    Question

# If f:R→R be a function defined as f(x)=ex+1∫0(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 solutionf(x)=ex+1∫0(x+yex)f(y) dy⇒f(x)=ex+x1∫0f(y) dy+ex1∫0yf(y) dy⇒f(x)=ex⎛⎜⎝1+1∫0yf(y) dy⎞⎟⎠+x1∫0f(y) dy⇒f(x)=aex+bx ⇒f(y)=aey+by ...(1) Now, a=1+1∫0yf(y) dy From eqn (1), a=1+1∫0y(aey+by) dy ⇒a=1+a1∫0yey dy+b1∫0y2 dy Using by parts ⇒a=1+a[ey⋅y−ey]10+b3⇒a=1+a+b3⇒b=−3 Now, b=1∫0f(y) dy⇒b=1∫0(aey+by) dy⇒b=[aey+by22]10⇒b=a(e−1)+b2⇒a=−32(e−1) ∴f(x)=−3[ex2(e−1)+x] When x→−∞⇒f(x)→∞x→∞⇒f(x)→−∞ Differentiating the function w.r.t. x, f′(x)=−3[ex2(e−1)+1]⇒f′(x)<0, ∀ x∈R ∴f(x) is decreasing function. Hence, f(x) is both one-one and onto. f(x)=0 has only one solution.  Suggest Corrections  0      Similar questions  Explore more