For every real number x \neq-1, let {\f}(x)=\frac{x}{x+1}. Write {\f}_1(x)= {\f}(x) & for {\n} \geq 2, {\f}_{ {\n}}(x)= {\f} ( {\f}_{ {\n}-1}(x)). Then {\f}_1(-2) \cdot {\f}_2(-2) . . . . {\f}_{ {\n}}(-2) must be

Correct Answer is : 2n1.3.5......(2n−1)

{2^{ {\n}}}/{1.3 .5 . . . . . .(2 {\n}-1)}

WBJEE 2024 Math Paper

Question : 43 of 75

Marks: +1, -0

For every real number x≠−1, let f(x)=‌

x+1

. Write f1(x)=f(x) & for n≥2,fn(x)=f(fn−1(x)). Then f1(−2)⋅f2(−2)....fn(−2) must be

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