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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 D0 Apply Langranges mean value Theorem f′(c)=f(a)−f(b)a−b where cϵ(a,b) ϕ′(c)=ϕ(a)−ϕ(b)a−b φ′(c)=φ(a)−φ(b)a−b So R1→R1−R2
∣∣
∣
∣∣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 (a−b), we get, ∣∣
∣
∣∣f′(c)ϕ′(c)φ′(c)f(b)ϕ(b)φ(b)f′(c)ϕ′(c)φ′(c)∣∣
∣
∣∣=0