x∈R, then find ab. Suppose f(x)=eax+ebx where a≠b and f′′(x)−2f1(x)−15f(x)=0 for all

f(x)=aex+bex⇒f′(x)=aeax+bebx,f"(x)=a2eax+b2ebxa2eax+b2ebx−2aeax+bebx−15eax+ebx=0a2−2a−15eax+b2−2b−15ebx=0 ∀xa2−2a−15=0 b2−2b−15=0a=-5, b=5,b=−3a≠bab=5(−3)=−15

x∈R, then find ab. Suppose f(x)=eax+ebx where a≠b and f′′(x)−2f1(x)−15f(x)=0 for all

f(x)=aex+bex⇒f′(x)=aeax+bebx,f"(x)=a2eax+b2ebxa2eax+b2ebx−2aeax+bebx−15eax+ebx=0a2−2a−15eax+b2−2b−15ebx=0 ∀xa2−2a−15=0 b2−2b−15=0a=-5, b=5,b=−3a≠bab=5(−3)=−15

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

$x \in R, then find ab.$ $Suppose f (x) = e^{a x} + e^{b x} where a \neq b and f^{''} (x) - 2 f^{1} (x) - 15 f (x) = 0 for all$

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

-15

-10

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = a e^{x} + b e^{x} \Rightarrow f^{'} (x) = a e^{a x} + b e^{b x}, f^{"} (x) = a^{2} e^{a x} + b^{2} e^{b x} \\ (a^{2} e^{a x} + b^{2} e^{b x}) - 2 (a e^{a x} + b e^{b x}) - 15 (e^{a x} + e^{b x}) = 0 \end{array}$