Let f(x)= cases(1+ax)^1/x, & x0 cases be continuous at x=0. Then e^a bc is equal to:

Question

Let f(x)= cases(1+ax)^1/x, & x<0\ 1+b, & x=0\ (x+4)^1/2-2(x+c)^1/3-2, & x>0 cases be continuous at x=0. Then e^a bc is equal to:

Accepted Answer

Correct answer: B. (b) f (x )=( (1+ ax )^1/x ,x<01+b ,x=0 (x+4 )^1/2-2 (x+c )^1/3-2 ,x>0) Hence, f (0 )=1+6 RHL= limx 0^+ 0.05em (x+4 )^1/2-2 (x+c )^1/3-2 ( For 00 form ,c=8) = limx 0 0.05em 2(1+ x4)^1/2-2(1+ x8)^1/3-2 = limx 0 0.05em 1+ x8-11+ x8 13-1= 18 18 13=3 ∴1+b=3 ⇒b=2 LHL= limx 0 (1+ ax )^1/x= limx 0 0.05eme^ 1+ ax-1x=e^a=3 ∴e^a bc=3 2.8=48

$Let f (x) = ⎩ ⎨ ⎧ (1 + a x)^{1/ x}, 1 + b, \frac{( x + 4 ) ^{1/2} - 2}{( x + c ) ^{1/3} - 2}, x < 0 x = 0 x > 0 be continuous at x = 0 . Then e^{a} b c is equal to:$

Solution

More Limits PYQs