Let f,g:[−2,3]→R be continuous functions which are twice differentiable on the interval (−2,3). Let the values of f and g at the points −2,0,1 and 3 be as given in table
x=−2
x=0
x=1
x=3
f(x)
11
-4
6
16
g(x)
1
-2
0
2
In each of the intervals (−2,0),(0,1) and (1,3) the function (f−5g)′′ never vanishes. Then the correct statement(s) is/are
A
f′(x)−5g′(x)=0 has exactly three solutions in(−2,0)∪(0,1)∪(1,3)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
f′(x)−5g′(x)=0 has exactly one solution in (0,1)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
f′(x)−5g′(x)=0 has atleast one solution in (1,3)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
f′(x)−5g′(x)=0 has atleast two solutions in (−2,0)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Cf′(x)−5g′(x)=0 has atleast one solution in (1,3) Let h(x)=f(x)−5g(x) h(x) is continuous, differentiable and f′(x) is also continuous, differentiable h(−2)=h(0)=h(3)=h(1)=6
and (f−5g)′′ never vanishes ⇒h′′(x)≠0 ⇒h′(x) never changes its nature in (−2,0),(0,1),(1,3).
So using rolle's theorem there exists exactly one solution for h′(x)=0 in each of the three intervals (−2,0),(0,1) and (1,3).