Let f (x) be a polynomial with positive degree satisfying the relation
f (x) f (y) = f (x) + f (y) + f(xy) – 2 and f (4) = 65, then
(A) f (2) = 9
(B) f (3/3) = 27
(C) f (5) = 126
(D) The roots of the equation f (x) = 2x2 are all real
Answers
Answer:
Step-by-step explanation:
Given, 2+f(x)f(y)=f(x)+f(y)+f(xy)
or 1−f(x)−f(y)+f(x)f(y)=f(xy)−1
or (1−f(x))(1−f(y))=f(xy)−1
The above result holds if and only if,
f(x)=1+x
n
if f(x)=a
n
x
n
+a
n−1
x
n−1
+...+a
0
Then, consider (1+f(x))(1−f(y))=f(xy)−1
Compare constant term on either side, we have
1−a
0
=a
0
−1⇒a
0
=1
Comparing coefficient of x
n
y
n
, we get
a
n
2
=a
n
⇒a
n
=1 or otherwise polynomial would not be of n degree.
Comparing coefficient of x,x
1
,....,x
n−1
on either sides, we have
a
1
=a
2
=...=a
n−1
=0
⇒a
n
=1 and f(x)=x
n
+1
Given, f(2)=5 ie, 2
n
+1=5
⇒n=2
Thus, f(x)=x
2
+1
f(f(2))=f(5)=5
2
+1=26Given, 2+f(x)f(y)=f(x)+f(y)+f(xy)
or 1−f(x)−f(y)+f(x)f(y)=f(xy)−1
or (1−f(x))(1−f(y))=f(xy)−1
The above result holds if and only if,
f(x)=1+x
n
if f(x)=a
n
x
n
+a
n−1
x
n−1
+...+a
0
Then, consider (1+f(x))(1−f(y))=f(xy)−1
Compare constant term on either side, we have
1−a
0
=a
0
−1⇒a
0
=1
Comparing coefficient of x
n
y
n
, we get
a
n
2
=a
n
⇒a
n
=1 or otherwise polynomial would not be of n degree.
Comparing coefficient of x,x
1
,....,x
n−1
on either sides, we have
a
1
=a
2
=...=a
n−1
=0
⇒a
n
=1 and f(x)=x
n
+1
Given, f(2)=5 ie, 2
n
+1=5
⇒n=2
Thus, f(x)=x
2
+1
f(f(2))=f(5)=5
2
+1=26
Answer:
Option (A), (B), (C) and (D) are all correct.
Explanation:
Polynomials are the type of algebraic expressions that contains both variables and coefficients. Variables are those terms that can have different values and coefficients are those terms which are written with the variable.
In an polynomial, the expression does not have a negative exponent. Also the polynomial only involves the basic operation of mathematics that is addition, subtraction, multiplication and division.
1 − f(x) − f(y) + f(x) f(y)= f(x y) − 1
or (1 − f(x)) (1 − f(y)) = f(x y) − 1
This means that:
F(x)=1+xⁿ
Now, f(4)=65
65=1+4ⁿ
64=4ⁿ
n=3
F(x)=1+x³
So, now we can see that it satisfies all the options.
Therefore, all the options are correct.
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