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If `a le 0 f(x) =e^(ax)+e^(-ax)` and S={x:f(x) is monotonically increasing then S equalsA. `f {x:x gt 0 } `B. `{x:x lt 0}`C. `{x:x lt 1}`D. `{x:x lt 1}` |
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Answer» Correct Answer - A We have `f(x)=e^(ax)+e^(-ax) rArr f(x) =a(e^(ax)-e^(-ax))` For f(x) to be increasing we must have `f(x) gt 0 ` `rArr a(e^ax-e^(-ax)) gt 0 ` `rArr a(e^(ax)-e^(-ax)) gt 0` `rArr e^(-bx)-e^(bx) lt 0 ` where a = -b and `b gt 0 ` `rArr e^(bx)-e^(-bx) gt 0 rArr x lt 0 ` Hence `S= {x: x gt 0 }` |
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