The domain of f(x)=16log10(5x−x2) is

f(x)=16log10(5x−x2) is defined ifflog10(5x−x2)≥0⇒ 5x−x2≥100⇒ 5x−x2≥1⇒ x2−5x+1≤0∴ x=5±25−42=5±212∴ x∈[5−212, 5+212 ]

The domain of f(x)=16log10(5x−x2) is

f(x)=16log10(5x−x2) is defined ifflog10(5x−x2)≥0⇒ 5x−x2≥100⇒ 5x−x2≥1⇒ x2−5x+1≤0∴ x=5±25−42=5±212∴ x∈[5−212, 5+212 ]

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$T h e d o m a i n o f f (x) = \frac{1}{6} \sqrt{\log_{10} (5 x - x^{2})} i s$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$(0, 5)$

$[\frac{5 - \sqrt{21}}{2}, \frac{5 + \sqrt{21}}{2}]$

$(1, 3)$

$(\frac{5 - \sqrt{21}}{2}, \frac{5 + \sqrt{21}}{2})$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = \frac{1}{6} \sqrt{\log_{10} (5 x - x^{2})} i s d e f i n e d i f f \\ \log_{10} (5 x - x^{2}) \geq 0 \\ \Rightarrow 5 x - x^{2} \geq 10^{0} \\ \Rightarrow 5 x - x^{2} \geq 1 \\ \Rightarrow x^{2} - 5 x + 1 \leq 0 \\ ∴ x = \frac{5 \pm \sqrt{25 - 4}}{2} = \frac{5 \pm \sqrt{21}}{2} \\ ∴ x \in [\frac{5 - \sqrt{21}}{2}, \frac{5 + \sqrt{21}}{2}] \end{array}$