Question

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

Answer 1

Hint: In this question, we need to determine the value of the term (a+b) such that $ f(x) = \dfrac{{\sin \{ x\} }}{{{x^2} + ax + b}} $ and $ f\left( {{5^ + }} \right) $ and $ 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) = \dfrac{{\sin \{ x\} }}{{{x^2} + ax + b}} $ such that $ f\left( {{5^ + }} \right) $ and $ f\left( {{3^ + }} \right) $ exists finitely and are not zero. So, substituting the value of ‘x’ as 5 in the given function, we get
 $
\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
 $
\Rightarrow f({5^ + }) = \dfrac{{\sin \{ 5\} }}{{{{(5)}^2} + 5a + b}} \\
  = \dfrac{0}{{25 + 5a + b}} \\
  $
According to the question, $ f\left( {{5^ + }} \right) $ exists finitely and are not zero, so from the above equation we can write:
 $ 25 + 5a + b = 0 - - - (i) $
Similarly,
Substituting the value of ‘x’ as 3 in the given function, we get
 $
\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
 $
\Rightarrow f({3^ + }) = \dfrac{{\sin \{ 3\} }}{{{{(3)}^2} + 3a + b}} \\
  = \dfrac{0}{{9 + 3a + b}} \\
  $
According to the question, $ f\left( {{3^ + }} \right) $ exists finitely and are not zero, so from the above equation we can write:
 $ 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) \\
  $
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 \\
  $
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 \\
  $
Hence, the value of the constant ‘b’ is -65.
Now, calculating the sum of the constants:
 $
  (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.