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Question

Let f(x),g(x) and h(x) be polynomials of degree 4 and satisfy the equation ∣ ∣f(x)g(x)h(x)abcpqr∣ ∣=lx4+mx3+nx2+4x+1 where a,b,c,p,q,r,l,m,n are real constants. Then the value of ∣ ∣f′′′(0)f′′(0)g′′′(0)g′′(0)h′′′(0)h′′(0)abcpqr∣ ∣ is

A
3mn
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B
2(3mn)
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C
3(3m+n)
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D
3m+n
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Solution

The correct option is B 2(3mn)
∣ ∣f(x)g(x)h(x)abcpqr∣ ∣=lx4+mx3+nx2+4x+1

Differentiating w.r.t. x, we get
∣ ∣f(x)g(x)h(x)abcpqr∣ ∣=4lx3+3mx2+2nx+4

Again differentiating w.r.t. x, we get
∣ ∣f′′(x)g′′(x)h′′(x)abcpqr∣ ∣=12lx2+6mx+2n (1)

Again differentiating w.r.t. x, we get
∣ ∣f′′′(x)g′′′(x)h′′′(x)abcpqr∣ ∣=24lx+6m (2)

From (1) and (2), we have
∣ ∣f′′′(x)f′′(x)g′′′(x)g′′(x)h′′′(x)h′′(x)abcpqr∣ ∣

=(24lx+6m)(12lx2+6mx+2n)

At x=0,
∣ ∣f′′′(0)f′′(0)g′′′(0)g′′(0)h′′′(0)h′′(0)abcpqr∣ ∣

=6m2n=2(3mn)

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