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1

### AIEEE 2002

The value of $$\,{2^{1/4}}.\,\,{4^{1/8}}.\,{8^{1/16}}...\infty$$ is
A
1
B
2
C
3/2
D
4

## Explanation

The product is $$p = {2^{1/4}}{.2^{2/8}}{.2^{3/16}}........$$

$$= {2^{1/4 + 2/8 + 3/16 + .......\infty }}$$

Now let

$$S = {1 \over 4} + {2 \over 8} + {3 \over {16}} + .......\infty \,\,\,\,........\left( 1 \right)$$

$${1 \over 2}S = {1 \over 8} + {2 \over {16}} + .......\infty \,\,\,\,........\left( 2 \right)$$

Subtracting $$(2)$$ from $$(1)$$

$$\Rightarrow {1 \over 2}S = {1 \over 4} + {1 \over 8} + {1 \over {16}} + .......\infty$$

or $${1 \over 2}S = {{1/4} \over {1 - 1/2}} = {1 \over 2} \Rightarrow S = 1$$

$$\therefore$$ $$P = {2^S} = 2$$
2

### AIEEE 2002

l, m, n are the $${p^{th}}$$, $${q^{th}}$$ and $${r^{th}}$$ term of a G.P all positive, $$then\,\left| {\matrix{ {\log \,l} & p & 1 \cr {\log \,m} & q & 1 \cr {\log \,n} & r & 1 \cr } } \right|\,equals$$
A
- 1
B
2
C
1
D
0

## Explanation

$$l = A{R^{p - 1}}$$

$$\Rightarrow \log 1 = \log A + \left( {p - 1} \right)\log R$$

$$m = A{R^{q - 1}}$$

$$\Rightarrow \log m = \log A + \left( {q - 1} \right)\log R$$

$$n = A{R^{r - 1}}$$

$$\Rightarrow \log n = \log A + \left( {r - 1} \right)\log R$$

Now, $$\left| {\matrix{ {\log l} & p & 1 \cr {\log m} & q & 1 \cr {\log n} & r & 1 \cr } } \right|$$

$$= \left| {\matrix{ {\log A + \left( {p - 1} \right)\log R} & p & 1 \cr {\log A + \left( {q - 1} \right)\log R} & q & 1 \cr {\log A + \left( {r - 1} \right)\log R} & r & 1 \cr } } \right|$$

Operating $${C_1} - \left( {\log R} \right){C_2} + \left( {\log R - \log A} \right){C_3}$$

$$= \left| {\matrix{ 0 & p & 1 \cr 0 & q & 1 \cr 0 & r & 1 \cr } } \right| = 0$$
3

### AIEEE 2002

If 1, $${\log _9}\,\,({3^{1 - x}} + 2),\,\,{\log _3}\,\,({4.3^x} - 1)$$ are in A.P. then x equals
A
$${\log _3}\,4\,\,\,$$
B
$$1 - \,{\log _3}\,4\,$$
C
$$1 - \,{\log _4}\,3$$
D
$${\log _4}\,3$$

## Explanation

$$1,\,{\log _9}\left( {{3^{1 - x}} + 2} \right),{\log _3}\left( {{{4.3}^x} - 1} \right)$$ are in $$A.P.$$

$$\Rightarrow 2{\log _9}\left( {{3^{1 - x}} + 2} \right)$$

$$\,\,\,\,\,\,\,\,\,$$ $$= 1 + {\log _3}\left( {{{4.3}^x} - 1} \right)$$

$$\Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)$$

$$\,\,\,\,\,\,\,\,\,$$ $$= {\log _3}3 + {\log _3}\left( {{{4.3}^x} - 1} \right)$$

$$\Rightarrow {\log _3}\left( {{3^{1 - x}} + 2} \right)$$

$$\,\,\,\,\,\,\,\,\,$$ $$= {\log _3}\left[ {3\left( {{{4.3}^x} - 1} \right)} \right]$$

$$\Rightarrow {3^{1 - x}} + 2 = 3\,\left( {{{4.3}^x} - 1} \right)$$

$$\Rightarrow {3.3^{ - x}} + 2 = {12.3^x} - 3.$$

Put $${3^x} = t$$

$$\Rightarrow {3 \over t} + 2 = 12t - 3$$

or $$12{t^2} - 5t - 3 = 0;$$

Hence $$t = - {1 \over 3},{3 \over 4} \Rightarrow {3^x} = {3 \over 4}$$

(as $${3^x}\,\, \ne \,\, - ve$$ )

$$\Rightarrow x = {\log _3}\left( {{3 \over 4}} \right)$$

or $$x = {\log _3}3 - {\log _3}4$$

$$\Rightarrow x = 1 - {\log _3}4$$

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