If f ( x ) = x 2 + a x + 3 and g ( x ) = x + b , where F ( x ) = lim n f ( x ) + ( x 2 ) n g ( x ) 1 + ( x 2 ) n . If F ( x ) is continuous at x = 1 and x = - 1 , then find the value of ( a 2 + b 2 )

Question

If $f (x) = x^{2} + a x + 3$ and $g (x) = x + b$ , where $F (x) = \lim_{n \to \infty} \frac{f (x) + {(x^{2})}^{n} g (x)}{1 + {(x^{2})}^{n}}$ . If $F (x)$ is continuous at $x = 1$ and $x = - 1$ , then find the value of $(a^{2} + b^{2})$

Smart Achievers · Accepted Answer

(17)

$F (x) = f (x)$ $x > 1 = \frac{f (x) + g (x)}{2}$ $x = 1$

$= f (x)$ $- 1 < x < 1 = \frac{f (x) + g (x)}{2}$ $x = - 1$

$= g (x)$ $(a^{2} + b^{2})$

If $f (x)$ is continuous at $x = 1$ , $F (1^{+}) = F (1) = F (1^{-})$ $b = a + 3$

$F (- 1^{-}) = F (- 1) = F (- 1^{+})$ $a + b = 5$

Solution

Q.
If $f (x) = x^{2} + a x + 3$ and $g (x) = x + b$ , where $F (x) = \lim_{n \to \infty} \frac{f (x) + {(x^{2})}^{n} g (x)}{1 + {(x^{2})}^{n}}$ . If $F (x)$ is continuous at $x = 1$ and $x = - 1$ , then find the value of $(a^{2} + b^{2})$

Ans.
(17)

$F (x) = f (x)$ $x > 1 = \frac{f (x) + g (x)}{2}$ $x = 1$

$= f (x)$ $- 1 < x < 1 = \frac{f (x) + g (x)}{2}$ $x = - 1$

$= g (x)$ $x < - 1$

If $f (x)$ is continuous at $x = 1$ , $F (1^{+}) = F (1) = F (1^{-})$ $b = a + 3$

$F (- 1^{-}) = F (- 1) = F (- 1^{+})$ $a + b = 5$

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Solution

Q. If f(x)=x2+ax+3 and g(x)=x+b, where F(x)=limn→∞f(x)+(x2)ng(x)1+(x2)n. If F(x) is continuous at x=1 and x=-1, then find the value of (a2+b2)

Ans. (17) F(x)=f(x) x>1=f(x)+g(x)2 x=1 =f(x) -1<x<1=f(x)+g(x)2 x=-1 =g(x) x<-1 If f(x) is continuous at x=1, F(1+)=F(1)=F(1-) b=a+3 F(-1-)=F(-1)=F(-1+) a+b=5

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Q.
If $f (x) = x^{2} + a x + 3$ and $g (x) = x + b$ , where $F (x) = \lim_{n \to \infty} \frac{f (x) + {(x^{2})}^{n} g (x)}{1 + {(x^{2})}^{n}}$ . If $F (x)$ is continuous at $x = 1$ and $x = - 1$ , then find the value of $(a^{2} + b^{2})$

Ans.
(17)

$F (x) = f (x)$ $x > 1 = \frac{f (x) + g (x)}{2}$ $x = 1$

$= f (x)$ $- 1 < x < 1 = \frac{f (x) + g (x)}{2}$ $x = - 1$

$= g (x)$ $x < - 1$

If $f (x)$ is continuous at $x = 1$ , $F (1^{+}) = F (1) = F (1^{-})$ $b = a + 3$

$F (- 1^{-}) = F (- 1) = F (- 1^{+})$ $a + b = 5$