Let $$f(x)$$ be a polynomial satisfying limx→∞ $$x2f(x)2x5+3=6$$ and $$f(1)=3$$,$$f(3)=7$$ and $$f(5)=11$$ Then the value of |$$f(0)$$| is $$_______$$ .

Since limx→∞ $$x2f(x)2x5+3=6$$ which is finite and $$non-zero$$, $$f(x)$$ must be polynomial of degree '$$3$$'.Also $$f(1)$$ $$=$$ $$3$$, $$f(3)$$ $$=$$ $$7$$ and $$f(5)$$ $$=$$ $$11$$∴ $$f(x)=$$λ(x$$−$$1)(x$$−$$3)(x$$−$$5)+(2x+1)⇒limx→∞ $$x2($$λ(x$$−$$1)(x$$−$$3)(x$$−$$5)+2x+1)2x5+3=6$$⇒ λ$$2=6$$⇒λ$$=12$$∴ $$f(x)=12(x$$−$$1)(x$$−$$3)(x$$−$$5)+(2x+1)So$$, $$f(0) $$=$$ $$12(-1)(-3)(-5)+1=$$ $$179$$

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Let f(x) be a polynomial satisfying $lim_{x \to \infty} \frac{x^{2} f (x)}{2 x^{5} + 3} = 6$ and $f (1) = 3, f (3) = 7 and f (5) = 11$ Then the value of $| f (0) |$ is _______ .

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Solution

Since $lim_{x \to \infty} \frac{x^{2} f (x)}{2 x^{5} + 3} = 6$ which is finite and non-zero, f(x) must be polynomial of degree '3'.
Also f(1) = 3, f(3) = 7 and f(5) = 11
$\begin{array}{ll} ∴ f (x) = λ (x - 1) (x - 3) (x - 5) + (2 x + 1) \\ \Rightarrow lim_{x \to \infty} \frac{x^{2} (λ (x - 1) (x - 3) (x - 5) + 2 x + 1)}{2 x^{5} + 3} = 6 \\ \Rightarrow \frac{λ}{2} = 6 \\ \Rightarrow λ = 12 \\ ∴ f (x) = 12 (x - 1) (x - 3) (x - 5) + (2 x + 1) \end{array}$
So, f(0) = 12(-1)(-3)(-5)+1= 179

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Column -I	Column -II
A. If $L = lim_{x \to - 1} \frac{\sqrt[3]{(7 - x)} - 2}{(x + 1)}$ , then 12L =	P. -2
B. If $L = lim_{x \to π / 4} \frac{\tan^{3} x - \tan x}{\cos (x + \frac{π}{4})}$ then -L/4=	Q. 2
C. If $L = lim_{x \to 1} \frac{(2 x - 3) (\sqrt{x} - 1)}{2 x^{2} + x - 3}$ , then 20L=	R. 1
D. If $L = lim_{x \to \infty} \frac{\log x^{n} - [x]}{[x]}$ , where $n \in N$ ([x] denotes greatest integer less than or equal to x),then -2L=	S. -1

Introduction to limits

Remember concepts with our Masterclasses.

Let f(x) be a polynomial satisfying limx→∞ x2f(x)2x5+3=6 and f(1)=3,f(3)=7 and f(5)=11 Then the value of |f(0)| is _______ .

Ready to Test Your Skills?

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Let f(x) be a polynomial satisfying $lim_{x \to \infty} \frac{x^{2} f (x)}{2 x^{5} + 3} = 6$ and $f (1) = 3, f (3) = 7 and f (5) = 11$ Then the value of $| f (0) |$ is _______ .