Let the tangent to the curve y = x3 − 2x2 − x + 5 at the points P and Q be parallel to the line y = -2x + 3. Find the coordinates of the points P and Q.

Here are the solutions to the questions from the image:

Question 1: Let the tangent to the curve $y = x^3 - 2x^2 - x + 5$ at the points P and Q be parallel to the line $y = -2x + 3$. Find the coordinates of the points P and Q.

Step 1: Find the derivative of the curve to get the slope of the tangent.

$$\frac{dy}{dx} = \frac{d}{dx}(x^3 - 2x^2 - x + 5) = 3x^2 - 4x - 1$$

Step 2: The tangent lines are parallel to $y = -2x + 3$, so their slope is $-2$. Set the derivative equal to $-2$.

$$3x^2 - 4x - 1 = -2$$ $$3x^2 - 4x + 1 = 0$$

Step 3: Solve the quadratic equation for $x$.
We can factor the quadratic: $(3x-1)(x-1) = 0$.
So, $3x-1=0 \implies x = \frac{1}{3}$ or $x-1=0 \implies x = 1$.

Step 4: Find the corresponding $y$-coordinates for each $x$ value using the original curve equation $y = x^3 - 2x^2 - x + 5$.
For $x = 1$:

$$y = (1)^3 - 2(1)^2 - (1) + 5 = 1 - 2 - 1 + 5 = 3$$

So, one point is $(1, 3)$.

For $x = \frac{1}{3}$:

$$y = \left(\frac{1}{3}\right)^3 - 2\left(\frac{1}{3}\right)^2 - \left(\frac{1}{3}\right) + 5$$ $$y = \frac{1}{27} - 2\left(\frac{1}{9}\right) - \frac{1}{3} + 5$$ $$y = \frac{1}{27} - \frac{2}{9} - \frac{1}{3} + 5$$

To combine these, find a common denominator, which is 27:

$$y = \frac{1}{27} - \frac{6}{27} - \frac{9}{27} + \frac{135}{27}$$ $$y = \frac{1 - 6 - 9 + 135}{27} = \frac{121}{27}$$

So, the other point is $\left(\frac{1}{3}, \frac{121}{27}\right)$.

The coordinates of the points P and Q are $\left(\frac{1}{3}, \frac{121}{27}\right)$ and $(1, 3)$.
Comparing with the options, this matches option (B).

The final answer is $\boxed{\text{(B) P}\left(\frac{1}{3}, \frac{121}{27}\right), \text{Q}(1, 3)}$.

Question 2: The radius of a circle is increasing at a rate of $2.0 \text{ mm/sec}$. At what rate is the area increasing when radius is $100 \text{ mm}$?

Step 1: Write down the formula for the area of a circle.

$$A = \pi r^2$$

Step 2: Differentiate the area $A$ with respect to time $t$ using the chain rule.

$$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$$

Step 3: Substitute the given values: $\frac{dr}{dt} = 2.0 \text{ mm/sec}$ and $r = 100 \text{ mm}$.

$$\frac{dA}{dt} = 2\pi (100 \text{ mm}) (2.0 \text{ mm/sec})$$ $$\frac{dA}{dt} = 400\pi \text{ mm}^2\text{/sec}$$

Comparing with the options, this matches option (B).

The final answer is $\boxed{\text{(B) } 400\pi \text{ mm}^2\text{/sec}}$.

Question 3: Suppose the distance $S$ in centimetre travelled by a body after time $t$ in seconds is given by $S = t^2 - 3t + 7$. Find the time at which the body comes to rest?

Step 1: The body comes to rest when its velocity is zero. Velocity $v$ is the derivative of distance $S$ with respect to time $t$.

$$v = \frac{dS}{dt} = \frac{d}{dt}(t^2 - 3t + 7) = 2t - 3$$

Step 2: Set the velocity to zero and solve for $t$.

$$2t - 3 = 0$$ $$2t = 3$$ $$t = \frac{3}{2} = 1.5 \text{ sec}$$

Comparing with the options, this matches option (B).

The final answer is $\boxed{\text{(B) } 1.5 \text{ sec}}$.

Question 4: The turning point of the function $y = x(x-1)^2$, at $x=1$.

Step 1: Expand the function.

$$y = x(x^2 - 2x + 1) = x^3 - 2x^2 + x$$

Step 2: Find the first derivative $\frac{dy}{dx}$.

$$\frac{dy}{dx} = 3x^2 - 4x + 1$$

Step 3: Evaluate the first derivative at $x=1$.

$$\frac{dy}{dx}\Big|_{x=1} = 3(1)^2 - 4(1) + 1 = 3 - 4 + 1 = 0$$

Since $\frac{dy}{dx} = 0$ at $x=1$, it is a critical point.

Step 4: Find the second derivative $\frac{d^2y}{dx^2}$.

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 - 4x + 1) = 6x - 4$$

Step 5: Evaluate the second derivative at $x=1$.

$$\frac{d^2y}{dx^2}\Big|_{x=1} = 6(1) - 4 = 2$$

Since $\frac{d^2y}{dx^2} > 0$ at $x=1$, the turning point is a minimum point.
Comparing with the options, this matches option (A).

The final answer is $\boxed{\text{(A) is a minimum point}}$.

Question 5: If $3x^2 - 5y^2 + 2xy = 7$, find $\frac{dy}{dx}$?

Step 1: Differentiate both sides of the equation with respect to $x$ using implicit differentiation.

$$\frac{d}{dx}(3x^2) - \frac{d}{dx}(5y^2) + \frac{d}{dx}(2xy) = \frac{d}{dx}(7)$$ $$$$