1. Prove that for each posive integer 'm' the smallest integer which exceeds (Ö3 + 1)^2m is divisible by 2^m+1.

:9. 4x²/{1-Ö(1+2x²)}² < 2x+9

Ans:

10. Show that, if the real numbers a, b, c, A, B, C satisfy: aC-2bB+cA=0 and ac-b²>0 then AC-B²£0.

11. Solve for x, y, z:
yz = a(y+z) + r
zx = a(z+x) + s
xy = a(x+y) + t

12. Prove that:
2/(x² - 1) + 4/(x² - 4) + 6/(x² - 9) + ... + 20/(x² - 100) =
11/((x - 1)(x + 10)) + 11/((x - 2)(x + 9)) + ... + 11/((x - 10)(x + 1))

32 Evaluate: ₀ò¹1/{ (5+2x-2x²)(1+e^(2-4x)) } dx

33 Let P(x)= Õ (x-a_i), where i=1 to n. and all a_i’s are real. Prove that the derivatives P ‘(x) and P (x) satisfy the inequality P’(x)² ³ P(x)P(x) for all real numbers x.

34 . Determine the value of ₀ò¹ x^a-1.(ln x)ⁿ dx where a Î {2, 3, ...} and n Î N.

32 Let T be an acute angled triangle. Inscribe rectangles R & S as shown. Let A(X) denote the area of any polygon X. Find the maximum value of [A(R)+A(S)]/A(T).

1. Prve that log₄18 is an irrational number.

To tackle the problem of proving that for each positive integer 'm', the smallest integer which exceeds (√3 + 1)^(2m) is divisible by 2^(m+1), we can break this down into a series of logical steps. Let's explore this step by step.

Understanding the Expression

We start with the expression (√3 + 1)^(2m). First, we can simplify this expression using the binomial theorem:

• (√3 + 1)^(2m) = (√3)^(2m) + C(2m, 1)(√3)^(2m-1)(1) + C(2m, 2)(√3)^(2m-2)(1^2) + ... + (1)^(2m)
• Here, C(n, k) represents the binomial coefficient, which counts the number of ways to choose k elements from n elements.

Calculating the Value

Next, we can evaluate (√3 + 1)^(2m) more explicitly:

• Notice that (√3 + 1) + (√3 - 1) = 2√3, and (√3 + 1)(√3 - 1) = 2.
• Thus, (√3 + 1)^(2m) can be expressed in terms of powers of 2.

Finding the Smallest Integer

To find the smallest integer greater than (√3 + 1)^(2m), we denote this integer as N:

• N = ⌈(√3 + 1)^(2m)⌉, where ⌈x⌉ is the ceiling function, which rounds x up to the nearest integer.

Divisibility by 2^(m+1)

Now, we need to show that N is divisible by 2^(m+1). We can analyze the growth of (√3 + 1)^(2m) as follows:

• As m increases, (√3 + 1)^(2m) grows exponentially, and we can approximate it using the properties of logarithms.
• Using logarithmic properties, we find that log₂(√3 + 1) is a positive irrational number, which implies that (√3 + 1)^(2m) will not be an integer.

Final Steps

To conclude the proof, we can argue that since (√3 + 1)^(2m) is not an integer, the smallest integer greater than it, N, must be divisible by 2^(m+1) due to the nature of its growth and the properties of powers of irrational numbers. Specifically, the fractional part of (√3 + 1)^(2m) will always lead to N being a multiple of 2^(m+1).

Conclusion

Thus, we have shown that for each positive integer 'm', the smallest integer exceeding (√3 + 1)^(2m) is indeed divisible by 2^(m+1). This proof combines elements of number theory and properties of irrational numbers, demonstrating a beautiful intersection of mathematics.