If f : R → R satisfies f x + y = f x + f y , for all x , y ∈ R and f 1 = 2 , then ∑ r = 1 7 f r is

Given, f x + y = f x + f y , for all x , y ∈ R And, f 1 = 2 If we put x = y = 0 , we get f 0 + 0 = f 0 + f 0 ⇒ f 0 = 2 f 0 ⇒ f 0 = 0 . . . 1 And, f 1 = 2 × 1 Again we substitute x = 1 , y = 1 , we get f 1 + 1 = f 1 + f 1 ⇒ f 2 = 2 f 1 ⇒ f 2 = 2 × 2 Again we substitute x = 1 , y = 2 , we get ⇒ f 1 + 2 = f 1 + f 2 ⇒ f 3 = 2 + 2 f 1 ⇒ f 3 = 2 × 3 And so on. Therefore, ∑ r = 1 7 f r = f 1 + f 2 + f 3 + · · · · · · + f 7 = 2 1 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7 = 2 1 + 2 + 3 + 4 + 5 + 6 + 7 = 2 7 × 7 + 1 2 = 56

Let a function f : R → R be given by f ( x + y ) = f ( x ) f ( y ) for all x , y ∈ R and f ( x ) ≠ 0 for any x ∈ R . If the function f ( x ) is differentiable at x = 0 , show that f ' ( x ) = f ' ( 0 ) f ( x ) for all x ∈ R . Also, determine f ( x ) .

Given, f x + y = f x f y , where f : R → R And, f x is differentiable at x = 0 and f x ≠ 0 . Since, it is differentiable at ' x ' = 0 , we differentiate the given relation with respect to x . When we differentiate f x + y = f x f y with respect to x , f y will be considered as a constant function. Now, f ' x + y = f y f ' x Put x = 0 , f ' 0 + y = f y f ' 0 And if we replace y → x , we get the equation f ' x = f ' 0 f x . . . 1 . Hence, proved. Now we find the equation of function f x , for this substitute x = y = 0 in the given relation f x + y = f x f y , then we get f 0 + 0 = f 0 f 0 ⇒ f 0 = f 0 f 0 ⇒ f 0 f 0 - 1 = 0 ⇒ f 0 = 0 & f 0 = 1 But we have given the condition that f x ≠ 0 . Therefore, f 0 = 1 Hence, by 1 , we get f ' x = f ' 0 f x ⇒ f ' x f x = f ' 0 ⇒ ∫ f ' x f x d x = ∫ f ' 0 d x ⇒ ln f x = f ' 0 x + c Substitute x = 0 , then we get, ln f 0 = f ' 0 × 0 + c ⇒ ln 1 = 0 + c ⇒ c = 0 Hence, final equation will be ln f x = f ' 0 x ⇒ f x = e f ' 0 x

Let f x be a polynomial function satisfying the relation f x · f 1 x = f x + f 1 x ∀ x ∈ R - { 0 } and f 3 = - 26 . Determine f ' 1 .

Given: f x · f 1 x = f x + f 1 x Let ' f x be a polynomial function of degree n . Then, we can write the equation f x = a 0 x n + a 1 x n - 1 + a 2 x n - 2 + a 3 x n - 3 + . . . . . . . . . . . . . . . . . . . . + a n - 1 x + a n . Therefore, f 1 x = a 0 x n + a 1 x n - 1 + a 2 x n - 2 + a 3 x n - 3 + . . . . . . . . . . . . . . + a n - 1 x + a n Now, according to question f x + f 1 x = f x × f 1 x . Here, L.H.S = f x + f 1 x and R.H.S = f x × f 1 x We can write R.H.S = f x × f 1 x = a 0 x n + a 1 x n - 1 + a 2 x n - 2 + a 3 x n - 3 + . . . . . . . . . . . . . . . . . . . . + a n - 1 x + a n × a 0 x n + a 1 x n - 1 + a 2 x n - 2 + a 3 x n - 3 + . . . . . . . . . . . . . . + a n - 1 x + a n When we multiply and expand above equation, we find the equation be like = a 0 2 + x a 0 a 1 + a 1 a 2 + a 2 a 3 + a 3 a 4 + . . . . . . . . . . . . + a n - 2 a n - 1 + a n - 1 a n + x 2 a 0 a 2 + a 1 a 3 + a 2 a 4 + a 3 a 5 + . . . . . . . . . . . . . . . . . . + a n - 3 a n - 1 + a n - 2 a n + + . . . . . . . . . . . . . . . . . . + x n a 0 a n + a n 2 + 1 x a 1 a 0 + a 2 a 1 + a 3 a 2 + . . . . + a n a n - 1 + 1 x 2 a 2 a 0 + a 3 a 1 + . . . . . . . . . . . . + a n - 1 a n - 3 + a n a n - 2 + . . . . . . . . . . . . . . . . . . . + 1 x n a 0 a n And, L.H.S = a 0 x n + a 1 x n - 1 + a 2 x n - 2 + a 3 x n - 3 + . . . . . . . . . . . . . . . . . . . . + a n - 1 x + a n . + a 0 x n + a 1 x n - 1 + a 2 x n - 2 + a 3 x n - 3 + . . . . . . . . . . . . . . + a n - 1 x + a n = a 0 x n + a 1 x n - 1 + a 2 x n - 2 + a 3 x n - 3 + . . . . . . . . . . . . . . . . . . . . + a n - 1 x + a 0 x n + a 1 x n - 1 + a 2 x n - 2 + a 3 x n - 3 + . . . . . . . . . . . . . . + a n - 1 x + 2 a n Now when we compare the coefficient of variable ' x ' of the expanded equation of R.H.S from L.H.S equation, we get the value, a 0 2 + a n 2 = 2 a n ; (On comparing constant terms) a 0 a n = a 0 (On comparing coefficients of x n and 1 x n ) And, a 1 = a 2 = a 3 = a 4 . . . . . . . . . . . . . . a n - 3 = a n - 2 = a n - 1 = 0 ⇒ a 0 a n = a 0 ⇒ a 0 a n - 1 = 0 ⇒ a 0 = 0 & a n = 1 Here we cannot take the value of a 0 = 0 , because our assumption of f x is the polynomial of degree ' n ' will be failed. So, we take the value of a n = 1 only. When we substitute a n = 1 in equation a 0 2 + a n 2 + = 2 a n , we get the value = a 0 2 + 1 = 2 ⇒ a 0 2 = 2 - 1 ⇒ a 0 2 = 1 ⇒ a 0 = ± 1 Therefore, the equation f x = a n + a 0 x n ⇒ f x = 1 ± x n We have given f 3 = - 26 Then, f x = 1 ± x n ⇒ 1 + 3 n = - 26 & 1 - 3 n = - 26 ⇒ 3 n = - 27 & - 3 n = - 27 ⇒ 3 n = - 27 & 3 n = 27 Here, we can observe 3 n = - 27 cannot be possible. So, 3 n = 27 = 3 3 ⇒ n = 3 Hence, the final equation, we have f x = 1 - x 3 . Differentiating w.r.t. x , we get f ' x = - 3 × x 2 ⇒ f ' 1 = - 3 × 1 2 ⇒ f ' 1 = - 3