { 16 }{ 35 } \)

\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } \)
\(\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-4 } } =\int _{ 3 }^{ 4 }{ \frac { dx }{ { x }^{ 2 }-{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| +c\right] \)
\(=\frac { 1 }{ 4 } \left[ log\left( \frac { 4-2 }{ 4+2 } \right) -log\left( \frac { 3-2 }{ 3+2 } \right) \right] \)
\(=\frac { 1 }{ 4 } log\left[ \left( \frac { 2 }{ 6 } \right) - log \ \frac { 1 }{ 5 } \right] \\ =\frac { 1 }{ 4 } log\left( \frac { 1 }{ 3 } \times 5 \right) \)
\(=\frac { 1 }{ 4 } log\left( \frac { 5 }{ 3 } \right) \)

Volume of sphere = \(\frac43\) πr³
Given r = 5 mm
⇒ dr = (5.3 - 5) = 0.3 mm
\(\text { Approximate volume }=\frac{4}{\not 3} \pi \cdot \not 3 r^{2} d r\)
= 4π (5²) (0.3)
= 100 π (0.3)
= 30π mm³

Let f (x) = x cos x
Then f (−x) = (−x) cos(−x) = −x cos x = − f (x).
So f (x) = x cos x is an odd function.
Hence, applying the property, for odd function f(x), \(\int _{ -a }^{ a }{ f(x)dx=0 } \)
\(\therefore\) we get \(\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ xcosx } dx=0\)

x = 3 and dx = 0.02
When x = 3 and dx = 0.02,
df = (6 + 3) (0.02)
= 9(0.02) = 0.18

Given \(f(x)=\frac { { x }^{ 2 }-6x-1 }{ x+3 } \)
\(\underset { x\rightarrow -3^{ + } }{ lim } \frac { { x }^{ 2 }-6x-1 }{ x+3 } =\underset { h\rightarrow { 0 }^{ + } }{ lim } \frac { \left( -3+h \right) ^{ 2 }-6(-3+h)-1 }{ -3+h+3 } \) 
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { 9+{ h }^{ 2 }-6h+18-6h-1 }{ h } \)
= \(\underset { h\rightarrow 0^{ + } }{ lim } \frac { { h }^{ 2 }-12h+26 }{ h } =\infty \)
and \(\underset { x\rightarrow 3^{ - } }{ lim } \frac { { x }^{ 2 }-6x-1 }{ x+3 } =\underset { h\rightarrow { o }^{ + } }{ lim } \frac { \left( -3-h \right) ^{ 2 }-6(-3-h)-1 }{ -3-h+3 } \)
= -∞
ஃx = -3 is the vertical asymptote,
Also

ஃ y = x - 9 is the slanting asymptote,

Given \(f(x)=x-2logx, x\in [2,7]\)
(i) f(x) is continuous in [2, 7]
(ii) f(x) is differentiable in (2, 7)
f(2) = 2 - 2 log 2
= 2 - log 2² = 2 - log 4
f (7) = 7 - 2 log 7
= 7 - log 7² = 7 - log 49
Since f(2) ≠ f (7), Rolle's theorem is not applicable.

Given = e^mx is the solution of
y' + 2y = 0 ...(1)
y = e^mx  ...... (2) 
\(\frac{dy}{dx} = e^{mx}. m\)
\(\frac{dy}{dx} = ym\)
\(\frac{dy}{dx} - my=0\)
⇒ y' - my = 0 ...(3)
Comparing equation (1) & (3),
we get m = -2

\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +\int { ydx } ={ x }^{ 3 }\)
The given differential equation is
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { dy }{ dx } +\int { ydx } ={ x }^{ 3 }.\)
Differentiating again with respect to 'x' we get.
\(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +5\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } +y=3{ x }^{ 2 }\)
The highest derivative is 3 and its power is 1.
∴ Order is 3 and degree is 1.

\( \lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}} \)         \(\left[1^{\infty}\right. \) form
\(\text {Let } y =(\cos x)^{\frac{1}{x}} \)
\(\log y =\frac{1}{x} \log \cos x \)
\(\lim _{x \rightarrow 0} \log y =\lim _{x \rightarrow 0} \frac{\log \cos x}{x}\)       \(\left[\frac{0}{0}\right. \) form
Applying L' Hopital's rule
\( =\lim _{x \rightarrow 0}-\frac{\sin x}{\frac{\cos x}{1}} \)
\( =\lim _{x \rightarrow 0}-\tan x=0 \)
\(\therefore \log \left(\lim _{x \rightarrow 0} y\right) =0 \)
\(\lim _{x \rightarrow 0} y =e^{0}=1 \)
\(ie., \lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=1\)

u = sin 3x cos 4y
\( \frac{\partial u}{\partial x} =3 \cos 3 \mathrm{x}\ \cos 4 \mathrm{y} \)
\(\frac{\partial^{2} u}{\partial y \partial x} =-12 \cos 3 \mathrm{x} \ \sin 4 \mathrm{y} \)..........(1)
\(\frac{\partial u}{\partial y} =-4 \sin 3 \mathrm{x} \sin 4 \mathrm{y} \) 
\(\frac{\partial^{2} u}{\partial x \partial y} =-12 \cos 3 \mathrm{x} \sin 4 \mathrm{y}\) ......(2)
From (1) and (2),
\(\frac{\partial^{2} u}{\partial x \partial y}=\frac{\partial^{2} u}{\partial y \partial x}\)

First let us find w_x, w_y, and w_z
Now w_x = 2xy + z², w_y = 2yz +x² and w_z = 2zx + y².
Thus,by (15), the differential is
dw = (2xy + z² )dx + (2yz + x² )dy+ (2zx + y² )dz.

\(\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } } \)
\(=\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+1+4 } } =\int _{ -1 }^{ 1 }{ \frac { dx }{ { (x+1) }^{ 2 }{ 2 }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } } =\frac { 1 }{ 2a } log\left| \frac { x-a }{ x+a } \right| \right] \)
\(={ \left[ \frac { 1 }{ 2 } { tan }^{ -1 }\left( \frac { x+1 }{ 2 } \right) \right] }_{ -1 }^{ 1 }\)
\(=\frac { 1 }{ 2 } \left[ { tan }^{ -1 }(1)-{ tan }^{ -1 }(0) \right] \)
\(\\ =\frac { 1 }{ 2 } \left[ \frac { \pi }{ 4 } \right] =\frac { \pi }{ 8 } \)

Let I = \(\int _{ 0 }^{ \frac { \pi }{ 3 } }{ \frac { sec\ x\ tan\ x }{ 1+{ sec }^{ 2 }x } dx } \)
Put sec x = u. Then, sec x tan x dx = du.
When x = 0, u = sec0 = 1. When x = \(\frac{\pi}{3}, u=sec\frac{\pi}{2}=2\)
\(\therefore I=\int _{ 1 }^{ 2 }{ \frac { du }{ 1+{ u }^{ 2 } } ={ [{ tan }^{ -1 }u] }_{ 1 }^{ 2 }={ tan }^{ -1 }1={ tan }^{ -1 } } (2)-\frac { \pi }{ 4 } \)

Let f(x) = \(f(x)={ x }^{ \frac { 2 }{ 3 } },{ x }_{ 0 }=125,\triangle x=-2\)
∴ (123)^\(\frac23\) = f(125) +1'(125) (-2) ... (1)
\(f(125)={ (125) }^{ \frac { 2 }{ 3 } }={ { (5 }^{ 3 }) }^{ \frac { 2 }{ 3 } }\) = 5² = 25
\({ f }^{ ' }(x)={ \frac { 2 }{ 3 } x }^{ \frac { 2 }{ 3 } -1 }={ \frac { 2 }{ 3 } x }^{ \frac { 1 }{ 3 } }=\frac { 2 }{ { 3x }^{ \frac { 1 }{ 3 } } } \)
\({ f }^{ ' }(125)=\frac { 2 }{ { 3(125)x }^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ { 3{ (5 }^{ 3 } })^{ \frac { 1 }{ 3 } } } =\frac { 2 }{ 3(5) } =\frac { 2 }{ 15 } \)
∴ \({ (123) }^{ \frac { 2 }{ 3 } }=25+\frac { 2 }{ 15 } (-2)\)
\(=25-\frac { 4 }{ 15 } =25-0.27\)
\({ (123) }^{ \frac { 2 }{ 3 } }=24.73\)

Function and its derivatives	log (1-x) cos x and its derivatives	Value at x = 0
f(x)	log (1-x)	log 1 = 0
fI(x)	\(\frac{-1}{1-x}\) = -1(1 - x)-1	\(\frac{-1}{1}\) = -1
fIl(x)	-1(1-x)-2	-1
fIIl(x)	-2(1-x)-3	-2
fIV(x)	-6 (1 - x)-4	-6
fV(x)	-24 (1 - x)-5	-24

Meclaurin's expansion
\(f(x)=f(0)+\frac { { f }^{ 1 }(0) }{ 1! } x+\frac { { f }^{ II }(0) }{ 2! } { x }^{ 2 }+\frac { { f }^{ III }(0) }{ 3! } { x }^{ 3 }+\).................
log(1-x) = \(0-\frac { 1 }{ 1! } x-\frac { { x }^{ 2 } }{ 2! }- \frac { 2 }{ 3! } { x }^{ 3 }-\frac { { 6x }^{ 4 } }{ 4! } +.....\)
= \(-x-\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } \)+ .....
log(1 - x) = \(-\left( x+\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 4 } }{ 4 } +.... \right) \)

Given that \((1+{ x }^{ 2 })\frac { dy }{ dx } =1+{ y }^{ 2 }\) ..(1)
The given equation is written in the variables separable form 
\(\frac { dy }{ { 1+y }^{ 2 } } =\frac { dx }{ { 1+x }^{ x } } \) ...(2)
Integrating both sides of (2), we get tan⁻¹ tan⁻¹x +C.
But tan^-1 y - tan^-1 x = tan^-1 \(\left( \frac { y-x }{ 1+xy } \right) .\) ...(4)
Using (4) in (3) leads to tan^-1 \(\left( \frac { y-x }{ 1+xy } \right)\) = C, which implies \(\frac { y-x }{ 1+xy } \) = tan C = a (say).
Thus, y − x = a(1+ xy) gives the required solution

Given curve is y = x² − 5x + 4 and the line is 3x + y = 7
Slope of the tangent to the curve
\({ m }_{ 1 }=\frac { dx }{ dt } \) = 2x - 5
Slope of the line = \({ m }_{ 2}=\frac { dx }{ dt } \) = -3
\(\left[ \because m=\frac { co-efficient \ of \ x }{ co-efficient \ of \ y } \right] \)
Since the tangent of the curve and the lines are parallel, their slopes are equal.
∴ m₁ = m₂
⇒ 2x - 5 = -3
⇒ 2x = 2
⇒ x = 1
Substituting x = 1 in y = x² - 5x + 4 we get
y = 12-5(1)+4 = 0
∴ The required point is (1, 0).

The equation of the family of ellipses having centre at the origin & foci on the y-axis, is given
\(\frac { { x }^{ 2 } }{ { b }^{ 2 } } +\frac { { y }^{ 2 } }{ { a }^{ 2 } } =1\) ...(1)
where b >a & a, b are the parameters or a,b are arbitrary constant.
Differentiating equation (1) twice successively, because we have two arbitrary constant) we get
\( \frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \frac{d y}{d x} =0 \)
\(2\left(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}\right) =0 \)
\(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x} =0\) ............(2)
Again differentiating equation (2)
\(\frac{1}{a^{2}}+\frac{y}{b^{2}} \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x} \frac{d y}{d x b^{2}}=0\)
\(\frac{1}{a^{2}}+\frac{y}{b^{2}} \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2} \frac{1}{b^{2}}=0\)
multiply by x
\(\frac{x}{a^{2}}+\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2} \frac{x}{b^{2}}=0\) .........(3) 
Equation (3)-(2)
\( \frac{x}{a^{2}}+\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2}\left(\frac{x}{b^{2}}\right) -\left(\frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}\right) =0 \)
\(\frac{x y}{b^{2}}\left(\frac{d^{2} y}{d x^{2}}\right)+\left(\frac{d y}{d x}\right)^{2} \frac{x}{b^{2}}-\frac{y}{b^{2}} \frac{d y}{d x} =0 \)
Taking \(\frac{1}{b^{2}}\) outside, we get
\( \frac{1}{b^{2}}\left[x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}\right]=0 \\ x y \frac{d^{2} y}{d x^{2}}+x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}=0 \)
is the required differential equation.

Note that the function u is not homogeneous. So we cannot apply Euler’s Theorem for u.
However, note that f(x,y) = \(\frac { x+y }{ \sqrt { x } +\sqrt { y } }\) = sin u is homogeneous; because
f(tx,ty) = \(\frac { tx+ty }{ \sqrt { tx } +\sqrt { ty } } \) = t^1/2 f(x, y), \(\forall \) x, y, t\(\ge \)0
Thus f is homogeneous with degree \(\frac { 1 }{ 2 } \) and so by Euler’s Theorem we have
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =\frac { 1 }{ 2 } f(x,y)\).
Now substituting f = sin u in the above equation, we obtain
\(x\frac { \partial (sinu) }{ \partial x } +y\frac { \partial (sinu) }{ \partial y } =\frac { 1 }{ 2 } sin \ u\)
\(x\quad cosu\frac { \partial u }{ \partial x } +y\quad cosu\frac { \partial u }{ \partial x } =\frac { 1 }{ 2 } sin \ u\)  ...(19)
Dividing both sides by cosu we obtain
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 1 }{ 2 } tan \ u\)
Note:
Solving this problem by direct calculation will be possible; but will involve lengthy calculations.

The ellipse is symmetric about both the axes. The major axis lies along x-axis. The region to be revolved is sketched
Hence, the required volume is given by
\(V=\pi \int _{ -a }^{ a }{ { y }^{ 2 }dx } =\pi \int _{ -a }^{ a }{ \frac { { b }^{ 2 } }{ { a }^{ 2 } } \left( { a }^{ 2 }-{ x }^{ 2 } \right) dx } \)
\(=\frac { 2\pi { b }^{ 2 } }{ { a }^{ 2 } } \int _{ 0 }^{ a }{ ({ a }^{ 2 }-{ x }^{ 2 })dx } \) since the integrand is an even function
\(=\frac { { 2\pi b }^{ 2 } }{ { a }^{ 2 } } { \left( { a }^{ 2 }x-\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=\frac { 2\pi { b }^{ 2 } }{ 3 } ={ \left( { a }^{ 2 }-\frac { { a }^{ 3 } }{ 3 } \right) }_{ 0 }^{ a }=\frac { 2\pi { b }^{ 2 } }{ 3 } \left( \frac { { 2a }^{ 3 } }{ 3 } \right) =\frac { 4\pi { ab }^{ 2 } }{ 3 } \)

Given U (x, y) = e^x sin y ; x = st² ; y = s²t
\(\frac { \partial U }{ \partial x } \) = e^x sin y ; \(\frac { \partial U }{ \partial y } \) = e^x cos y
\(\frac { \partial U }{ \partial x } \) = \({ e }^{ { st }^{ 2 } }\) sin (s²t)
\(\frac { \partial U }{ \partial y } \) = \({ e }^{ { st }^{ 2 } }\) cos (s²t)
\(\frac{dx}{dt}\) = 2st; \(\frac{dy}{dt}\) = s²
\(\frac{dx}{ds}\) = t²; \(\frac{dy}{ds}\) = 2 st
By chain rule
\(\frac { dU }{ ds } =\frac { \partial U }{ \partial x } .\frac { dx }{ ds } +\frac { \partial U }{ \partial y } .\frac { dy }{ ds } \)
= \({ e }^{ { st }^{ 2 } }\). sin (s²t) (t²) + \({ e }^{ { st }^{ 2 } }\) cos(s^2t).(2^st)
∴ \({ \left( \frac { \partial U }{ \partial s } \right) }_{ (s=t=1) }\) = e¹ sin (1) + 2e¹ cos (1)
= e [sin (1) + 2 cos (1)] and
\(\frac { dU }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } \)
=  \({ e }^{ { st }^{ 2 } }\) . sin (s²t)(2^st) + \({ e }^{ { st }^{ 2 } }\) cos (s²t). (s²)
∴ \({ \left( \frac { \partial U }{ \partial t } \right) }_{ (s=t=1) }\) = 2e¹ sin (1) + e¹ cos (1)
= e [2 sin (1) + cos (1)]

The region is sketched. The upper boundary of the region is y = sin x for \(\\ \\ \frac { \pi }{ 4 } \le x\le \frac { 5\pi }{ 4 } \) and the lower boundary of the region is y = cos x for \(\frac { \pi }{ 4 } \le x\le \frac { 5\pi }{ 4 } \). So the required area A is given by

\(A=\int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ ({ y }_{ U }-{ y }_{ L })dx= } \int _{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } }{ (sinx-cosx)dx={ [-cosx=sinx] }_{ \frac { \pi }{ 4 } }^{ \frac { 5\pi }{ 4 } } } \)
\(=\left( -sin\frac { 5\pi }{ 4 } -cos\frac { 5\pi }{ 4 } \right) -\left( -sin\frac { \pi }{ 4 } -cos\frac { \pi }{ 4 } \right) \)
\(=\left( -\left( -\frac { 1 }{ \sqrt { 2 } } \right) -\left( -\frac { 1 }{ \sqrt { 2 } } \right) -\left( -\left( \frac { 1 }{ \sqrt { 2 } } \right) -\left( \frac { 1 }{ \sqrt { 2 } } \right) \right) \right) \)
\(=\frac { 2 }{ \sqrt { 2 } } +\frac { 2 }{ \sqrt { 2 } } =2\sqrt { 2 } \)

First, we get the points of intersection of the parabolas. For this, we solve y² x = 4 and x² y = 4 simultaneously Eliminating y between them, we get x⁴ = 64x and so x = 0 and x = 4. Then the points of intersection are (0, 0) and (4, 4). The required region is sketched.

Viewing in the direction of y -axis, the equation of the upper boundary is y = 2\(\sqrt x\) for 0\(\le x \le\) 4 and the equation of the lower boundary is \(y =\frac {x^2}{4}\)for \(0 \leq x \leq 4\). So, the required area \(\Delta\) is
\(A=\int_{0}^{4}\left(y_{U}-Y_{L}\right) d x=\int_{0}^{4}\left(2 \sqrt{x}-\frac{x^{2}}{4}\right) d x=\left[2\left(\frac{2 x^{3 / 2}}{3}\right)-\frac{x^{3}}{12}\right]_{0}^{4}=\left[2\left(\frac{2 \times 8}{3}\right)-\frac{64}{12}\right]-0=\frac{16}{3}\)

Put u = tan \(\frac{x}{2}\) 
Then, sin x = \(\frac{2 tan \frac {x}{2}}{1 + tan^2 \frac{x}{2}}\) = \(\frac{2u}{1+u^2}\), du = \(\frac{1}{2}\) sec² \(\frac{x}{2}\) dx  \(\Rightarrow \) dx = \(\frac {2du}{1+u^2}\)
When x = 0,u = tan 0 = 0
When x = \(\frac{\pi}{2},u=tan\frac{\pi}{4}=1\)
∴ I =\(\int ^\frac{\pi}{2}_0\) \(\frac{dx}{4+5 sin x}\) =\(\int ^{1}_0\) \(\frac{\frac {2du}{1+u^2}}{4+5 (\frac{2u}{1+u^2})}\) =\(\int ^{1}_0\) \(\frac {du}{2u^2+5u +2}\)=\(\frac{1}{2}\)\(\int ^{1}_{0}\) \(\frac {du}{u^2+\frac{5}{2}u +1}\)
\(\frac { 1 }{ 2 } \int _{ 0 }^{ 1 } \frac { du }{ (u+\frac { 5 }{ 4 } )^2-(\frac { 3 }{ 4 } )^{ 2 } } { \left[ \frac { 1 }{ 2 } \times \frac { 1 }{ 2\times \left( \frac { 3 }{ 4 } \right) } log\left( \frac { (u+\frac { 5 }{ 4 } )-\frac { 3 }{ 4 } }{ (u+\frac { 5 }{ 4 } )+\frac { 3 }{ 4 } } \right) \right] }_{ 0 }^{ 1 } = \frac{1}{3} [log (\frac{u+\frac{1}{2}}{u+2})] =\frac { 1 }{ 3 } log2\)

Let P be denote the population of a city
Given that \(\frac{dP}{dt}\infty\)
\(\Rightarrow P=\frac{dP}{dt}kP\)
\(\Rightarrow \frac{dP}{P}=kdt\)
\(\Rightarrow \int { \frac { dP }{ P } = } k\int { dt } \)
\(\Rightarrow logP=kt+logc\)
\(\Rightarrow log\left( \frac { P }{ c } \right) =kt\)
\(\Rightarrow \frac { P }{ c } ={ e }^{ kt }\)
\(_{ }^{ c }{ P }={ c.e }^{ kt } ..(1)\)
Given when t = 0, P = 3,00,000
\(\therefore(1)\rightarrow\) = ce⁰ \(\Rightarrow\) c = 3,00,000
\(\therefore\) P = 3,00,000 e^kt ...(2)
Again when t = 40, P = 4,00,000
\(\therefore\) (2) \(\Rightarrow\) 4,00,000 = 3,00,000e^40k
\(\Rightarrow \frac { 4 }{ 3 } ={ e }^{ 40k }\)
\(\Rightarrow log\left( \frac { 4 }{ 3 } \right) =40K\)
\(\Rightarrow K=\frac { 1 }{ 40 } log\left( \frac { 4 }{ 3 } \right) \)
\(\Rightarrow k=log{ \left( \frac { 4 }{ 3 } \right) }^{ \frac { 1 }{ 40 } }...(3)\)
\(\therefore\) (2) becomes, P = 3,00,000\({ e }^{ log{ \left( \frac { 4 }{ 3 } \right) }^{ \frac { 1 }{ 40 } t } }\)
\(\Rightarrow\) P = 3,00,000 \(^{ { \left( \frac { 4 }{ 3 } \right) }^{ \frac { 1 }{ 40 } } }\)

\((x+a)\frac { dy }{ dx } -2y={ (x+a) }^{ 4 }\)
\(\div (x+a)\)we get,
\(\frac { dy }{ dx } -\frac { 2 }{ x+a } y={ (x+a) }^{ 3 }\)
This is a linear differential equation
\(\therefore P=\frac { -2 }{ x+a } ;Q={ (x+a) }^{ 3 }\)
\(\int { pdx } =\int { \frac { -2 }{ x+a } } dx=-2log(x+a)\)
\(=log{ \left( \frac { 1 }{ x+a } \right) }^{ 2 }\)
\(\therefore I.F.={ e }^{ \int { pdx } }={ e }^{ log{ \left( \frac { 1 }{ x+a } \right) }^{ 2 } }=\frac { 1 }{ { (x+a) }^{ 2 } } \)
The solution is \(\Rightarrow { ye }^{ \int { pdx } }=\int { Q{ e }^{ \int { pdx } }dx+c } \)
\(\Rightarrow y\left( \frac { 1 }{ { (x+a) }^{ 2 } } \right) =\int { { (x+a) }^{ 3 }\frac { 1 }{ (x+a{ ) }^{ 2 } } dx+c } \)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\int { (x+a)dx+c } \)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\int { (x+a)dx+c } \)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\frac { { (x+a) }^{ 2 } }{ 2 } +c\)
\(\Rightarrow \frac { y }{ { (x+a) }^{ 2 } } =\frac { { (x+a) }^{ 2 }+2c }{ 2 } \)
\(2y={ (x+a) }^{ 4 }{ +2c(x+a) }^{ 2 }\)

Given AC = b and CD = a
Let hand r be the height and radius of the cylinder
ஃ AB = AC - BC = b - h
In the diagram

ΔABE - ΔACD
\(\therefore \frac { AB }{ BE } =\frac { AC }{ CD } \) 
\(\Rightarrow \frac { b-h }{ r } =\frac { b }{ a } \)
 \(\Rightarrow r=\frac { a\left( b-h \right) }{ b } \)
Volume of the cylinder = \({ \pi r }^{ 2 }h\)
= \(\pi \left( \frac { a }{ b } \left( b-b \right) ^{ 2 }h \right) \) 
= \(\pi \frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( b-h \right) ^{ 2 }h\)
= \(\pi \frac { { a }^{ 2 } }{ { b }^{ 2 } } \left( { b }^{ 2 }+{ h }^{ 2 }-2bh \right) h\)
 \(V=\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( { b }^{ 2 }{ h }^{ 2 }+{ h }^{ 3 } \right) \)
Differentiating with respect to 'h',
\(\frac { dV }{ dh } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( { b }^{ 2 }+{ 3h }^{ 2 }-4bh \right) \)
\(\frac { dV }{ dh } =0\Rightarrow { b }^{ 2 }-4bh+{ 3j }^{ 2 }=0\)
\(\Rightarrow (b-3h)+(b-h)=0\Rightarrow b-3h=0\quad \)or b = h
\(\Rightarrow h=\frac { b }{ 3 } asb\neq h\) 
\(\frac { dV }{ { dh }^{ 2 } } =\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( 6\left( \frac { b }{ 3 } \right) -4b \right) \)
\(\left( \frac { { d }^{ 2 }V }{ { dh }^{ 2 } } \right) _{ at\ h=\frac { b }{ 3 } }=\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi \left( 6\left( \frac { b }{ 3 } \right) -4b \right) \)
= \(\frac { { a }^{ 2 } }{ { b }^{ 2 } } \pi (-2b)<0\) 
ஃ V is maximum at \(h=\frac { b }{ 3 } \)
ஃ Maximum volume of the cylinder
\(V=\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( b-h \right) ^{ 2 }h\) 
= \(\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( b-\frac { b }{ 3 } \right) ^{ 2 }\left( \frac { b }{ 3 } \right) \left[ \therefore h=\frac { b }{ 3 } \right] \)
= \(\frac { { \pi a }^{ 2 } }{ { b }^{ 2 } } \left( \frac { 2b }{ 3 } \right) ^{ 2 }\left( \frac { b }{ 3 } \right) \)
\(=\frac{\pi a^{2}}{\not b^{2}}\left(\frac{4\not b^{2}}{9}\right)\left(\frac{b}{3}\right)=\frac{4}{9}\left(\frac{1}{3} \pi a^{2} b\right)\)
= \(\frac { 4 }{ 9 } \)(Volume of the cone)
Hence, volume of the largest cylinder that can be hidden underneath is \(\frac { 4 }{ 9 } \) times volume of the cone.

Given that the equation is \(\frac{dv}{dx}+2y\ cot\ x=3x^2 cosec^2x\)
This is a linear differential equation. Here, P = 2 cot x ; Q = 3x²cosec²x.
\(\int { Pdx=\int { 2cot\quad xdx=2log|sin\quad x|=log|sin\quad x{ | }^{ 2 } } =log{ sin }^{ 2 }x } \)
Thus, \(I.F={ e }^{ \int { Pdx } }={ e }^{ log{ sin }^{ 2 } }x={ sin }^{ 2 }x\)
Hence, the solution is \({ ye }^{ \int { Pdx } }={ \int { Qe } }^{ \int { Pdx } }dx+C\)
That is, \(y{ sin }^{ 2 }x=\int { { 3x }^{ 2 }cose{ c }^{ 2 }x.{ sin }^{ 2 }xdx+C=\int { { 3x }^{ 2 }dx+C={ x }^{ 3 }+C } } \)
Hence, \(y{ sin }^{ 2 }x={ x }^{ 3 }+C\) is the required solution

Givenf(x) = sin x + cos x, 0 f'(x) = cos x - sin x
f"(x) = sin x - cos x
\(\therefore\) f"(x) = 0
\(\Rightarrow\) sin x - cos X = 0
\(\Rightarrow\) -sin x = cosx
\(\Rightarrow\) sin (-x) = cos x
\(\Rightarrow\) \(x=\frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } ,2\pi \)
ஃ The possible intervals are \(\left( 0,\frac { 3\pi }{ 4 } \right) \left( \frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \left( \frac { 7\pi }{ 4 } ,2\pi \right) \) 

Interval	\(\left( 0,\frac { 3\pi }{ 4 } \right) \)	\(\left( \frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \)	\(\left( \frac { 7\pi }{ 4 } ,2\pi \right) \)
	Say x = -1 \(f''\left( x \right) =-\frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ \sqrt { 2 } } =-\frac { 2 }{ \sqrt { 2 } } \)	Say x = π f''(x) = -sinπ-cosπ = 0-(1) = 1 +ve	Say x = 320° f" (x) = - sin 320 - cos 320 = -sin (270 + 60) - cos (270 + 60) = cos 60 - sin 60 = \(\frac { 1 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } \) -ve
Concavity	Concave down	Concave up	Concave down

ஃf (x) is concave upward in \(\left( \frac { 3\pi }{ 4 } ,\frac { 7\pi }{ 4 } \right) \) and concave downward in \(\left( 0,\frac { 3\pi }{ 4 } \right) \left( \frac { 7\pi }{ 4 } ,2\pi \right) \) 
Since f" (x) changes its sign from negative to positive at \(\frac { 3\pi }{ 4 } \) and positive at \(\frac { 7\pi }{ 4 } \) f(x) has point of inflection at \(\left( \frac { 3\pi }{ 4 } ,f\left( \frac { 3\pi }{ 4 } \right) \right) \) and 

\(\left( \frac { 7\pi }{ 4 } ,f\left( \frac { 7\pi }{ 4 } \right) \right) \) 
\(\therefore f\left( \frac { 3\pi }{ 4 } \right) =sin\frac { 3\pi }{ 4 } -cos\frac { \pi }{ 4 } \) 
= \(sin\left( \pi -\frac { \pi }{ 4 } \right) +cos\left( \frac { \pi }{ 4 } \right) \) 
= \(\frac { 1 }{ \sqrt { 2 } } -\frac { 1 }{ \sqrt { 2 } } =0\) 
ஃ The points of inflection are \(\left( \frac { 3\pi }{ 4 } ,0 \right) \) and \(\left( \frac { 7\pi }{ 4 } ,0 \right) \)

The given differential equation may be written
\(\Rightarrow \frac { dy }{ dx } =\frac { -3{ e }^{ \frac { y }{ x } }\left( 1-\frac { y }{ x } \right) }{ 1+3{ e }^{ \frac { y }{ x } } } ...(1)\)
This is a homogeneous differential equation
\(\therefore put\quad y=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(v+x\frac { dv }{ dx } =\frac { -3{ e }^{ v }(1-v) }{ 1+3{ e }^{ v } } \)
\(\Rightarrow x\frac { dv }{ dx } =\frac { -3{ e }^{ v }(1-v) }{ 1+3{ e }^{ v } } -v\)

\(=\frac { -3{ e }^{ v }-v }{ 1+3{ e }^{ v } } =-\left( \frac { -3{ e }^{ v }+v }{ 1+3{ e }^{ v } } \right) \)
Separating the variables we get,
\(\frac { 1+3{ e }^{ v } }{ 3{ e }^{ v }+v } dv=-\frac { dx }{ x } \)
\(\Rightarrow log(3{ e }^{ v }+v)=-log\quad x+log\quad c\)
\(\Rightarrow log(3{ e }^{ v }+v)=log\left( \frac { c }{ x } \right) \Rightarrow { 3e }^{ v }+v=\frac { c }{ x } \)
\({ 3e }^{ \frac { y }{ x } }+\frac { y }{ x } =\frac { c }{ x } \)
\(\Rightarrow \frac { 3x{ e }^{ \frac { y }{ x } }+y }{ x } =\frac { c }{ x } \)
\(\Rightarrow y+3x{ e }^{ \frac { y }{ x } }=c\)
Given that y = 0 when x = 1
3(1)e⁰+ 0 = c \(\Rightarrow\) c = 3
\(\therefore\) (2) becomes, 3x\({ e }^{ \frac { y }{ x } }\)+ y = 3

Let us now find the slopes of the curves.
The slope m₁ for the curve y = x².
\(m_{1}=\frac{dy}{dx}=2x\).
and slope m₂ for the curve x = y² 
\(m_{2}=\frac{dy}{dx}=\frac{1}{2y}\).
Let \(\theta \)₁ and \(\theta _2\) be the acute angles at (0,0) and (1,1) respectively.
At (0,0), we come across the indeterminate form of 0×∞ in the denominator of
 tan \(\theta \)₁ = \(|\frac{2x-\frac{1}{2y}}{1+(2x)(\frac{1}{2y})}|\) and so we follow the limiting process

\(tan\theta_{1}=\underset{(x,y)\rightarrow(0,0)}{lim}|\frac{2x-\frac{1}{2y}}{1+(2x)(\frac{1}{2y})}|\)
=\(\underset{(x,y)\rightarrow(0,0)}{lim} {|\frac{4xy-1}{2(y-x)}|}\)
= \( \infty \)
which gives \(\theta_{1}=tan^{-1}(\infty)=\frac{\pi}{2}\).
At, (1,1) \(m_{1}=2, m_{2}=\frac{1}{2}\)
\(tan\theta_{2}=|\frac{2-\frac{1} {2}}{1+(2)(\frac{1}{2})}|\)
=\(\frac{3}{4}\)
which gives \(\theta_{2}=tan^{-1}(\frac{3}{4})\).

Given h = 12 m
r = 5 m
∴ tan α = \(\frac { OA }{ VO } =\frac { 5 }{ 12 } =\frac { r }{ h } \)
⇒ 12r = 5h 
⇒ r = \(\frac{5h}{12}\) ...(1)
Given rate of changeof volume = \(\frac{dv}{dt}=10\)
Volume of cone V = \(\frac13\) πr²h
= \(\frac { 1 }{ 3 } \)π\({ \left( \frac { 5h }{ 12 } \right) }^{ 2 }.h\) [From (1)]
\(v=\frac { 1 }{ 3 } \pi { \left( \frac { 25h }{ 144 } \right) }^{ 2 }=\frac { { 25\pi h }^{ 3 } }{ 3\times 144 } \)
Differentiating with respect to 't' we get,
\(\frac { dv }{ dt } =\frac { 25\pi }{ 3\times 144 } 3{ h }^{ 2 }\frac { dh }{ dt } \)
= \(\frac { 25\pi }{ 144 } { h }^{ 2 }\frac { dh }{ dt } \)
[∵ Given h = 8, \(\frac { dh }{ dt } \) = 10m³/min]
\(\Rightarrow \frac { 10\times 144 }{ 25\pi \times 64 } =\frac { dh }{ dt } \)
\(\Rightarrow \frac { dh }{ dt } =\frac { 9 }{ 10\pi } \)
The depth of the water increasing at the rate of \(=\frac { 9 }{ 10\pi } \)m / min.