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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Limits

NTA Abhyas 2020 · Mathematics · Limits

If f(x) is a polynomial satisfying f(x)f((1)/(x)) = f(x) +...

If f(x)f\left(x\right) is a polynomial satisfying f(x)f(1x)=f(x)+f(1x)f\left(x\right)f\left(\dfrac{1}{x}\right) = f\left(x\right) + f\left(\dfrac{1}{x}\right) and f(2)>1,f\left(2\right) > 1, then lim⁡x→1f(x)\lim_{x \to 1} f \left(x\right) is

Correct answer

2

Step-by-step solution

As given f(x).f(1x)=f(x)+f(1x),f(2),f\left(x\right) . f\left(\dfrac{1}{x}\right) = f\left(x\right) + f\left(\dfrac{1}{x}\right), f\left(2\right),

⇒f(x)f(1x)−f(x)−f(1x)=0⇒ f ( x ) f( \dfrac{1}{x} ) - f ( x ) - f( \dfrac{1}{x} ) = 0

⇒(f(x)−1)(f(1x)−1)=1⇒ ( f ( x ) - 1 )( f( \dfrac{1}{x} ) - 1 ) = 1

⇒h(x)h(1x)=1⇒ h ( x ) h( \dfrac{1}{x} ) = 1 as it f(x)f( x ) is a polynomial,hence h(x)h( x ) will also be a polynomial

⇒h(x)=±xn⇒ h( x ) = \pm {x}^{n}
⇒f(x)=±xn+1⇒ f( x ) = \pm {x}^{n} + 1

Hence f(x)=±xn+1f( x ) = \pm {x}^{n} + 1

Now f(2)>1f( 2 ) > 1

⇒f(x)=xn+1⇒ f ( x ) = {x}^{n} + 1

hence lim⁡x→1f(x)=lim⁡x→1(xn+1)=2\lim_{x \to 1} f( x ) = \lim_{x \to 1} ( {x}^{n} + 1 ) = 2
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