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  •  Let the given function is differentiable at . <img alt="f(x)= \begin{cases}\lim _{n \rightarrow \infty} \fr

 Let the given function is differentiable at x=1.
f(x)= \begin{cases}\lim _{n \rightarrow \infty} \frac{a x(x-1)\left(\cot \frac{\pi x}{4}\right)^n+\left(p x^2+2\right)}{\left(\cot \frac{\pi x}{4}\right)^n+1}, & x \in(0,1) \cup(1,2) \\ 0, & x=1\end{cases}
Then the value of |a+p| is

Option: 1

4


Option: 2

6


Option: 3

8


Option: 4

10


Answers (1)

best_answer

\begin{aligned} & f(x)=\left\{\begin{array}{l} \lim _{n \rightarrow \infty} \frac{\operatorname{ax}(x-1)\left(\cot \frac{\pi x}{4}\right)^n+\left(p x^2+2\right)}{\left(\cot \frac{\pi x}{4}\right)^n+1}, \\ 0, \end{array}\right. \\ & x \in(0,1) \cup(1,2) \\ & \therefore \quad f(x)= \begin{cases}a x(x-1), & 0<x<1 \\ 0, & x=1 \\ p x^2+2, & 1<x<2\end{cases} \\ & \end{aligned}


\therefore f(x) must be continuous at x=1

\begin{array}{ll} \therefore & \text { L.H.L. }=\text { R.H.L }=f(1) \\ \Rightarrow & p+2=0 \Rightarrow p=-2 \end{array}

Also f^{\prime}(x)= \begin{cases}a(2 x-1), & 0<x<1 \\ -4 x, & x>1\end{cases}
\because \quad f(x) is differentiable at x=1, so
L.H.D. = R.H.D. \Rightarrow a=-4
Hence |a+p|=6

Posted by

Devendra Khairwa

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