Let $$f(x)=limn$$→∞ $$ax(x$$−$$1)$$$$cot$$⁡π$$x4n+px2+2cot$$⁡π$$x4n+1$$, if x∈(0$$,$$1)∪$$(1$$,$$2)0$$ , if $$x=1$$ . If f is differentiable at $$x=1$$ , then the value of |$$a+p$$| is

as n→∞ If x>$$1$$ then xn→∞ If $$0$$

Differentiability

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$Let f (x) = \{\begin{array}{ll} lim_{n \to \infty} \frac{a x (x - 1) {(\cot \frac{π x}{4})}^{n} + (p x^{2} + 2)}{{(\cot \frac{π x}{4})}^{n} + 1}, if x \in (0, 1) \cup (1, 2) \\ 0, if x = 1 \end{array} . If f is differentiable$ $at x = 1, then the value of | a + p | is$

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$\begin{array}{l} as n \to \infty \\ If x > 1 then x^{n} \to \infty \\ If 0 < x < 1 then x^{n} \to 0 \\ Now, f (1^{+}) = \frac{P x^{2} + 2}{1} = 0 \\ \Rightarrow P = - 2 \\ And f (1^{-}) = a x (x - 1) = 0 \\ L \cdot H \cdot D = lim_{h \to 0} \frac{f (1 - h) - f (1)}{- h} \\ = lim_{h \to 0} \frac{a (1 - h) (- h)}{- h} = a \end{array}$
$\begin{array}{l} R \cdot H \cdot D = \lim_{h \to 0} \frac{f (1 + h) - f (1)}{h} = \lim_{h \to 0} \frac{P (1 + h)^{2} + 2}{h} \\ = \lim_{h \to 0} \frac{- 2 (1 + 2 h + h^{2}) + 2}{h} = - 4 \\ ∴ a = - 4 \\ ∴ | a + p | = | - 4 - 2 | = 6 \end{array}$

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Differentiability

Remember concepts with our Masterclasses.

Let f(x)=limn→∞ ax(x−1)cot⁡πx4n+px2+2cot⁡πx4n+1, if x∈(0,1)∪(1,2)0 , if x=1 . If f is differentiable at x=1 , then the value of |a+p| is

as n→∞ If x>1 then xn→∞ If 0<x<1 then xn→0 Now, f1+=Px2+21=0⇒P=−2 And f1−=ax(x−1)=0L⋅H⋅D=limh→0 f(1−h)−f(1)−h=limh→0 a(1−h)(−h)−h=aR⋅H⋅D=limh→0f(1+h)−f(1)h=limh→0P(1+h)2+2h=limh→0−21+2h+h2+2h=-4∴a=−4∴|a+p|=|−4−2|=6

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$Let f (x) = \{\begin{array}{ll} lim_{n \to \infty} \frac{a x (x - 1) {(\cot \frac{π x}{4})}^{n} + (p x^{2} + 2)}{{(\cot \frac{π x}{4})}^{n} + 1}, if x \in (0, 1) \cup (1, 2) \\ 0, if x = 1 \end{array} . If f is differentiable$ $at x = 1, then the value of | a + p | is$