f(x) is continuous function on [1, 3] and f(1)=2,f(3)=−2, then which of the following not necessarily hold good?
A
f(2)⩾0
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B
x2f(x)=0 has a root in (1,3)
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C
−2≤f(x)≤2∀x∈[1,3]
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D
f(x)−x2=0 has a root in (1,3)
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Solution
The correct option is C−2≤f(x)≤2∀x∈[1,3] f(2)⩾0 is not necessarily true as from the figure we can see, f(2) may be negative. Letg(x)=x2f(x) g(1)=f(1)=2g(3)=9×f(3)=−18g(1)g(3)<0 x2f(x)=0 has a root in (1,3)
Graph of f(x):
It is clear from the above figure, f(x) may be greater than 2 or less than -2.
Leth(x)=f(x)−x2 h(1)=2−1=1h(3)=−2−9=−11h(1)h(3)<0
So f(x)−x2 has a root in
(1, 3)