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
3m−n
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B
2(3m−n)
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C
3(3m+n)
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D
3m+n
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Solution
The correct option is B2(3m−n) ∣∣
∣∣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∣∣
∣∣