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Question

Number of solution of the equation 3tanx+x3=2 in (0,π4) is

A
0
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B
1
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C
2
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D
3
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Solution

The correct option is C 1
Number of solutions of the equation 3tanx+x3=2 in (0,π4):
Let,f(x)=3tanx+x32
f(x)=3sec2x+3x2
f(x)=3[sec2x+x2]>0xϵ(0,π4)
[sec2x+x20], there it is always increasing
f(x) is strictly increasing function.
Now f(0)=2<0 and f(π4)=3tanπ4+(π4)32
=3+(π4)32
=1+(π4)3>0
f(0)<ve and f(π4)>+ve
Therefore, f willl cross the xaxis at least once.
And since, in [0,π4],x is an increasing graph, there number of roots of f(x) will be 1.

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