Solveeit Logo
Login

Question

Question: Let \[f(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}}\]. If \[f({5^ + })\]and \[f({3^ + })\] exist...

Let f(x)=sin⁡xx2+ax+bf(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}}. If f(5+)f({5^ + })and f(3+)f({3^ + }) exists finitely and are not zero, then the value of (a+b) is (where {.} represents fractional part function)
a) 7
b) 10
c) 11
d) 20

Explanation

Solution

If it is provided that the given equation is non zero and but it is found that in the given equation the numerator part is zero then it implies that the denominator part of the given equation will also be equal to zero.
Given: f(x)=sin⁡xx2+ax+bf(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}}with f(5+)f({5^ + }) and f(3+)f({3^ + }) exists
To find: value of (a+b)

Complete step by step answer:
Step 1: We have f(x)=sin⁡xx2+ax+bf(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}} such that f(5+)f({5^ + })and f(3+)f({3^ + }) exists. So substituting the value of x as 5 in the given function, we get
f(x)=sin⁡xx2+ax+bf(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}}
f(5+)=sin⁡552+a5+bf({5^ + }) = \dfrac{{\sin \\{ 5\\} }}{{{5^2} + a5 + b}}
Now in the question, it is given that {.} denotes fractional part, therefore sin⁡5\sin \\{ 5\\} = 0 so we have
f(5+)=sin⁡552+a5+bf({5^ + }) = \dfrac{{\sin \\{ 5\\} }}{{{5^2} + a5 + b}}
f(5+)=sin⁡552+a5+b=052+a5+bf({5^ + }) = \dfrac{{\sin \\{ 5\\} }}{{{5^2} + a5 + b}} = \dfrac{0}{{{5^2} + a5 + b}}
But it is also provided in the question that f(5+)f({5^ + }) exists finitely and non zero. Therefore we have
52+a5+b=0{5^2} + a5 + b = 0
25+5a+b=025 + 5a + b = 0
Step 2: now substituting the value of x as 3 in the given equation f(x)=sin⁡xx2+ax+bf(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}}, we have
f(x)=sin⁡xx2+ax+bf(x) = \dfrac{{\sin \\{ x\\} }}{{{x^2} + ax + b}}
f(3+)=sin⁡332+a3+bf({3^ + }) = \dfrac{{\sin \\{ 3\\} }}{{{3^2} + a3 + b}}
Now in the question, it is given that {.} denotes fractional part, therefore sin⁡3\sin \\{ 3\\} = 0 so we have
f(3+)=sin⁡332+a3+bf({3^ + }) = \dfrac{{\sin \\{ 3\\} }}{{{3^2} + a3 + b}}
f(3+)=sin⁡332+a3+b=032+a3+bf({3^ + }) = \dfrac{{\sin \\{ 3\\} }}{{{3^2} + a3 + b}} = \dfrac{0}{{{3^2} + a3 + b}}
Here also it is provided in the question that f(3+)f({3^ + }) exists finitely and non zero. Therefore, we have
32+a3+b=0{3^2} + a3 + b = 0
9+3a+b=09 + 3a + b = 0
Step 3: now we have two unknowns a and b, and two equations 25+5a+b=025 + 5a + b = 0 & 9+3a+b=09 + 3a + b = 0
Determining the value of b from the equation 25+5a+b=025 + 5a + b = 0
25+5a+b=025 + 5a + b = 0
b=−25−5ab = - 25 - 5a
Now substituting the above value of b in the equation 9+3a+b=09 + 3a + b = 0, we get
9+3a+b=09 + 3a + b = 0
9+3a+(−25−5a)=09 + 3a + ( - 25 - 5a) = 0
9−25+3a−5a=09 - 25 + 3a - 5a = 0
−16−2a=0- 16 - 2a = 0
2a=−162a = - 16
a=−8a = - 8
Now substituting the above value of an inequation 9+3a+b=09 + 3a + b = 0, we get
9+3a+b=09 + 3a + b = 0
9+3×(−8)+b=09 + 3 \times ( - 8) + b = 0
9−24+b=09 - 24 + b = 0
b=15b = 15
Now adding the value of a and b we get
a+b=−8+15a + b = - 8 + 15
a+b=7a + b = 7
Hence, the value of (a+b) is 7

So, the correct answer is “Option a”.

Note: For finding the value of ‘n’ number of unknown values we require ‘n’ number of equations, like here we have two unknowns a and b so to find the values we generated two equations 25+5a+b=025 + 5a + b = 0 & 9+3a+b=09 + 3a + b = 0
If it is provided that the value of a given equation is non zero but the numerator is found to be equal to zero then the denominator will also be equal to zero.