Get Answers to all your Questions

header-bg qa

If \dpi{100} f(x)=\left ( \frac{3}{5} \right )^{x}+\left ( \frac{4}{5} \right )^{x}-1,x\; \epsilon\; R,  then the equation f(x)=0 has :

Option: 1

no solution


Option: 2

one solution


Option: 3

two solutions


Option: 4

more than two solutions


Answers (1)

best_answer

\text { derivative of } y=\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}-1

\\=\left ( \frac{3}{5}\right )^{x} \ln \left(\frac{3}{5}\right)+\left(\frac{4}{5}\right)^{x} \ln \left(\frac{4}{5}\right)-0\\\\\\ =\ln \left(\frac{3}{5}\right)\left(\frac{3}{5}\right)^{x}+\ln \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)^{x}

\text { derivative of } y>0

Hence it has only one root or none

y=\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}-1=0

\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}=1

\left({3}\right)^{x}+\left({4}\right)^{x}=5^x

From here x=2

Hence B option is true

Posted by

Ajit Kumar Dubey

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE