If f(x) = e^{(x+1)^n}; (n in N),then the valu | vedclass

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(D) Given $f(x) = e^{(x+1)^n}$.First derivative: $f'(x) = e^{(x+1)^n} \cdot n(x+1)^{n-1} = f(x) \cdot n(x+1)^{n-1}$.Second derivative: $f''(x) = f'(x)

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$4$

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(D) Given $f(x) = e^{(x+1)^n}$.First derivative: $f'(x) = e^{(x+1)^n} \cdot n(x+1)^{n-1} = f(x) \cdot n(x+1)^{n-1}$.Second derivative: $f''(x) = f'(x) \cdot n(x+1)^{n-1} + f(x) \cdot n(n-1)(x+1)^{n-2}$.Substituting $f'(x)$: $f''(x) = f(x) \cdot [n(x+1)^{n-1}]^2 + f(x) \cdot n(n-1)(x+1)^{n-2}$.At $x=1$,$f(1) = e^{2^n}$.$f''(1) = e^{2^n} \cdot [n(2)^{n-1}]^2 + e^{2^n} \cdot n(n-1)(2)^{n-2}$.$f''(1) = e^{2^n} \cdot [n^2 \cdot 2^{2n-2} + n(n-1) \cdot 2^{n-2}]$.$f''(1) = e^{2^n} \cdot 2^{n-2} [n^2 \cdot 2^n + n(n-1)]$.Given $f''(1) = 67 \cdot 2^n \cdot e^{2^n}$.So,$2^{n-2} [n^2 \cdot 2^n + n^2 - n] = 67 \cdot 2^n$.Dividing by $2^{n-2}$: $n^2 \cdot 2^n + n^2 - n = 67 \cdot 2^2 = 268$.For $n=4$: $16 \cdot 2^4 + 16 - 4 = 16 \cdot 16 + 12 = 256 + 12 = 268$.Thus,$n=4$ is the correct value.

If $f(x) = e^{(x+1)^n}; (n \in N)$,then the value of $n$ for which $f''(1) = 67(2^n e^{2^n})$ is

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