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Question

If f(x), ϕ(x),φ(x) are continuous on [a, b] and differentiable on (a,b) cϵ(a,b), then

∣ ∣ ∣f(a)ϕ(a)φ(a)f(b)ϕ(b)φ(b)f(c)ϕ(c)φ(c)∣ ∣ ∣=

A
f(c)
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B
ϕ(c)
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C
φ(c)
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D
0
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Solution

The correct option is D 0
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Langranges mean value Theorem
f(c)=f(a)f(b)ab where cϵ(a,b)
ϕ(c)=ϕ(a)ϕ(b)ab
φ(c)=φ(a)φ(b)ab
So R1R1R2
∣ ∣ ∣f(a)f(b)ϕ(a)ϕ(b)φ(a)φ(b)f(b)ϕ(b)φ(c)f(c)ϕ(c)φ(c)∣ ∣ ∣
Dividing First Row Elements by (ab), we get,
∣ ∣ ∣f(c)ϕ(c)φ(c)f(b)ϕ(b)φ(b)f(c)ϕ(c)φ(c)∣ ∣ ∣=0
Since, rows 1 and 3 are identical.

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