Q.Maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is:
(A) $0$
(B) $12$
(C) $16$
(D) $32$

Step 1 — Write the slope as a function

For the curve $y = -x^3 + 3x^2 + 2x - 27$, the slope of the tangent at any point is

The question asks for the maximum value of $m(x)$ — so $m$ is now the function being optimised, not $y$.

Step 2 — Critical point of $m$

Step 3 — Confirm it is a maximum

$m''(x) = \frac{d^3y}{dx^3} = -6 < 0$ …

Step 1 — Set up the constraint

For a right circular cone with radius $r$, height $h$ and slant height $l$, the axial cross-section is a right triangle, so

Step 2 — Express the volume in one variable

(Reducing to a single variable is the whole point of the constraint — only then can we use $dV/dh = 0$.)

Step 3 — Find the critical point

$\frac{dV}{dh} = \frac{\pi}{3}\left(36 - 3h^2\right) = 0 \;\Longrightarrow\; h^2 = 12 \;\Longrightarrow\; h = 2\sqrt3$ …

A maximum at $x=2$ means the vertex is $(2,20)$ and the parabola opens downward ($a<0$).

Step-by-step reasoning

1. Vertex form.

2. Use $f(1)=19$.

3. Expand to standard form.

4. Match to the options. This is exactly option (D): $f(x) = -x^2+4x+16$. …

Step 1 — Read the circle.

Centre $(-1, 3)$, radius $r = \sqrt{64} = 8$. (The centre is irrelevant — area is translation-invariant.)

Step 2 — The key geometric fact.

A rectangle inscribed in a circle has all four vertices on the circle, so its diagonal is a diameter:

If the sides are $a$ and $b$, then by Pythagoras

Step 3 — Maximise the area $A = ab$ subject to $a^2+b^2 = 256$.

Parametrise $a = 16\cos\theta$, $b = 16\sin\theta$:

$A = 256\sin\theta\cos\theta = 128\sin 2\theta$ …

The key idea is that when a variable appears both in the base and the exponent, the natural logarithm is our best friend. It turns the messy expression into a product we can differentiate easily. We want the maximum of $f(x) = \left(\frac{1}{x}\right)^x$, which is defined for $x > 0$ (since raising a positive number to any real power is fine).

Let $f(x) = x^{-x}$. Taking natural logs:

Now we maximise $\log f(x)$ instead of $f(x)$ itself — because $\log$ is a strictly increasing function, the $x$ that maximises $\log f(x)$ also maximises $f(x)$. This is a standard trick in optimisation problems with exponentials.

1. Differentiate $\log f(x)$ with respect to $x$:

2. Set the derivative to zero to find critical points:

3. Check that this is a maximum. The second derivative of $\log f(x)$ is $-\frac{1}{x}$, which is negative for all $x > 0$. So the function is concave down everywhere, and the critical point is indeed a global maximum.

4. Compute the maximum value of $f(x)$ at $x = 1/e$: …

Step 1 — Identify the standard form. The focus $(6, 0)$ lies on the positive $x$-axis and the directrix $x = -6$ is a vertical line on the other side of the origin, equidistant from it. The vertex is the midpoint of the perpendicular from the focus to the directrix, i.e. $\left(\frac{6 + (-6)}{2}, 0\right) = (0,0)$. So this is the standard right-opening parabola

Step 2 — Find $a$. Matching: $a = 6$. Hence

Step 3 — Derive it from the definition (check). A parabola is the locus of points equidistant from the focus and the directrix. For $P(x, y)$: …

The key idea here is that the arithmetic mean (A.M.) and geometric mean (G.M.) of the roots of a quadratic equation directly give us the sum and product of those roots. Once we have the sum and product, we can immediately write the quadratic.

For a quadratic equation $x^2 - Sx + P = 0$, where $S$ is the sum of the roots and $P$ is the product of the roots, the roots themselves are the two numbers we are averaging. So if the roots are $\alpha$ and $\beta$, then:

• A.M. of $\alpha$ and $\beta$ = $\frac{\alpha + \beta}{2}$
• G.M. of $\alpha$ and $\beta$ = $\sqrt{\alpha \beta}$

The problem gives us these means directly. We don't need to find the individual roots — just the sum and product.

1. Find the sum of the roots.

   The A.M. is 5, so $\frac{\alpha + \beta}{2} = 5$.

   Multiplying both sides by 2 gives $\alpha + \beta = 10$.

   So the sum $S = 10$.

2. Find the product of the roots.

   The G.M. is 4, so $\sqrt{\alpha \beta} = 4$.

   Squaring both sides gives $\alpha \beta = 16$.

   So the product $P = 16$.

3. Write the quadratic equation. …