answer:Part (a) 1. Let 1 leq p leq q leq r leq n be indices such that: [ d = d_q, quad a_p = max left{a_j : 1 leq j leq qright}, quad a_r = min left{a_j : q leq j leq nright} ] Thus, d = a_p - a_r. Note that these indices are not necessarily unique. 2. For arbitrary real numbers x_1 leq x_2 leq ldots leq x_n, we particularly consider the quantities |x_p - a_p| and |x_r - a_r|. 3. Considering the following sum: [ (a_p - x_p) + (x_r - a_r) = (a_p - a_r) + (x_r - x_p) geq a_p - a_r = d, ] since x_r geq x_p due to the non-decreasing order of x_i. 4. Consequently, we have: [ text{Either} quad a_p - x_p geq frac{d}{2} quad text{or} quad x_r - a_r geq frac{d}{2}. ] 5. This leads to the following inequality: [ max left{ |x_i - a_i| : 1 leq i leq n right} geq max left{ |x_p - a_p|, |x_r - a_r| right} geq max left{ a_p - x_p, x_r - a_r right} geq frac{d}{2}. ] Part (b) 1. Define the sequence (x_k) as follows: [ x_1 = a_1 - frac{d}{2}, quad x_k = max left{ x_{k-1}, a_k - frac{d}{2} right} quad text{for } 2 leq k leq n. ] 2. By definition, the sequence (x_k) is non-decreasing, and for all 1 leq k leq n, we have: [ x_k - a_k geq -frac{d}{2}. ] 3. Now we need to show that: [ x_k - a_k leq frac{d}{2} quad text{for all } 1 leq k leq n. ] 4. Consider an arbitrary index 1 leq k leq n. Let ell leq k be the smallest index such that x_k = x_ell. There are two cases: - If ell = 1, then x_ell = a_1 - frac{d}{2}. - If ell geq 2 and x_ell > x_{ell-1}, then x_ell = a_ell - frac{d}{2}. 5. Thus, we have: [ x_k = x_ell = a_ell - frac{d}{2}. ] 6. Since: [ a_ell - a_k leq max left{ a_j : 1 leq j leq k right} - min left{ a_j : k leq j leq n right} = d_k leq d, ] it follows that: [ x_k - a_k = a_ell - a_k - frac{d}{2} leq d - frac{d}{2} = frac{d}{2}. ] 7. Consequently, we have: [ -frac{d}{2} leq x_k - a_k leq frac{d}{2} quad text{for all } 1 leq k leq n. ] 8. Therefore: [ max left{ |x_i - a_i| : 1 leq i leq n right} leq frac{d}{2}. ] 9. Equality is achieved because: [ |x_1 - a_1| = frac{d}{2}. ] # 2 1. Define: [ M_i = max left{ a_j : 1 leq j leq i right}, quad m_i = min left{ a_j : i leq j leq n right} ] 2. Note that M_i leq M_{i+1} and m_i leq m_{i+1} which means the sequences (M_i) and (m_i) are non-decreasing. 3. Furthermore, for all 1 leq i leq n: [ m_i leq a_i leq M_i. ] 4. Set: [ x_i = frac{M_i + m_i}{2}. ] 5. This sequence (x_i) is non-decreasing since both M_i and m_i are non-decreasing. 6. Using the definition d_i = M_i - m_i for all 1 leq i leq n, we get: [ -frac{d_i}{2} = frac{m_i - M_i}{2} = x_i - M_i leq x_i - a_i leq x_i - m_i = frac{M_i - m_i}{2} = frac{d_i}{2}. ] 7. Thus: [ max left{ |x_i - a_i| : 1 leq i leq n right} leq max left{ frac{d_i}{2} : 1 leq i leq n right} = frac{d}{2}. ] 8. Since the opposite inequality has been established in part (a), this confirms that equality is achieved. blacksquare

answer:If Jason needs 7 shelves and each shelf holds 45 books, then the total number of books Jason has is: 7 shelves * 45 books per shelf = 315 books Jason has boxed{315} books.

answer:Start by re-writing the given equation using the identities for sec x and tan x, so: [ frac{1}{cos x} - frac{sin x}{cos x} = frac{4}{3} implies frac{1 - sin x}{cos x} = frac{4}{3} ] Multiply both sides by cos x to get: [ 1 - sin x = frac{4}{3} cos x ] Now, squaring both sides: [ (1 - sin x)^2 = left(frac{4}{3} cos xright)^2 implies 1 - 2sin x + sin^2 x = frac{16}{9} cos^2 x = frac{16}{9}(1 - sin^2 x) ] After simplifying: [ 1 - 2sin x + sin^2 x = frac{16}{9} - frac{16}{9} sin^2 x ] Re-arranging and combining all terms involving sin x: [ frac{25}{9} sin^2 x - 2 sin x - frac{7}{9} = 0 ] Solving this quadratic in sin x, transform and solve for cos x: [ sin x = frac{9 pm sqrt{361}}{50} = frac{9 pm 19}{50} ] So, we have sin x = frac{28}{50} = frac{14}{25} or sin x = -frac{10}{50} = -frac{1}{5}. Now, cos x = sqrt{1 - sin^2 x} hence: [ cos x = sqrt{1 - left(frac{14}{25}right)^2} = sqrt{frac{625 - 196}{625}} = sqrt{frac{429}{625}} = frac{sqrt{429}}{25} ] For sin x = -frac{1}{5}, cos x = sqrt{1 - left( -frac{1}{5} right)^2} = sqrt{1 - frac{1}{25}} = sqrt{frac{24}{25}} = frac{sqrt{24}}{5}. Thus, the possible values of cos x are boxed{frac{sqrt{429}}{25}} and boxed{frac{sqrt{24}}{5}}.

answer:1. First, determine the common difference d of the sequence. The difference between consecutive terms is 94 - 100 = -6. 2. The n^text{th} term of the sequence can be expressed as a_n = 100 - 6(n - 1) = 106 - 6n. 3. Set a_n = 4 and solve for n. Thus, we have: [ 106 - 6n = 4 ] [ 6n = 106 - 4 = 102 ] [ n = frac{102}{6} = 17 ] Therefore, 4 is the 17^text{th} term in this sequence. 4. To find how many terms appear before 4, calculate 17 - 1 = 16. Hence, boxed{16} terms appear before 4 in this sequence.