If q^2-4pr=0 , p0 Then the domain of the function f(x)=log(px^3+(p+q)x^2+(q+r)x+r) is

Question

If q^2-4pr=0 , p>0 Then the domain of the function f(x)=log(px^3+(p+q)x^2+(q+r)x+r) is

Arun · Answer

Dear student f(x)=log{(px^2+qx+r)(x+1)}=log(px^2+qx+r)+log(x+1) we Know that px^2+qx+r>0 & x+1>0 since p>0 & q^2-4pr=0 therefore px^2+qx+r always greater than zero except -q/2p and x>-1 so domain of the function x>-1/{q/2p} RegardsArun (askIITians forum expert)

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If q^2-4pr=0 , p>0 Then the domain of the function f(x)=log(px^3+(p+q)x^2+(q+r)x+r) is

If q^2-4pr=0 , p>0
Then the domain of the function f(x)=log(px^3+(p+q)x^2+(q+r)x+r) is

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If q^2-4pr=0 , p>0 Then the domain of the function f(x)=log(px^3+(p+q)x^2+(q+r)x+r) is

If q^2-4pr=0 , p>0Then the domain of the function f(x)=log(px^3+(p+q)x^2+(q+r)x+r) is

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If q^2-4pr=0 , p>0
Then the domain of the function f(x)=log(px^3+(p+q)x^2+(q+r)x+r) is