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Question: Let \( f(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}} \) . If \( f\left( {{5^ + }} \right) \) and...

Let f(x)=sin⁡xx2+ax+bf(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}} . If f(5+)f\left( {{5^ + }} \right) and f(3+)f\left( {{3^ + }} \right) exists finitely and are not zero, then the value of (a+b) is equal to? (where {⋅} represents fractional part function).

Explanation

Solution

Hint : In this question, we need to determine the value of the term (a+b) such that f(x)=sin⁡xx2+ax+bf(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}} and f(5+)f\left( {{5^ + }} \right) and f(3+)f\left( {{3^ + }} \right) exists finitely and are not zero. For this, we will follow the general arithmetic equations.

Complete step-by-step answer :
The given function is f(x)=sin⁡xx2+ax+bf(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}} such that f(5+)f\left( {{5^ + }} \right) and f(3+)f\left( {{3^ + }} \right) exists finitely and are not zero. So, substituting the value of ‘x’ as 5 in the given function, we get
⇒f(x)=sin⁡xx2+ax+b ⇒f(5+)=sin⁡5(5)2+5a+b  \Rightarrow f(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}} \\\ \Rightarrow f({5^ + }) = \dfrac{{\sin \\{ 5\\} }}{{{{(5)}^2} + 5a + b}} \\\
As {.} denotes the fractional part, so the above equation can also be written as
⇒f(5+)=sin⁡5(5)2+5a+b =025+5a+b  \Rightarrow f({5^ + }) = \dfrac{{\sin \\{ 5\\} }}{{{{(5)}^2} + 5a + b}} \\\ = \dfrac{0}{{25 + 5a + b}} \\\
According to the question, f(5+)f\left( {{5^ + }} \right) exists finitely and are not zero, so from the above equation we can write:
25+5a+b=0−−−(i)25 + 5a + b = 0 - - - (i)
Similarly,
Substituting the value of ‘x’ as 3 in the given function, we get
⇒f(x)=sin⁡xx2+ax+b ⇒f(3+)=sin⁡3(3)2+3a+b  \Rightarrow f(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}} \\\ \Rightarrow f({3^ + }) = \dfrac{{\sin \\{ 3\\} }}{{{{(3)}^2} + 3a + b}} \\\
As {.} denotes the fractional part, so the above equation can also be written as
⇒f(3+)=sin⁡3(3)2+3a+b =09+3a+b  \Rightarrow f({3^ + }) = \dfrac{{\sin \\{ 3\\} }}{{{{(3)}^2} + 3a + b}} \\\ = \dfrac{0}{{9 + 3a + b}} \\\
According to the question, f(3+)f\left( {{3^ + }} \right) exists finitely and are not zero, so from the above equation we can write:
9+3a+b=0−−−(ii)9 + 3a + b = 0 - - - (ii)
Solving the equations (i) and (ii) for the values of the ‘a’ and ‘b’.
From the equation (i), we can write
25+5a+b=0 b=−25−5a−−−−(iii)  25 + 5a + b = 0 \\\ b = - 25 - 5a - - - - (iii) \\\
Substituting the value from equation (iii) in the equation (ii), we get
9+3a+b=0 9+3a+(−25−5a)=0 9+3a−25−5a=0 \-2a−16=0 2a=−16 a=−8  9 + 3a + b = 0 \\\ 9 + 3a + ( - 25 - 5a) = 0 \\\ 9 + 3a - 25 - 5a = 0 \\\ \- 2a - 16 = 0 \\\ 2a = -16 \\\ a = -8 \\\
Hence, the value of the constant ‘a’ is 8.
Substituting the value of the constant ‘a’ in the equation (iii) to determine the value of the constant ‘b’.
b=−25−5a =−25−5(−8) =−25+40 =15  b = - 25 - 5a \\\ = - 25 - 5(-8) \\\ = - 25 + 40 \\\ = 15 \\\
Hence, the value of the constant ‘b’ is -65.
Now, calculating the sum of the constants:
(a+b)=(15−8) =7  (a + b) = (15-8) \\\ = 7 \\\
Hence, the sum of and b is 7.

Note : {.} denotes the fractional part and so, for the proper (or exact) integer the fractional part is zero. Therefore, we have taken the fractional part in our solution as zero.