CAT 2017 Question Paper Slot 2

It is given that $$9^{x-\frac{1}{2}}-2^{2x-2}=4^{x}-3^{2x-3}$$

Let us try to reduce them to powers of $$3$$ and $$2$$
The given equation can be reduced to $$3^{2x-1} + 3^{2x-3} = 2^{2x} + 2^{2x-2}$$

Hence, $$3^{2x-3} \times 10 = 2^{2x-2} \times 5$$
Therefore, $$3^{2x-3} = 2^{2x-3}$$

This is possible only if $$2x-3=0$$ or $$x=3/2$$

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{2}\times3^{3}\times5) + log(2^{6}\times3\times5^{7} ) + log(2 \times3^{2}\times5^{4} ) }{3} $$

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{2+6+1}\times3^{3+1+2}\times5^{1+7+4}) }{3} $$

$$log(2^{a}\times3^{b}\times5^{c} )$$ = $$ \frac{log ( 2^{9}\times3^{6}\times5^{12}) }{3} $$

$$3log(2^{a}\times3^{b}\times5^{c} )$$ = $$ log ( 2^{9}\times3^{6}\times5^{12}) $$
Hence, 3a = 9 or a = 3

Sum of the sequence of even numbers is $$2a_{3} + (2a_{3} - 2) + (2a_{3} - 4)$$ $$+ (2a_{3} - 6) + (2a_{3} - 8) = 450$$
=> $$10a_{3} - 20 = 450$$
=> $$a_{3} = 47$$
Hence $$a_{5} = 47 + 4 = 51$$

$$\frac{1}{a}+\frac{1}{b}=\frac{1}{9}$$
=> $$ab = 9(a + b)$$
=> $$ab - 9(a+b) = 0$$
=> $$ ab - 9(a+b) + 81 = 81$$
=> $$(a - 9)(b - 9) = 81, a > b$$
Hence we have the following cases,
$$ a - 9 = 81, b - 9 = 1$$ => $$(a,b) = (90,10)$$
$$ a - 9 = 27, b - 9 = 3$$ => $$(a,b) = (36,12)$$
$$ a - 9 = 9, b - 9 = 9$$ => $$(a,b) = (18,18)$$
Hence there are three possible positive integral values of (a,b)

After Amal, Bimal and Kamal are given their minimum required pens, the pens left are 8 - (1 + 2 + 3) = 2 pens
Now these two pens have to be divided between three persons so that each person can get zero pens = $$^{2+3-1}C_{3-1}$$ = $$^4C_2$$  = 6

For the number to be divisible by 6, the sum of the digits should be divisible by 3 and the units digit should be even. Hence we have the digits as

Case I: 2, 3, 4, 6
Now the units place can be filled in three ways (2,4,6), and the remaining three places can be filled in 3! = 6 ways.
Hence total number of ways = 3*6 = 18

Case II: 0, 2, 3, 4
case II a: 0 is in the units place => 3! = 6 ways
case II b: 0 is not in the units place => units place can be filled in 2 ways( 2,4) , thousands place can be filled in 2 ways ( remaining 3 - 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 2 * 2 * 2 = 8
Total number of ways in this case = 6 + 8 = 14 ways.

Case III: 0, 2, 4, 6
case III a: 0 is in the units place => 3! = 6 ways

case II b: 0 is not in the units place => units place can be filled in 3 ways( 2,4,6) , thousands place can be filled in 2 ways (remaining 3 - 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 3 * 2 * 2 = 12
Total number of ways in this case = 6 + 12= 18 ways.

Hence the total number of ways = 18 + 14  + 18 = 50 ways

f(1 * 1) = f(1)f(1)
=> f(1) = f(1)f(1)
=> f(1) = 0 or f(1) = 1
Hence maximum value of f(1) is 1

$$|f(x)+ g(x)| = |f(x)| + |g(x)|$$ if and only if

case 1: $$f(x) \geq 0$$ and $$g(x) \geq 0$$
<=> $$ 2x-5 \geq 0 $$ and $$7-2x \geq 0$$
<=> $$ x \geq \frac{5}{2}$$ and $$ \frac{7}{2} \geq x$$ 

<=> $$\frac{5}{2}\leq x\leq\frac{7}{2}$$

case 2: $$f(x) \leq 0$$ and $$g(x) \leq 0$$

<=> $$ 2x-5 \leq 0 $$ and $$7-2x \leq 0$$
<=> $$ x \leq \frac{5}{2}$$ and $$ \frac{7}{2} \leq x$$
 So x<=5/2 and x>=7/2 which is not possible.

Hence, answer is 

<=> $$\frac{5}{2}\leq x\leq\frac{7}{2}$$

Let the common ratio of the G.P. be r.
Hence we have $$a_n= 3(a_{n+1}+ a_{n+2} + ...)$$

The sum up to infinity of GP is given by $$\frac{a}{1-r}$$ where a here is $$a_{n+1}$$

=> $$a_n= 3(\frac{a_{n+1}}{1-r})$$
=> $$a_n= 3(\frac{a_{n}\times r}{1-r})$$
=> $$ r = \frac{1}{4}$$
Now, $$a_1+a_2+a_2...+=32$$
=> $$\frac{a_1}{1-r} = 32$$
=> $$\frac{a_1}{3/4} = 32$$
=> $$a_1 = 24$$

$$a_5 = a_1 \times r^4$$
$$a_5 = 24 \times (1/4)^4 = \frac{3}{32}$$

$$a_{100} = \frac{1}{ (3\times100 -1) \times (3\times100 + 2)}= \frac{1}{ 299 \times 302}$$

$$\frac{1}{2\times5} = \frac{1}{3} \times (\frac{1}{2} - \frac{1}{5})$$

$$\frac{1}{5\times8} = \frac{1}{3} \times (\frac{1}{5} - \frac{1}{8})$$

$$\frac{1}{8\times11} = \frac{1}{3} \times (\frac{1}{8} - \frac{1}{11})$$
....

$$\frac{1}{299\times302} = \frac{1}{3} \times (\frac{1}{299} - \frac{1}{302})$$

Hence $$a_{1}+a_{2}+a_{3}+...+a_{100}$$ = $$\frac{1}{3} \times (\frac{1}{2} - \frac{1}{5})$$ + $$\frac{1}{3} \times (\frac{1}{5} - \frac{1}{8})$$ + $$\frac{1}{3} \times (\frac{1}{8} - \frac{1}{11})$$ + ... + $$\frac{1}{3} \times (\frac{1}{299} - \frac{1}{302})$$

= $$\frac{1}{3} \times (\frac{1}{2} - \frac{1}{302})$$

= $$\frac{25}{151}$$