
Let f(x) be a real valued function not identically zero satisfy the equation \[f\left( {x + {y^n}} \right) = f\left( x \right) + {\left\{ {f\left( y \right)} \right\}^n}\]for all real x, y and $f'(0) \geqslant 0$ , where n is an odd natural number. F (10) = 10K.Find the value of K.
Answer
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Hint: Here, find some pattern of function by putting different values of x and y. Put values of x and y carefully and values should be least and increase logically in a pattern so that we can get the value of function at a particular point. Here, given that n is an odd number so two values are possible.
Complete step-by-step answer:
\[f\left( {x + {y^n}} \right) = f\left( x \right) + {\left\{ {f\left( y \right)} \right\}^n}\]
Putting x = 0 and y = 0
\[f\left( {0 + {0^n}} \right) = f\left( 0 \right) + {\left\{ {f\left( 0 \right)} \right\}^n}\]
\[ \Rightarrow f(0) = f\left( 0 \right) + {\left\{ {f\left( 0 \right)} \right\}^n}\]
\[ \Rightarrow {\left\{ {f\left( 0 \right)} \right\}^n} = 0 \Rightarrow f(0) = 0\]
Putting x = 0 and y = 1
\[f\left( {0 + {1^n}} \right) = f\left( 0 \right) + {\left\{ {f\left( 1 \right)} \right\}^n}\]
\[ \Rightarrow f(1) = f\left( 0 \right) + {\left\{ {f\left( 1 \right)} \right\}^n}\]
\[ \Rightarrow f(1)\left[ {{{\{ f(1)\} }^{n - 1}} - 1} \right] = 0 \Rightarrow {\{ f(1)\} ^{n - 1}} - 1= 0 \Rightarrow {\{ f(1)\} ^{n - 1}} = 1 \Rightarrow f(1) = \pm 1\]
Since, n is odd, n – 1 is even.
Now put x = 1 and y = 1
\[ \Rightarrow f(2) = f\left( 1 \right) + {\left\{ {f\left( 1 \right)} \right\}^n}\]
So, F(2) = 0 or 2
Similarly, we can find different values and see that the pattern, that the output is either sum or difference of values of x and y.
Putting x = 2 and y = 1
F(3) = 1 or 3
F(4) = 2 or 4
F(5) = 3 or 5
F(6) = 4 or 6
F(7) = 5 or 7
F(8) = 6 or 5
F(9) = 7 or 9
F(10) = 8 or 10
So, F(10) = 10 = 10K ⇒ K = 1.
Note: In these types of questions, function given does not contain one variable hence not possible to find the value of function at a particular point. So by putting random values of x and y we try to find values of function with some pattern so that we can recognize the value of function at a particular point. Be careful while solving or simplifying equations as n is odd, so values of function at a point may be more than one. At last compare the value obtained and value given.
Complete step-by-step answer:
\[f\left( {x + {y^n}} \right) = f\left( x \right) + {\left\{ {f\left( y \right)} \right\}^n}\]
Putting x = 0 and y = 0
\[f\left( {0 + {0^n}} \right) = f\left( 0 \right) + {\left\{ {f\left( 0 \right)} \right\}^n}\]
\[ \Rightarrow f(0) = f\left( 0 \right) + {\left\{ {f\left( 0 \right)} \right\}^n}\]
\[ \Rightarrow {\left\{ {f\left( 0 \right)} \right\}^n} = 0 \Rightarrow f(0) = 0\]
Putting x = 0 and y = 1
\[f\left( {0 + {1^n}} \right) = f\left( 0 \right) + {\left\{ {f\left( 1 \right)} \right\}^n}\]
\[ \Rightarrow f(1) = f\left( 0 \right) + {\left\{ {f\left( 1 \right)} \right\}^n}\]
\[ \Rightarrow f(1)\left[ {{{\{ f(1)\} }^{n - 1}} - 1} \right] = 0 \Rightarrow {\{ f(1)\} ^{n - 1}} - 1= 0 \Rightarrow {\{ f(1)\} ^{n - 1}} = 1 \Rightarrow f(1) = \pm 1\]
Since, n is odd, n – 1 is even.
Now put x = 1 and y = 1
\[ \Rightarrow f(2) = f\left( 1 \right) + {\left\{ {f\left( 1 \right)} \right\}^n}\]
So, F(2) = 0 or 2
Similarly, we can find different values and see that the pattern, that the output is either sum or difference of values of x and y.
Putting x = 2 and y = 1
F(3) = 1 or 3
F(4) = 2 or 4
F(5) = 3 or 5
F(6) = 4 or 6
F(7) = 5 or 7
F(8) = 6 or 5
F(9) = 7 or 9
F(10) = 8 or 10
So, F(10) = 10 = 10K ⇒ K = 1.
Note: In these types of questions, function given does not contain one variable hence not possible to find the value of function at a particular point. So by putting random values of x and y we try to find values of function with some pattern so that we can recognize the value of function at a particular point. Be careful while solving or simplifying equations as n is odd, so values of function at a point may be more than one. At last compare the value obtained and value given.
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