Tip:A–D to answerE for explanationV for videoS to reveal answer
If f(x) = \frac{(\tan 1^\circ)x + \log_e(123)}{x \log_e(1234) - (\tan 1^\circ)}, x > 0 then the least value of f(f(x))+f(f(x4)) is:
- A.
2
- B.
4
(Correct Answer) - C.
8
- D.
0
Explanation
Since, tan1∘,loge123 and loge1234 are constants, so let a=tan1∘,b=loge123 and c=loge1234.
So
f(x)=xloge1234−(tan1∘)(tan1∘)x+loge123=cx−aax+b
Now
f(f(x))=c(cx−aax+b)−aa(cx−aax+b)+b=acx+bc−acx+a2a2x+ab+bcx−ab
and
So using A.M. ≥ G.M.
2f(f(x))+f(f(x4))≥f(f(x))×f(f(x4))
⇒f(f(x))+f(f(x4))≥2x×x4=4
Hence least value of
Explanation
Since, tan1∘,loge123 and loge1234 are constants, so let a=tan1∘,b=loge123 and c=loge1234.
So
f(x)=xloge1234−(tan1∘)(tan1∘)x+loge123=cx−aax+b
Now
f(f(x))=c(cx−aax+b)−aa(cx−aax+b)+b=acx+bc−acx+a2a2x+ab+bcx−ab
and
So using A.M. ≥ G.M.
2f(f(x))+f(f(x4))≥f(f(x))×f(f(x4))
⇒f(f(x))+f(f(x4))≥2x×x4=4
Hence least value of
