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Question

If a,b>0 and 0<p<1, then which of the following option is CORRECT?

A
(a+b)p>ap+bp
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B
(a+b)p<ap+bp
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C
(a+b)pap+bp
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D
(a+b)pap+bp
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Solution

The correct option is B (a+b)p<ap+bp
Let, f(x)=(1+x)p1xp, x>0
f(x)=p(1+x)p1pxp1(1)
Now,
1+x>x
(1+x)1p>x1p
1(1+x)p1>1xp1
(1+x)p1<xp1
(1+x)p1xp1<0(2)
From (1) and (2), we get
f(x)<0
Therefore, f(x) is a decreasing function
Now, f(0)=0
x>0f(x)<f(0)
(1+x)p1xp<0
(1+x)p<1+xp
Put x=ab,
Hence, (a+b)p<ap+bp

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