Let \( f: R \rightarrow R \) be defined by \( f(x)=\frac{x-1}{2} \) an

1.	Let \( f: R \rightarrow R \) be defined by \( f(x)=\frac{x-1}{2} \) and, \( g: R \rightarrow R \) be defined by \( g(x)=2 x+1 \). Then, \( \frac{d}{d x}[(g \circ f)(7)]= \)(a) 1(b) 0(c) \( -1 \)(d) None of these
Answer» f(7) = 7-1/2 = 6/2 = 3 (gof)(7) = g(f(7)) = g(3) = 2 x 3 + 1 = 7 Hence, option (d) is correct.

Let \( f: R \rightarrow R \) be defined by \( f(x)=\frac{x-1}{2} \) and, \( g: R \rightarrow R \) be defined by \( g(x)=2 x+1 \). Then, \( \frac{d}{d x}[(g \circ f)(7)]= \)(a) 1(b) 0(c) \( -1 \)(d) None of these

Answer»

f(7) = 7-1/2 = 6/2 = 3

(gof)(7) = g(f(7))

= g(3)

= 2 x 3 + 1

= 7

Hence, option (d) is correct.

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Let \( f: R \rightarrow R \) be defined by \( f(x)=\frac{x-1}{2} \) and, \( g: R \rightarrow R \) be defined by \( g(x)=2 x+1 \). Then, \( \frac{d}{d x}[(g \circ f)(7)]= \)(a) 1(b) 0(c) \( -1 \)(d) None of these

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