JEE Main 29 June 2022 Shift 2 Solved Paper

Section: Mathematics
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Question : 26 of 90
 
Marks: +1, -0
Let f and g be twice differentiable even functions on (−2,2) such that f(‌
1
4
)
=0
,f(‌
1
2
)
=0
, f(1)=1 and g(‌
3
4
)
=0
,g(1)=2
. Then, the minimum number of solutions of f(x)g′′(x)+f′(x)g′(x)=0 in (−2,2) is equal to____
[29-Jun-2022-Shift-2]
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