1. Problem: Evaluate $(f \circ g)'(1)$ given $f(1)=2$, $g(1)=2$, $f'(1)=3$, $g'(1)=2$, $f'(2)=1$, and $g'(2)=3$. Formula: The derivative of a composite function is given by the chain rule: $$ (f \circ g)'(x) = f'(g(x)) \cdot g'(x) $$ Step 1: Calculate $f'(g(1))$. Since $g(1) = 2$, we need $f'(2)$ which is given as 1. Step 2: Calculate $g'(1)$ which is given as 2. Step 3: Multiply these values: $$ (f \circ g)'(1) = f'(g(1)) \cdot g'(1) = 1 \times 2 = 2 $$ Answer: (A) 2 2. Problem: An object is dropped and hits the ground after 5 seconds. Find the height from which it was dropped. Given: acceleration $a(t) = -9.8$ m/s² (taking upward as positive), initial velocity $v_0 = 0$ (dropped), time $t=5$ s. Formula: Position function under constant acceleration: $$ s(t) = s_0 + v_0 t + \frac{1}{2} a t^2 $$ Since the object hits the ground at $t=5$, final position $s(5) = 0$. Step 1: Substitute known values: $$ 0 = s_0 + 0 \times 5 + \frac{1}{2} (-9.8) \times 5^2 $$ Step 2: Simplify: $$ 0 = s_0 - 4.9 \times 25 $$ $$ s_0 = 4.9 \times 25 = 122.5 $$ Answer: (E) 4.9 × (25)m 3. Problem: Population growth is proportional to the current population. Initial population $P_0=10000$, doubles in 15 years. Find time to triple. Formula: Exponential growth: $$ P(t) = P_0 e^{kt} $$ Step 1: Use doubling info to find $k$: $$ 2P_0 = P_0 e^{15k} \Rightarrow 2 = e^{15k} \Rightarrow \ln 2 = 15k \Rightarrow k = \frac{\ln 2}{15} $$ Step 2: Find $t$ when population triples: $$ 3P_0 = P_0 e^{kt} \Rightarrow 3 = e^{kt} \Rightarrow \ln 3 = kt = \frac{\ln 2}{15} t $$ Step 3: Solve for $t$: $$ t = \frac{15 \ln 3}{\ln 2} $$ Answer: (D) 15 ln(3)/ln(2) 4. Problem: Side of cube increases at 3 cm/s. Find rate of volume increase when side length is 2 cm. Formula: Volume of cube: $$ V = s^3 $$ Rate of change: $$ \frac{dV}{dt} = 3s^2 \frac{ds}{dt} $$ Step 1: Substitute $s=2$ cm and $\frac{ds}{dt} = 3$ cm/s: $$ \frac{dV}{dt} = 3 \times 2^2 \times 3 = 3 \times 4 \times 3 = 36 $$ Answer: (E) 36 cm³/s 5. Problem: Find minimum slope of tangent to curve $f(x) = x^5 + x^3 - 2x$. Step 1: Find derivative: $$ f'(x) = 5x^4 + 3x^2 - 2 $$ Step 2: Find critical points of $f'(x)$ by differentiating again: $$ f''(x) = 20x^3 + 6x $$ Set $f''(x) = 0$: $$ 20x^3 + 6x = 0 \Rightarrow x(20x^2 + 6) = 0 $$ Solutions: $$ x=0 \quad \text{or} \quad 20x^2 + 6=0 \Rightarrow x^2 = -\frac{6}{20} $$ No real roots from second term, so only $x=0$. Step 3: Check values of $f'(x)$ at critical points and limits: As $x \to \pm \infty$, $5x^4$ dominates and is positive, so $f'(x) \to +\infty$. Evaluate $f'(0) = -2$. Step 4: Since $f'(0) = -2$ is a minimum value of the slope. Answer: (C) -2