Show that 1 ! · 3 ! · 5 ! · … · 2 n - 1 ! > n ! n

To show that 1 ! 3 ! 5 ! … … 2 n - 1 ! > n ! n For proving this result, let us prove the following. r ! 2 n - r ! n ! > n ! where n > r Now, r ! 2 n - r ! n ! = 1 . 2 . 3 . . . . r 2 n - r 2 n - r - 1 … n + 1 n ! n ! = 1 . 2 . 3 . . . . r 2 n - r 2 n - r - 1 … . . n + 1 = 1 . 2 . 3 . . . . r 2 n - r 2 n - r - 1 … … n + 1 > 1 . 2 . 3 . . . . r r + 1 r + 2 . . . . . . n Since, n > r ⇒ n - r > 0 ⇒ 2 n - r > n ∴ 1 . 2 . 3 … r . r + 1 r + 2 … … > n ! ∴ r ! 2 n - r ! n ! > n ! Now giving r successively the values 1 , 3 , 5 … … 2 n - 1 , we get 1 ! 2 n - 1 ! > n ! 2 3 ! 2 n - 3 ! > n ! 2 Similarly, 2 n - 1 ! 1 ! > n ! 2 Multiplying both sides, we have 1 ! 3 ! 5 ! . . . . . . . 2 n - 1 ! 2 > n ! 2 n Taking square root of both sides, we get 1 ! 3 ! 5 ! … … 2 n - 1 ! > n ! n

If a , b , c , d , … are p positive integers, whose sum is equal to n , show that the least value of a ! b ! c ! d ! … is q ! p - r q + 1 ! r , where q is the quotient and r is the remainder when n is divided by p .

We need to find the least value of a ! b ! c ! d ! … Now, if any two of the quantities, say a and b are not equal, we can make a ! b ! smaller by taking a and b equal and keeping the sum of a and b to be the same, because when they are equal it ensures that the minimum possible value comes otherwise as the value of one of them increases, the overall product will also increase as numbers will keep on multiplying. Similarly, the value of a ! b ! c ! d ! . . . is the least when all the quantities a , b , c , d , . . . are equal. But, here it is given that n is not exactly divisible by p so a , b , c , d are not all equal. Also, given that q is the quotient and r is the remainder when n is divided by p , we have, n = p q + r = p - r q + r q + 1 ∵ adding , subtracting q r Therefore, p - r of the quantities a , b , c , d , . . . . will be equal to q and the remaining r will be equal to ( q + 1 ) , Hence, the least value of the expression is ( q ! ) p - r ( q + 1 ! ) r

If x is positive, show that log e 1 + x < x and > x 1 + x .

If x is positive then to show that log e ( 1 + x ) < x and > x 1 + x We know that, e x = 1 + x + x 2 2 ! + x 3 3 ! + … Since, x > 0 , 1 + x < e x Hence log e ( 1 + x ) < x ∵ log x x = 1 Now, log e 1 + x > x 1 + x , if 1 + x > e x 1 + x We have, 1 + x = 1 1 1 + x = 1 1 - x 1 + x = 1 - x 1 + x - 1 Since, 1 - x - 1 = 1 + x + x 2 + x 3 + . . . ∞ , therefore 1 - x 1 + x - 1 = 1 + x 1 + x + x 2 1 + x 2 + x 3 ( 1 + x ) 3 + … ∞ . . . 1 Also, e x 1 + x = 1 + x x + 1 + x 2 2 ! x + 1 2 + x 3 3 ! x + 1 3 . . . . . . 2 By 1 & 2 , we can observe that apart from the first two terms on the right-hand side, the terms in 2 , keep on getting divided by larger numbers and hence, 1 - x 1 + x - 1 > e x 1 + x that is 1 + x > e x 1 + x or log e 1 + x > x 1 + x .

E M B I B E

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. b.

H S Hall and S R Knight Solutions for Chapter: Inequalities, Exercise 2: EXAMPLES. XIX. b.

Author:H S Hall & S R Knight

H S Hall Mathematics Solutions for Exercise - H S Hall and S R Knight Solutions for Chapter: Inequalities, Exercise 2: EXAMPLES. XIX. b.

Attempt the free practice questions on Chapter 19: Inequalities, Exercise 2: EXAMPLES. XIX. b. with hints and solutions to strengthen your understanding. Higher Algebra solutions are prepared by Experienced Embibe Experts.

Questions from H S Hall and S R Knight Solutions for Chapter: Inequalities, Exercise 2: EXAMPLES. XIX. b. with Hints & Solutions

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. b.>Q 3

Show that $1! \cdot 3! \cdot 5! \cdot \dots \cdot (2 n - 1)! > {(n!)}^{n}$

HARD

JEE Main

IMPORTANT

Higher Algebra>Chapter 19 - Inequalities>EXAMPLES. XIX. b.>Q 5

If $a, b, c, d, \dots$ are $p$ positive integers, whose sum is equal to $n,$ show that the least value of $a! b! c! d! \dots$ is ${(q!)}^{p - r} {((q + 1)!)}^{r}$ , where $q$ is the quotient and $r$ is the remainder when $n$ is divided by $p$ .