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Question

If f:RR be a function defined as f(x)=ex+10(x+yex)f(y) dy, then

A
f(x) is decreasing function
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B
f(x) is one-one function
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C
f(x) is onto function
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D
f(x)=0 has only one solution
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Solution

The correct option is D f(x)=0 has only one solution
f(x)=ex+10(x+yex)f(y) dyf(x)=ex+x10f(y) dy+ex10yf(y) dyf(x)=ex1+10yf(y) dy+x10f(y) dyf(x)=aex+bx
f(y)=aey+by ...(1)

Now,
a=1+10yf(y) dy
From eqn (1),
a=1+10y(aey+by) dy
a=1+a10yey dy+b10y2 dy
Using by parts
a=1+a[eyyey]10+b3a=1+a+b3b=3

Now,
b=10f(y) dyb=10(aey+by) dyb=[aey+by22]10b=a(e1)+b2a=32(e1)


f(x)=3[ex2(e1)+x]

When
xf(x)xf(x)

Differentiating the function w.r.t. x,
f(x)=3[ex2(e1)+1]f(x)<0, xR
f(x) is decreasing function.
Hence, f(x) is both one-one and onto.
f(x)=0 has only one solution.

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