Vardaan Learning Institute

Find the equation of the common tangent to the curves $y = x^2$ and $y = -x^2 + 4x - 4$. Solution: Let the points of tangency be $(a, a^2)$ on the first curve and $(b, -b^2+4b-4)$ on the second.
Derivative of first curve: $y' = 2x$. At $x=a$, slope is $2a$.
Derivative of second curve: $y' = -2x + 4$. At $x=b$, slope is $-2b+4$.
Since it's a common tangent, slopes are equal: $2a = -2b + 4 \implies a = 2 - b \implies b = 2 - a$.
Equation of tangent from first curve: $y - a^2 = 2a(x - a) \implies y = 2ax - a^2$.
Equation of tangent from second curve: $y - (-b^2+4b-4) = (-2b+4)(x - b)$. Substituting $b=2-a$ and $-2b+4 = 2a$:
$y - (-(2-a)^2 + 4(2-a) - 4) = 2a(x - (2-a))$
$y - (-4 + 4a - a^2 + 8 - 4a - 4) = 2ax - 4a + 2a^2$
$y - (a^2) = 2ax - 4a + 2a^2 \implies y = 2ax + a^2 - 4a$.
Comparing intercepts of both equations: $-a^2 = a^2 - 4a \implies 2a^2 - 4a = 0 \implies 2a(a - 2) = 0$.
So, $a = 0$ or $a = 2$.
If $a = 0$, tangent is $y = 0$. If $a = 2$, tangent is $y = 4x - 4$.

Calculate the shortest distance between the parabola $y = x^2 + 3x + 2$ and the straight line $x - y - 2 = 0$. Solution: The shortest distance occurs along the common normal. Therefore, the tangent to the parabola at the closest point must be parallel to the given line.
Slope of given line $x - y - 2 = 0 \implies y = x - 2$ is $m = 1$.
Derivative of parabola: $\frac{dy}{dx} = 2x + 3$.
Set slope equal to 1: $2x + 3 = 1 \implies 2x = -2 \implies x = -1$.
When $x = -1$, $y = (-1)^2 + 3(-1) + 2 = 1 - 3 + 2 = 0$. The closest point is $P(-1, 0)$.
Perpendicular distance from $P(-1, 0)$ to the line $x - y - 2 = 0$ is $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.
$d = \frac{|-1 - 0 - 2|}{\sqrt{1^2 + (-1)^2}} = \frac{|-3|}{\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ units.

Find the value(s) of the parameter $a$ for which the function $f(x) = \sin x - ax + b$ is strictly decreasing on the entire real number line $\mathbb{R}$. Solution: For $f(x)$ to be strictly decreasing on $\mathbb{R}$, $f'(x) \le 0$ for all $x \in \mathbb{R}$ (and $f'(x)=0$ only at discrete points).
$f'(x) = \cos x - a$.
We require $\cos x - a \le 0 \implies a \ge \cos x$ for all $x \in \mathbb{R}$.
Since the maximum value of $\cos x$ is $1$, for $a$ to be greater than or equal to $\cos x$ everywhere, we must have $a \ge 1$.
Therefore, the values of $a$ are $[1, \infty)$.

Using the concepts of monotonicity, prove the double inequality: $\frac{x}{1+x} < \ln(1+x) < x$ for all $x > 0$. Solution: **Part 1:** Prove $\ln(1+x) < x$. Let $g(x) = x - \ln(1+x)$.
$g'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x}$.
For $x > 0$, $g'(x) > 0$, so $g(x)$ is strictly increasing.
Since $g(0) = 0 - \ln(1) = 0$, for $x > 0$, $g(x) > g(0) \implies x - \ln(1+x) > 0 \implies \ln(1+x) < x$.

**Part 2:** Prove $\frac{x}{1+x} < \ln(1+x)$. Let $h(x) = \ln(1+x) - \frac{x}{1+x}$.
$h'(x) = \frac{1}{1+x} - \frac{1(1+x) - x(1)}{(1+x)^2} = \frac{1}{1+x} - \frac{1}{(1+x)^2} = \frac{1+x-1}{(1+x)^2} = \frac{x}{(1+x)^2}$.
For $x > 0$, $h'(x) > 0$, so $h(x)$ is strictly increasing.
Since $h(0) = \ln(1) - 0 = 0$, for $x > 0$, $h(x) > h(0) \implies \ln(1+x) - \frac{x}{1+x} > 0 \implies \frac{x}{1+x} < \ln(1+x)$.
Combining both parts proves the double inequality.

A window is designed in the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the entire window is fixed at $12\text{ m}$, find the dimensions of the rectangular portion that will maximize the total area of the window. Solution: Let the base of the rectangle be $x$ and the height be $y$. The sides of the equilateral triangle are also $x$.
Perimeter of the window (excluding the internal base of the triangle) is $x + 2y + x + x = 3x + 2y = 12 \implies y = \frac{12 - 3x}{2} = 6 - 1.5x$.
Total Area $A = xy + \frac{\sqrt{3}}{4}x^2 = x(6 - 1.5x) + \frac{\sqrt{3}}{4}x^2 = 6x - 1.5x^2 + \frac{\sqrt{3}}{4}x^2$.
$A' = 6 - 3x + \frac{\sqrt{3}}{2}x = 0 \implies 6 = x\left(3 - \frac{\sqrt{3}}{2}\right) \implies 6 = x\left(\frac{6 - \sqrt{3}}{2}\right)$.
$x = \frac{12}{6 - \sqrt{3}}$.
$y = 6 - 1.5\left(\frac{12}{6 - \sqrt{3}}\right) = 6 - \frac{18}{6 - \sqrt{3}} = \frac{36 - 6\sqrt{3} - 18}{6 - \sqrt{3}} = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}}$.
$A'' = -3 + \frac{\sqrt{3}}{2} < 0$, verifying this is a maximum.
Dimensions: Base $x = \frac{12}{6 - \sqrt{3}}\text{ m}$, Height $y = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}}\text{ m}$.

An inverted conical tank has a base radius of $2\text{ m}$ and a height of $4\text{ m}$. Water is flowing into the tank at the constant rate of $2\text{ m}^3/\text{min}$. At what exact rate is the water level rising at the instant when the depth of the water is $3\text{ m}$? Solution: By similar triangles, $\frac{r}{h} = \frac{2}{4} = \frac{1}{2} \implies r = \frac{h}{2}$.
Volume of water $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi}{12}h^3$.
Differentiating w.r.t $t$: $\frac{dV}{dt} = \frac{\pi}{4}h^2 \frac{dh}{dt}$.
Given $\frac{dV}{dt} = 2\text{ m}^3/\text{min}$, and we want $\frac{dh}{dt}$ when $h = 3\text{ m}$.
$2 = \frac{\pi}{4}(3)^2 \frac{dh}{dt} \implies 2 = \frac{9\pi}{4}\frac{dh}{dt} \implies \frac{dh}{dt} = \frac{8}{9\pi}\text{ m/min}$.

Determine the absolute maximum and absolute minimum values of the trigonometric function $f(x) = \sin x + \frac{1}{2}\cos 2x$ over the closed interval $[0, \frac{\pi}{2}]$. Solution: $f'(x) = \cos x - \frac{1}{2}(2\sin 2x) = \cos x - \sin 2x = \cos x - 2\sin x \cos x = \cos x(1 - 2\sin x)$.
Set $f'(x) = 0$. Solutions in $[0, \pi/2]$: $\cos x = 0 \implies x = \pi/2$, and $1 - 2\sin x = 0 \implies \sin x = 1/2 \implies x = \pi/6$.
Evaluate at endpoints and critical points:
$f(0) = \sin(0) + \frac{1}{2}\cos(0) = 0 + 1/2 = 1/2$.
$f(\pi/6) = \sin(\pi/6) + \frac{1}{2}\cos(\pi/3) = 1/2 + \frac{1}{2}(1/2) = 1/2 + 1/4 = 3/4$.
$f(\pi/2) = \sin(\pi/2) + \frac{1}{2}\cos(\pi) = 1 + \frac{1}{2}(-1) = 1 - 1/2 = 1/2$.
Absolute Max $= 3/4$ (at $x = \pi/6$). Absolute Min $= 1/2$ (at $x=0$ and $x=\pi/2$).

Prove analytically that the family of ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and the family of hyperbolas $\frac{x^2}{a^2 + \lambda} + \frac{y^2}{b^2 + \lambda} = 1$ intersect orthogonally at all points of intersection. Solution: Curve 1 slope ($m_1$): Differentiating gives $\frac{2x}{a^2} + \frac{2y}{b^2}y' = 0 \implies y' = -\frac{b^2 x}{a^2 y}$.
Curve 2 slope ($m_2$): Differentiating gives $\frac{2x}{a^2+\lambda} + \frac{2y}{b^2+\lambda}y' = 0 \implies y' = -\frac{(b^2+\lambda)x}{(a^2+\lambda)y}$.
Product $m_1 m_2 = \frac{b^2(b^2+\lambda)x^2}{a^2(a^2+\lambda)y^2}$.
Subtracting the equations of the curves at intersection $(x, y)$:
$x^2\left(\frac{1}{a^2} - \frac{1}{a^2+\lambda}\right) + y^2\left(\frac{1}{b^2} - \frac{1}{b^2+\lambda}\right) = 0$
$x^2\left(\frac{\lambda}{a^2(a^2+\lambda)}\right) + y^2\left(\frac{\lambda}{b^2(b^2+\lambda)}\right) = 0 \implies \frac{x^2}{a^2(a^2+\lambda)} = -\frac{y^2}{b^2(b^2+\lambda)} \implies \frac{x^2}{y^2} = -\frac{a^2(a^2+\lambda)}{b^2(b^2+\lambda)}$.
Substitute $\frac{x^2}{y^2}$ into $m_1 m_2$ product:
$m_1 m_2 = \frac{b^2(b^2+\lambda)}{a^2(a^2+\lambda)} \times \left(-\frac{a^2(a^2+\lambda)}{b^2(b^2+\lambda)}\right) = -1$.
Since $m_1 m_2 = -1$, they intersect orthogonally.

For the polynomial function $f(x) = x^4 - 4x^3 + 4x^2$, find all points of local maxima, local minima, and the exact coordinates of any points of inflection. Solution: $f'(x) = 4x^3 - 12x^2 + 8x = 4x(x^2 - 3x + 2) = 4x(x-1)(x-2)$.
Critical points are $x = 0, 1, 2$.
$f''(x) = 12x^2 - 24x + 8 = 4(3x^2 - 6x + 2)$.
At $x=0$, $f''(0) = 8 > 0 \implies$ Local Min at $(0, 0)$.
At $x=1$, $f''(1) = -4 < 0 \implies$ Local Max at $(1, 1)$.
At $x=2$, $f''(2) = 8 > 0 \implies$ Local Min at $(2, 0)$.
Points of inflection occur where $f''(x) = 0$ (and changes sign): $3x^2 - 6x + 2 = 0 \implies x = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = 1 \pm \frac{1}{\sqrt{3}}$.
Both are legitimate points of inflection.

A manufacturer can sell $x$ items at a price of rupees $(5 - \frac{x}{100})$ each. The total cost price of producing $x$ items is rupees $(\frac{x}{5} + 500)$. Find the number of items he should produce and sell to maximize his overall profit. Solution: Revenue $R(x) = x(5 - x/100) = 5x - x^2/100$.
Cost $C(x) = x/5 + 500 = 0.2x + 500$.
Profit $P(x) = R(x) - C(x) = 5x - x^2/100 - (0.2x + 500) = 4.8x - x^2/100 - 500$.
$P'(x) = 4.8 - \frac{2x}{100} = 4.8 - \frac{x}{50} = 0 \implies \frac{x}{50} = 4.8 \implies x = 240$.
$P''(x) = -1/50 < 0$, which confirms it is a maximum.
The manufacturer should produce and sell 240 items.

Determine the precise intervals of strict monotonicity (intervals of increase and decrease) for the transcendental function $f(x) = x^x$ defined for $x > 0$. Solution: Let $y = x^x$. Taking natural logarithm: $\ln y = x\ln x$.
Differentiating w.r.t $x$: $\frac{1}{y} \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1 \implies f'(x) = x^x(\ln x + 1)$.
Since $x^x > 0$ for all $x > 0$, the sign of $f'(x)$ depends solely on $(\ln x + 1)$.
$f'(x) = 0 \implies \ln x + 1 = 0 \implies \ln x = -1 \implies x = e^{-1} = 1/e$.
For $0 < x < 1/e$, $\ln x < -1 \implies f'(x) < 0$. Function is strictly decreasing.
For $x > 1/e$, $\ln x > -1 \implies f'(x) > 0$. Function is strictly increasing.
Interval of strict decrease: $(0, 1/e)$. Interval of strict increase: $(1/e, \infty)$.

A camera is mounted on the ground $2000\text{ m}$ away from the base of a rocket launch pad. The rocket is rising vertically at a constant velocity of $50\text{ m/s}$. Find the rate of change of the angle of elevation of the camera at the exact moment when the rocket is $1500\text{ m}$ above the ground. Solution: Let $y$ be the height of the rocket and $\theta$ be the angle of elevation. We have $\tan\theta = \frac{y}{2000}$.
Differentiating w.r.t $t$: $\sec^2\theta \frac{d\theta}{dt} = \frac{1}{2000}\frac{dy}{dt}$.
Given $\frac{dy}{dt} = 50$, and at the instant $y = 1500$, $\tan\theta = \frac{1500}{2000} = \frac{3}{4}$.
Using identity $\sec^2\theta = 1 + \tan^2\theta = 1 + \frac{9}{16} = \frac{25}{16}$.
Substituting values: $\left(\frac{25}{16}\right) \frac{d\theta}{dt} = \frac{1}{2000}(50) = \frac{1}{40}$.
$\frac{d\theta}{dt} = \frac{1}{40} \times \frac{16}{25} = \frac{16}{1000} = 0.016\text{ rad/s}$.

Find the equations of the two distinct tangent lines to the quadratic curve $y = x^2 - 2x + 3$ that can be drawn from the external origin point $(0, 0)$. Solution: Let $(x_1, y_1)$ be the point of tangency on the curve. Then $y_1 = x_1^2 - 2x_1 + 3$.
Slope of tangent $m = \frac{dy}{dx} = 2x_1 - 2$.
Equation of tangent line: $y - y_1 = (2x_1 - 2)(x - x_1)$.
Since this tangent passes through $(0,0)$, plug in $x=0, y=0$: $-y_1 = (2x_1 - 2)(-x_1) \implies y_1 = 2x_1^2 - 2x_1$.
Equating $y_1$: $x_1^2 - 2x_1 + 3 = 2x_1^2 - 2x_1 \implies x_1^2 = 3 \implies x_1 = \pm\sqrt{3}$.
If $x_1 = \sqrt{3}$, $m = 2\sqrt{3} - 2$. Tangent equation is $y = (2\sqrt{3} - 2)x$.
If $x_1 = -\sqrt{3}$, $m = -2\sqrt{3} - 2$. Tangent equation is $y = -(2\sqrt{3} + 2)x$.

Find the points on the curve $y = x^3$ where the slope of the tangent is equal to the $x$-coordinate of the point. Subsequently, find the equation of the normal at the non-zero point. Solution: Let point be $(x, y)$. Slope $\frac{dy}{dx} = 3x^2$.
Given slope equals x-coordinate: $3x^2 = x \implies 3x^2 - x = 0 \implies x(3x - 1) = 0 \implies x = 0, x = 1/3$.
Points are $(0, 0)$ and $(1/3, 1/27)$.
At the non-zero point $(1/3, 1/27)$, the tangent slope $m_t = 1/3$. Thus, normal slope $m_n = -3$.
Equation of normal: $y - 1/27 = -3(x - 1/3) \implies y - 1/27 = -3x + 1 \implies 3x + y - 28/27 = 0 \implies 81x + 27y - 28 = 0$.

Show that the maximum volume of a cylinder which can be inscribed in a sphere of radius $R$ is $\frac{1}{\sqrt{3}}$ times the volume of the sphere itself. Solution: Let height of cylinder be $2x$. By Pythagoras, base radius $r = \sqrt{R^2 - x^2}$.
Volume of cylinder $V_c = \pi r^2(2x) = 2\pi x(R^2 - x^2) = 2\pi R^2 x - 2\pi x^3$.
$V_c' = 2\pi R^2 - 6\pi x^2 = 0 \implies x^2 = R^2/3 \implies x = R/\sqrt{3}$.
Max volume $V_c = 2\pi \left(\frac{R}{\sqrt{3}}\right) \left(R^2 - \frac{R^2}{3}\right) = \frac{2\pi R}{\sqrt{3}} \left(\frac{2R^2}{3}\right) = \frac{4\pi R^3}{3\sqrt{3}}$.
Volume of sphere $V_s = \frac{4}{3}\pi R^3$.
Ratio $\frac{V_c}{V_s} = \frac{\frac{4\pi R^3}{3\sqrt{3}}}{\frac{4}{3}\pi R^3} = \frac{1}{\sqrt{3}}$. Proved.