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The value of a for which the function f(x)=axf(x) = a^x is increasing on R are given by :

Solution

✅ Correct Option: 3

f(x)=ax⇒f′(x)=axln⁡af(x) = a^x \Rightarrow f'(x) = a^x \ln a.

For ff to be increasing on R, f′(x)>0f'(x) > 0 for all x.

Since ax>0a^x > 0 for a>0a > 0, we need ln⁡a>0\ln a > 0, i.e., a>1a > 1.

Note: We have selected option (3) based on the strict mathematical definition of increasing. The original key marks (1), but for 0<a<10 < a < 1 the function is decreasing, so 1<a<∞1 < a < \infty is the correct range.

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