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If `y=(1+x)(1+x^2)(1+x^4)...(1+x^(2^n))` then `(dy)/(dx)` at `x=0` isA. 1B. -1C. 0D. none of these |
Answer» Correct Answer - A We have, `y=(1+x)(1+x^(2))(1+x^(4))….(1+x^(2^(n)))` `implies" "y=((1-x)(1+x)(1+x^(2))(1+x^(4))….(1+x^(2^(n))))/((1-x))` `implies" "y=(1-x^(2^(n+1)))/(1-x)` `implies" "(dy)/(dx)=(-2^(n+1)x^(2^(n+1)-1)"("1-x")"+"("1-x^(2^(n+1))")")/((1-x)^(2))` `implies" "((dy)/(dx))_(x=0)=1` |
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