Solution 📝:

Let’s analyze the matching options:

• $ iv) $ refers to one of the composite functions.
• The match “c” and “2” goes together.

Given the answer key, the correct match is: Option 3: $ iv) - c) - 2) $

So the correct answer is Option 3.

Solution 📝:

1️⃣ The domain requires $ x+1 > 0 \implies x > -1 $, so domain is $ (-1, \infty) $.

2️⃣ $ (-\infty, 1) $ is not correct.

3️⃣ For $ x \in (-1,1) $, $ \log(x+1) $ is not one-one because the modulus will make it symmetric.

4️⃣ Thus, it is not one-one.

So, Options 1 and 3 are correct.

Solution 📝:

For $ x = 0 $:

• $ f(0) = 1 $
• The limit $ \lim_{x \to 0} \frac{\sin x}{x} = 1 $ (So, function is continuous)
• The derivative at 0 is:

Use Taylor expansion: $ \sin h = h - \frac{h^3}{6} + O(h^5) $, so $ \frac{\sin h}{h} - 1 \approx -\frac{h^2}{6} $, thus $ \frac{-h^2/6}{h} = -h/6 \to 0 $. So $ f’(0) = 0 $

So, true statements:

• $ f $ is differentiable everywhere, including at $ x = 0 $, and $ f’(0) = 0 $.

Correct options: 1 and 3.

Solution 📝:

Let’s analyze:

• $ f’(x) = 5a_5x^4 + 4a_4x^3 + 3a_3x^2 + 2a_2x + a_1 $
• At $ x=0 $, $ f’(0) = a_1 $ — correct.
• Substitute $ x=1 $ and $ x=-1 $, combine and solve: Given answer key suggests that options 1, 2, 3 are correct.

Correct options: 1, 2, 3.

Solution 📝:

Let $ x = \log_3 m $.

• $ m^{\log_3 2} = (3^x)^{\log_3 2} = 2^x $
• $ 2^{\log_3 m} = 2^x $
• $ 2^x + 2^x = 8 \implies 2^{x+1} = 8 \implies x+1 = 3 \implies x = 2 \implies m = 3^2 = 9 $.

Final Answer: $ m = 9 $

Solution 📝:

For continuity at $ x = -1 $:

• $ Ax - B $ at $ x=-1 $: $ -A - B $
• $ 2(-1)^2 + 3A(-1) + B $: $ 2 - 3A + B $
• Setting equal: $ -A - B = 2 - 3A + B \implies 2A - 2B = 2 \implies A - B = 1 $ …(1)

For continuity at $ x = 1 $:

• $ 2(1)^2 + 3A(1) + B = 2 + 3A + B $
• $ 4 $ So, $ 2 + 3A + B = 4 \implies 3A + B = 2 $ …(2)

Now solve (1) and (2):

(1): $ A - B = 1 $ (2): $ 3A + B = 2 $

Add: $ (A - B) + (3A + B) = 1 + 2 \implies 4A = 3 \implies A = 3/4 $ Plug into (1): $ 3/4 - B = 1 \implies B = 3/4 - 1 = -1/4 $

$ 6(A+B) = 6(3/4 + (-1/4)) = 6(2/4) = 6 \cdot 1/2 = 3 $

Final Answer: 3

Solution 📝:

• Slope of the tangent at $ x=9 $: $ a = f’(9) = 4 $
• Equation of tangent: $ y = 4x + b $ passes through $ (9, -14) $: $ -14 = 4 \cdot 9 + b \implies b = -14 - 36 = -50 $
• $ a - b = 4 - (-50) = 54 $

Final Answer: 54

Step-by-step Solution 📝:

1️⃣ For option 1: $ f \circ h = f(h(x)) = f(x-1) = (x-1)^2 $. This is not a one-one function because a square function is not one-one.

2️⃣ For option 2: Calculate $ f(f(h(x))) $:

• $ h(x) = x-1 $
• $ f(h(x)) = (x-1)^2 $
• $ f(f(h(x))) = ((x-1)^2)^2 = (x-1)^4 $

So $ f(f(h(x))) \times h(x) = (x-1)^4 \times (x-1) = (x-1)^5 $. This matches the statement.

So only the second option is correct.

Final Answer: Correct option: $ f(f(h(x))) \times h(x) = (x-1)^5 $

Step-by-step Solution 🌀:

Let’s compute: $ f(x) = x^2 $, $ h(x) = x-1 $ $ h(f(x)) = x^2 - 1 $ $ h(h(f(x))) = (x^2 - 1) - 1 = x^2 - 2 $

Set $ x^2 - 2 = 0 \implies x^2 = 2 \implies x = \pm \sqrt{2} $ There are only 2 distinct solutions, not 3.

Final Answer: False. There are only 2 distinct real solutions: $ x = \sqrt{2}, -\sqrt{2} $

Step-by-step Solution 📈:

Focus on the highest degree terms as $n \to \infty$:

Numerator dominant term: $ n^5 $ Denominator dominant term: $ 2n^5 $

So,

All lower-degree and oscillatory terms become negligible.

Final Answer: $ \boxed{0.5} $

Step-by-step Solution 🚀:

For very large $ n $:

• Numerator: $ e^{2n} $ dominates over $ n^4 $
• Denominator: $ e^{3n} $ dominates over $ n^5 $

So,

As $ n \to \infty $, $ e^{-n} \to 0 $

Final Answer: $ \boxed{0} $