If - a x -2-8- a x -5- then x - where a -1- -A- - empty set B- RC- 1- -D- -1- -

The correct option is A ϕ (empty set)Assume a^x=t Then |t 2|+|8 t| lt;5 nbsp; Minimum value of |a|+|b| is |a+b| ⇒min (|t 2|+|8 t|)=|t 2+8 t|=|6| gt;5 Hen ...

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Standard XI

Mathematics

Logarithms

If | a x -2|+...

Question

If $| a^{x} - 2 | + | 8 - a^{x} | < 5$ , then $x \in$ ____ , where $a \in (1, \infty)$

ϕ (empty set)

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(1, \infty)

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Solution

The correct option is A $ϕ (empty set)$
Assume $a^{x} = t$
Then $| t - 2 | + | 8 - t | < 5$
Minimum value of $| a | + | b | is | a + b |$
$\Rightarrow min (| t - 2 | + | 8 - t |) = | t - 2 + 8 - t | = | 6 | > 5$
Hence, no solution exists.

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Logarithms

Standard XI Mathematics

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If |ax−2|+|8−ax|<5 , then x∈____ , where a∈(1,∞)

The correct option is A ϕ (empty set)Assume ax=t Then |t−2|+|8−t|<5 Minimum value of |a|+|b| is |a+b| ⇒min(|t−2|+|8−t|)=|t−2+8−t|=|6|>5 Hence, no solution exists.

If $| a^{x} - 2 | + | 8 - a^{x} | < 5$ , then $x \in$ ____ , where $a \in (1, \infty)$

The correct option is A $ϕ (empty set)$
Assume $a^{x} = t$
Then $| t - 2 | + | 8 - t | < 5$
Minimum value of $| a | + | b | is | a + b |$
$\Rightarrow min (| t - 2 | + | 8 - t |) = | t - 2 + 8 - t | = | 6 | > 5$
Hence, no solution exists.