QUIZ 2 Foundational

Solution 📝: Let’s analyze $ f(x) = |x^2 - 4x + 3| + 17 $.

1️⃣ The quadratic $ x^2 - 4x + 3 = (x-1)(x-3) $ can take any real value and is defined for all real $ x $. So, option A is FALSE.

2️⃣ The function involves an absolute value, which is not one-to-one (e.g., $ x = a $ and $ x = b $ can give the same value due to the absolute value), hence not bijective. Option B is FALSE.

3️⃣ The minimum of $ |x^2 - 4x + 3| $ is 0 (attained when $ x^2 - 4x + 3 = 0,\ x=1,3 $), so minimum value of $ f(x) $ is 17, and the function takes all values $ f(x) \geq 17 $. So, the range is $[17, \infty)$, not $[0, \infty)$. Option C is FALSE.

4️⃣ As derived above, the minimum value is 17, attained at $ x = 1, 3 $. Option D is TRUE.

Answer: The minimum value of $ f $ is 17.

Solution 📝: $ y = x^3 - 1 $ is a cubic function, which is bijective and continuous over $ \mathbb{R} $; it has an inverse $ x = \sqrt[^1]{y+1} $, which is also defined for all $ y \in \mathbb{R} $.

So, the domain of the inverse function is all real numbers $ \mathbb{R} $.

Answer: $\mathbb{R}$

Solution 📝: Let’s analyze each option:

A. For $ 0 < b < 1 $, the logarithm function is decreasing. For $ 0 < x < 1 $, $ \log_b x > 0 $, so A is FALSE.

B. For $ 0 < b < 1, 0 < x < 1 $, and $ x > b $, $ \log_b x $ will not necessarily be greater than 1. FALSE.

C. For $ 0 < b < 1 $ and $ x < y $, $ \log_b x > \log_b y $ because the function is decreasing when base is between 0 and 1. TRUE.

D. $ \log_{10} 100 = 2 $, which is rational. TRUE.

Answers: C and D are correct.

Solution 📝: Curve 1 is continuous and differentiable, so Rolle’s theorem applies. There must be points where the derivatives are equal between -4 and 4. TRUE.

Curve 2 has a cusp (sharp corner) at the origin, so the derivative does not exist there. TRUE.

Curve 3: The derivative at the origin and (-2,0) are not necessarily equal. FALSE.

Curve 4 is a constant function (horizontal line), so the derivative exists everywhere (and is zero). FALSE.

Answers: A and B are correct.

Solution 📝: Let $ x = \log_3 m $, so $ m = 3^x $.

Then, $ m^{\log_3 2} = (3^x)^{\log_3 2} = 2^x $ And, $ 2^{\log_3 m} = 2^{x} $

So, $ m^{\log_3 2} + 2^{\log_3 m} = 2^x + 2^x = 2 \cdot 2^x = 8 \implies 2^{x+1} = 8 \implies 2^{x+1} = 2^3 \implies x+1=3 \implies x=2 $.

Therefore, $ m = 3^{2} = 9 $.

Answer: $ m = 9 $

Solution 📝: First, $ f(x) = \sqrt{9 - x^2} $.

Calculate $ f(2) = \sqrt{9 - 4} = \sqrt{5} $.

We need $ \lim_{x \to 2} \frac{\sqrt{9-x^2} - \sqrt{5}}{x-2} $.

Let $ L = \lim_{x \to 2} \frac{\sqrt{9-x^2} - \sqrt{5}}{x-2} $.

Multiply numerator and denominator by the conjugate:

But $ 4 - x^2 = (2 - x)(2 + x) $.

So numerator: $ (2-x)(2+x) = -(x-2)(2+x) $.

Thus,

As $ x \to 2 $, this becomes:

Value required: $ \sqrt{5} \times L = \sqrt{5} \times \left(-\frac{2}{\sqrt{5}}\right) = -2 $.

Answer: $-2$

Solution 📝:

Let’s analyze the piecewise function:

• The left-hand limit as $ x \to 0^- $ is $ [0+1] = [^1] = 1 $.
• The right-hand limit as $ x \to 0^+ $ is $ {0+1} = {1} = 0 $.

So LHL ≠ RHL, both exist but are not equal.

Answer: FALSE

Solution 📝:

From above:

• LHL at $ x=0 $ is 1.
• RHL at $ x=0 $ is 0.
• $ f(0) = 0 $.

Since LHL ≠ RHL, and even RHL $ = f(0) $, but for continuity both limits must be equal and equal to $ f(0) $.

Answer: FALSE

Solution 📝:

Discontinuity can occur at endpoints and where the floor or fractional part “jumps”, i.e., at integers for the floor function and fractional part.

• In (-3, 0): Discontinuity at $ x = -2, -1, 0 $ (from greatest integer function).
• In (0, 3): Discontinuity at $ x = 0, 1, 2, 3 $ (from fractional part function).
• So, discontinuities occur at $ x = -2, -1, 0, 1, 2 $.

Total: 5 points

Answer: 5

Solution 📝:

At $ x = -1 $, the derivative of $ |x+1| $ changes abruptly:

• For $ x < -1 $: $ |x+1| = -(x+1) $
• For $ x > -1 $: $ |x+1| = (x+1) $

Their derivatives at $ x = -1 $ are different, hence LHD ≠ RHD.

Answer: FALSE

Solution 📝:

At $ x = -1 $: $ |x+1| = 0 $, so $ |x+1| f(x) = 0 $ for any $ f(x) $. Both side-limits as $ x \to -1 $ approach 0.

Answer: TRUE

Solution 📝:

From earlier:

• LHL ($ x = 0^- $): Slope from left = derivative of [x+1] = 0 almost everywhere (except at discontinuities), but as left from $[-3,0)$, $[x+1]$ is constant near 0 so derivative = 0.
• RHL ($ x=0^+ $): Derivative of $ {x+1} $ with respect to $ x $ is 1 for non-integer $ x $, since $ {x+1} = x+1 - [x+1] $ and derivative of $ x+1 $ is 1.
• But at $ x=0 $, function is not continuous or differentiable; so derivative does not exist.

Answer: FALSE

Solution 📝:

Left-hand limit: As $ x \to 0^- $, $ e^{(0)^2} + 3 = 1 + 3 = 4 $.

Right-hand limit: As $ x \to 0^+ $, $ (0)^2 - 2 \times 0 + 4 = 4 $.

Both limits = 4

Answer: 4

Solution 📝:

Substitute $ x=1 $: Numerator: $ 1^4 - 3 \cdot 1^3 + 2 = 1 - 3 + 2 = 0 $ Denominator: $ 1^4 - 5 \cdot 1^3 + 3 \cdot 1^2 + 1 = 1 - 5 + 3 + 1=0 $

We get $ 0/0 $, so use L’Hospital’s Rule: Differentiate numerator and denominator:

Numerator derivative: $ 4x^3 - 9x^2 $, at $ x=1 $: $ 4-9 = -5 $ Denominator derivative: $ 4x^3 - 15x^2 + 6x $, at $ x=1 $: $ 4-15+6= -5 $ So, limit = $ \frac{-5}{-5} = 1 $

Answer: 1

Step-by-step Solution 🧮:

1️⃣ Numerator simplification: $6 + 6 \cdot 2^2 + 6 \cdot 3^2 + \dots + 6 \cdot n^2 = 6 \sum_{k=1}^{n} k^2$ Recall: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$

So, numerator = $6 \cdot \frac{n(n+1)(2n+1)}{6} = n(n+1)(2n+1)$

2️⃣ Denominator: $\sqrt{4n^6 + 5}$

3️⃣ Limit:

The numerator is a cubic polynomial: $n(n+1)(2n+1) \approx 2n^3$ for large $n$

Denominator: $\sqrt{4n^6 + 5} \approx 2n^3$

So the limit is:

Final Answer: 1

Step-by-step Solution 🧮:

1️⃣ For large $ n $, the highest power dominates.

• Numerator: $100n^2 - 11 \approx 100n^2$
• Denominator: $100n^3 + 7 \approx 100n^3$

2️⃣ So,

Final Answer: 0

Step-by-step Solution 🧑🏫:

• Linear approximation at $a$: $L(x) = f(a) + f’(a)(x - a)$

Given $f(x) = x e^x - 1$

• $f(a) = a e^a - 1$
• $f’(x) = e^x + x e^x$, so $f’(a) = e^a + a e^a = e^a(a+1)$

Thus,

Expand:

Which matches the given formula.

Final Answer: TRUE

Step-by-step Solution 🚦:

• The tangent at $ (3,0) $ passes through $ (5,4) $.
• Slope of tangent = $ m = \frac{4-0}{5-3} = 2 $

By definition, the slope of the tangent to the graph at $ x=3 $ is $ f’(3) $.

Final Answer: 2

Step-by-step Solution 📝:

Let’s match:

• $ f \circ g (x) = f(g(x)) = \sqrt{\sqrt{3-x}} $ → $ \sqrt{3-x} = d $, domain for $ 3-x \geq 0 \implies x \leq 3 $, so domain is $ ( -\infty, 3 ] $.
• $ g \circ f (x) = g(f(x)) = \sqrt{3-\sqrt{x}} $, which is $ a $, and the inside must be $ \geq 0 $: $ 3-\sqrt{x} \geq 0 \implies x \leq 9 $, so domain is $ $.
• $ f \circ f (x) = f(f(x)) = \sqrt{\sqrt{x}} = \sqrt[^1]{x} $, which is $ b $, and $ x \geq 0 $, so domain is $ [0, \infty) $.

So, the matching is:

• i) – d) – 3)
• ii) – a) – 4)
• iii) – b) – 1)

Answers: i) – d) – 3) ii) – a) – 4) iii) – b) – 1)