menu search
brightness_auto
Feel free to answer or ask any questions.
If you are new here please see how to use or FAQ.
more_vert
If a function f(x) defined by

$f(x) = \begin{cases}ae^x+be^{-x}&,&-1\leq x<1\\cx^2&,&1\leq x\leq3\\ ax^2+2cx&,&3<x\leq4\end{cases}$

be continuous for some a, b, c $\in$ R and f'(0) + f'(2) = e, then the value of a is :

(1) $1\over e^2-3e+13$

(2) $e\over e^2-3e+13$

(3) $e\over e^2-3e-13$

(4) $e\over e^2+3e+13$
thumb_up_off_alt 3 like thumb_down_off_alt 0 dislike

1 Answer

more_vert
 
done_all
Best answer

Ans. (2) $e\over e^2-3e+13$

Sol. Continuous at x = 1, 3

f(1) = f(1+) $\implies$ ae + be–1 = c     .....(1)

f(3) = f(3+) $\implies$ 9c = 9a + 6c

$\implies$ c = 3a      .....(2)

From (1) and (2)

b = ae(3 – e)      .....(3)

$f'(x) = \begin{cases}ae^x-be^{-x}&-1<x<1\\2cx&1<x<3\\ 2ax+2c&3<x<4\end{cases}$

$f'(0)=a-b, f'(2)=4c$

Given f'(0) + f'(2) = e

a – b + 4c = e       .....(4)

by using eq. (1), (2), (3) & (4)

$a={e\over e^2-3e+13}$

thumb_up_off_alt 0 like thumb_down_off_alt 0 dislike
Welcome to Jee Neet QnA, where you can ask questions and receive answers from other members of the community.

1.1k questions

772 answers

366 comments

77 users

...