Q.Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height $h$ and semi vertical angle $\alpha$ is one-third that of the cone and the greatest volume of cylinder is $\frac{4}{27}\pi h^3 \tan^2 \alpha$.

The problem gives a cone with radius $\sqrt{3}$ and semivertical angle $\frac{\pi}{3}$. The semivertical angle is the angle between the axis and the slant edge. That means in a vertical cross-section through the axis, the cone looks like an isosceles triangle with base $2\sqrt{3}$ and height $h$ such that $\tan(\pi/3) = \frac{\text{radius}}{\text{height}} = \frac{\sqrt{3}}{h}$. Since $\tan(\pi/3) = \sqrt{3}$, we get $\sqrt{3} = \frac{\sqrt{3}}{h}$, so $h = 1$. The cone’s height is $1$.

Now imagine a right circular cylinder inscribed in this cone — its axis coincides with the cone’s axis, and its top face touches the slant surface. In the cross-section, the cylinder appears as a rectangle inside the triangle. Let the cylinder’s radius be $r$ and its height be $H$. The key is that the top corners of the rectangle lie on the slant edges of the triangle.

1. Relate $r$ and $H$ using similar triangles. In the cross-section, consider the smaller triangle above the cylinder: its base is $r$ (half the cylinder’s diameter) and its height is $1 - H$ (the distance from the cylinder’s top to the cone’s apex). This small triangle is similar to the whole triangle (base $\sqrt{3}$, height $1$). So:

Hence $r = \sqrt{3}\,(1 - H)$.

2. Write the volume of the cylinder.

   Volume $V = \pi r^2 H = \pi \left[\sqrt{3}(1-H)\right]^2 H = 3\pi (1-H)^2 H$.

3. Maximise $V$ with respect to $H$. …

We are asked for the absolute maximum — the highest value the function reaches anywhere on the given closed interval. A cubic polynomial is continuous everywhere, so on a closed interval it must attain both a maximum and a minimum. The candidates are the endpoints and any points where the derivative is zero (critical points) inside the interval.

The derivative is $f'(x) = 6x^2 - 6x - 36$. Factor it: $6(x^2 - x - 6) = 6(x-3)(x+2)$. So $f'(x)=0$ at $x=3$ and $x=-2$. Both lie in $[-3,3]$, so they are valid critical points.

Now evaluate $f$ at all candidates:

1. At $x=-3$ (left endpoint):

   $f(-3) = 2(-27) - 3(9) - 36(-3) + 9 = -54 - 27 + 108 + 9 = 36$.

2. At $x=-2$ (critical point):

   $f(-2) = 2(-8) - 3(4) - 36(-2) + 9 = -16 - 12 + 72 + 9 = 53$.

3. At $x=3$ (critical point and right endpoint):

   $f(3) = 2(27) - 3(9) - 36(3) + 9 = 54 - 27 - 108 + 9 = -72$.

The values are $36$, $53$, and $-72$. The largest is $53$ — but wait, that's not among the options. Let's check again.

$f(-2) = 2(-8) - 3(4) - 36(-2) + 9 = -16 - 12 + 72 + 9$. …

Concept & Intuition

$f$ is decreasing where $f'(x)\le 0$ (with equality only at isolated points). We differentiate, simplify the sign, and take the longest interval on which $f$ never increases. The log term $\log\frac{1+x}{1-x}$ requires $\frac{1+x}{1-x}>0$, so the domain is $(-1,1)$; the answer must lie inside it.

Step-by-step

1. Domain. $\dfrac{1+x}{1-x}>0$ and $1-x^2\neq 0$ give $x\in(-1,1)$.

2. Differentiate.

The first two terms combine to $\dfrac{2}{1-x^2}$.

3. Last term (quotient rule).

4. Combine over $(1-x^2)^2$.

The numerator is

So

$f'(x)=\frac{-x^2(1+x^2)}{(1-x^2)^2}.$ …

Ellipse. $\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\Rightarrow a=5,\ c^2=25-16=9,\ c=3.$ Foci $(\pm3,0)$ and $e_1=\dfrac{3}{5}$, so $e_1^2=\dfrac{9}{25}.$

Hyperbola (same foci). $\dfrac{x^2}{A^2}-\dfrac{y^2}{B^2}=1$ with $A^2+B^2=c^2=9$ and $e_2=\dfrac{3}{A}$, so $e_2^2=\dfrac{9}{A^2}.$

Condition. $e_2^2-e_1^2\ge 1$: …

A quadratic expression of the form $f(x) = Ax^2 + Bx + C$ represents a parabola when graphed. The sign of the coefficient $A$ determines the direction the parabola opens:

• If $A > 0$, the parabola opens upwards, and its vertex is the lowest point, representing the minimum value of the expression.
• If $A < 0$, the parabola opens downwards, and its vertex is the highest point, representing the maximum value of the expression.

In this problem, the expression is $x^2 + 5x - 2$. Here, the coefficient of $x^2$ is $A=1$, which is positive. Therefore, the parabola opens upwards, and the expression has a minimum value. This minimum value occurs at the vertex of the parabola.

The x-coordinate of the vertex for a quadratic $Ax^2 + Bx + C$ is given by a standard formula. Once we find this x-coordinate, which is $a$ in this problem, we can substitute it back into the expression to find the minimum value, $M$.

1. Identify the coefficients of the quadratic expression.

   The given quadratic expression is $x^2 + 5x - 2$.

   Comparing this to the standard form $Ax^2 + Bx + C$, we can identify the coefficients:

   $A = 1$

   $B = 5$

   $C = -2$

2. Find the x-coordinate where the minimum value exists.

   The problem states that the minimum value exists at $a$. This $a$ is the x-coordinate of the vertex of the parabola.

   For a quadratic expression $Ax^2 + Bx + C$, the x-coordinate of the vertex is given by $x = -\frac{B}{2A}$.

   Substitute the values of $A$ and $B$ into this formula:

   $a = -\frac{5}{2(1)}$

   $a = -\frac{5}{2}$

3. Find the minimum value of the expression.

   The problem states that the minimum value is $M$. This $M$ is the value of the expression when $x = a$.

   Substitute $a = -\frac{5}{2}$ into the expression $x^2 + 5x - 2$:

   $M = \left(-\frac{5}{2}\right)^2 + 5\left(-\frac{5}{2}\right) - 2$

   $M = \frac{25}{4} - \frac{25}{2} - 2$

   To combine these fractions, we find a common denominator, which is 4: …

We are told that $f(x) = \dfrac{ax+b}{(x-1)(x-4)}$ has a local maximum at the point $(2, -1)$. This gives us two pieces of information: the point lies on the graph, and the derivative is zero at $x = 2$. Let’s unpack why.

Concept and intuition:

A local maximum at a point means the function’s value there is higher than nearby values. For a differentiable function, this implies the tangent line is horizontal — so the derivative is zero. Also, the point must satisfy the function itself. So we have two equations in the unknowns $a$ and $b$. Solve them, then compute $a+b$.

1. Use the point on the curve Since $(2, -1)$ lies on $f(x)$, we have $f(2) = -1$.

Set equal to $-1$:

2. Use the derivative condition The derivative must be zero at $x = 2$. First, find $f'(x)$. Write $f(x) = \frac{ax+b}{x^2 - 5x + 4}$. Use the quotient rule:

We only need $f'(2) = 0$, so the numerator must be zero at $x = 2$ (denominator is nonzero there).

Numerator at $x = 2$:

Set equal to zero:

3. Solve for $a$ and $b$ …

A local maximum of a function $f(x)$ at a point $(x_0, y_0)$ provides two critical pieces of information:

1. The point lies on the curve: The function's value at $x_0$ is $y_0$. This means $f(x_0) = y_0$.
2. The slope of the tangent is zero: At a local maximum (or minimum), the tangent line to the curve is horizontal. The slope of this tangent is given by the first derivative, so $f'(x_0) = 0$.

We will use these two conditions to set up a system of equations for $a$ and $b$, and then solve for them.

1. Use the condition that the point $(2, -1)$ lies on the curve. Since the local maximum exists at $(2, -1)$, the point $(2, -1)$ must satisfy the function's equation. Substitute $x=2$ and $f(x)=-1$ into the given function $f(x) = \dfrac{ax+b}{(x-1)(x-4)}$:

Multiplying both sides by $-2$ gives our first equation:

2. Find the first derivative of $f(x)$. First, expand the denominator of $f(x)$:

We use the quotient rule for differentiation.
> [!FORMULA]
> If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.
Here, we have:
*   $u(x) = ax+b \implies u'(x) = a$
*   $v(x) = x^2-5x+4 \implies v'(x) = 2x-5$
Substituting these into the quotient rule formula:

3. Use the condition that the derivative is zero at the local maximum. At a local maximum, the first derivative $f'(x)$ must be zero. Since the local maximum is at $x=2$, we must have $f'(2)=0$. Substitute $x=2$ into the expression for $f'(x)$: …

The key idea: for a quadratic $ax^2 + bx + c$, the vertex (where the minimum or maximum occurs) is at $x = -\frac{b}{2a}$, and the value there is $f\left(-\frac{b}{2a}\right)$. Here the expression is $ax^2 - 7x + 3a$, so $b = -7$ and $c = 3a$. Since the minimum is given, $a$ must be positive (otherwise the parabola opens downward and has no minimum). We’ll use the vertex condition to find $a$, then use Vieta’s formulas to get the sum of the roots.

1. Find the vertex’s x-coordinate. For $f(x) = ax^2 - 7x + 3a$, the vertex is at

2. Compute the minimum value. Substitute $x = \frac{7}{2a}$ into $f(x)$:

Simplify term by term:

and the constant $+3a$ stays. So

3. Set this equal to the given minimum $-\frac{1}{8}$.

Multiply through by $8a$ (since $a > 0$, no sign issues):

Rearranging:

4. Solve for $a$. This quadratic in $a$ factors or use the quadratic formula:

Since $9409 = 97^2$ (check: $97^2 = 9409$), we have …

Left branch $(x\le1):\ f(x)=1+6x-3x^2$. Its derivative $f'(x)=6-6x=0$ at $x=1$, and the parabola opens downward, so on $(-\infty,1]$ the maximum is at $x=1$:

$f(1)=1+6-3=4.$ …

The key idea is that for a fixed number of Bernoulli trials, the probability of exactly $k$ successes is given by the binomial formula. The "most probable" $k$ is the mode of this distribution. Instead of computing all probabilities, we can find where the sequence $P(k)$ stops increasing and starts decreasing — that is, where the ratio $P(k+1)/P(k)$ becomes less than 1.

1. Set up the binomial probability For $n=7$ trials, success probability $p = \frac{2}{5}$, failure probability $q = 1-p = \frac{3}{5}$, the probability of exactly $k$ successes is

2. Consider the ratio of successive probabilities

This ratio tells us how $P(k)$ changes as $k$ increases. If it is greater than 1, $P(k+1) > P(k)$; if less than 1, $P(k+1) < P(k)$.

3. Find where the ratio crosses 1 Set the ratio equal to 1 to find the threshold:

Solve:

4. Interpret the result The ratio is >1 when $k < 2.2$ (so probabilities increase up to $k=2$), and <1 when $k > 2.2$ (so probabilities decrease after $k=3$). This means the maximum occurs at the integer just after the crossing point: $k = 3$. …