Let f be a twice differentiable function, graph of which touches the x−axis at x=1 and f′′(1)=4. If limn→∞⎛⎜⎜⎜⎜⎜⎝limx→0⎛⎜⎜⎜⎜⎜⎝n∑r=1f(1+x2r)ln(1+tan2x+tan4x)⎞⎟⎟⎟⎟⎟⎠⎞⎟⎟⎟⎟⎟⎠=pq, where p,q∈N, then the least value of (p+q) is

Let f be a twice differentiable function, graph of which touches the x

1.	Let f be a twice differentiable function, graph of which touches the x−axis at x=1 and f′′(1)=4. If limn→∞⎛⎜⎜⎜⎜⎜⎝limx→0⎛⎜⎜⎜⎜⎜⎝n∑r=1f(1+x2r)ln(1+tan2x+tan4x)⎞⎟⎟⎟⎟⎟⎠⎞⎟⎟⎟⎟⎟⎠=pq, where p,q∈N, then the least value of (p+q) is
Answer» Let $f$ be a twice differentiable function, graph of which touches the $x -$ axis at $x = 1$ and $f^{''} (1) = 4.$ If $lim n \to \infty ⎛ ⎜⎜⎜⎜⎜ ⎝ lim x \to 0 ⎛ ⎜⎜⎜⎜⎜ ⎝ \frac{n \sum r = 1 f (1 + \frac{x}{2^{r}})}{ln (1 + {tan}^{2} x + {tan}^{4} x)} ⎞ ⎟⎟⎟⎟⎟ ⎠ ⎞ ⎟⎟⎟⎟⎟ ⎠ = \frac{p}{q},$ where $p, q \in N,$ then the least value of $(p + q)$ is