Q.16. Let for some , be a function satisfying for all . If and , then the value of is :

Q: What is the correct answer for JEE Main Mathematics Q16?

The correct answer is 2. Sol. Given $f(x + y) = f(x) + 2y^2 + y + axy$ ...(1) Given $f(0) = -1$ and $f(1) = 2$. Put $x=0$ in (1): $f(y) = f(0) + 2y^2 + y + a(0)y$ $f(y) = -1 + 2y^2 + y$. Now, we use $f(1)=2$ to find $a$. (Wait, $a$ is not in this expression. This means $a$ must be determined from the original equation using $f(1)=2$ and $f(0)=-1$.) Let's use $f(y) = -1 + 2y^2 + y$ to find $f(1)$: $f(1) = -1 + 2(1)^2 + 1 =

Q: Q.16. Let for some , be a function satisfying for all . If and , then the value of is :

The correct answer is 2. Sol. Given $f(x + y) = f(x) + 2y^2 + y + axy$ ...(1) Given $f(0) = -1$ and $f(1) = 2$. Put $x=0$ in (1): $f(y) = f(0) + 2y^2 + y + a(0)y$ $f(y) = -1 + 2y^2 + y$. Now, we use $f(1)=2$ to find $a$. (Wait, $a$ is not in this expression. This means $a$ must be determined from the original equation using $f(1)=2$ and $f(0)=-1$.) Let's use $f(y) = -1 + 2y^2 + y$ to find $f(1)$: $f(1) = -1 + 2(1)^2 + 1 =

Q: What are the options for this JEE Main Mathematics question?

A. 110 · B. 140 · C. 150 · D. 170

Question

Q.16. Let for some ,  be a function satisfying  for all . If  and , then the value of  is :

FirstInTest · Accepted Answer

Sol. Given $f(x + y) = f(x) + 2y^2 + y + axy$ ...(1)
Given $f(0) = -1$ and $f(1) = 2$.
Put $x=0$ in (1):
$f(y) = f(0) + 2y^2 + y + a(0)y$
$f(y) = -1 + 2y^2 + y$.
Now, we use $f(1)=2$ to find $a$. (Wait, $a$ is not in this expression. This means $a$ must be determined from the original equation using $f(1)=2$ and $f(0)=-1$.)
Let's use $f(y) = -1 + 2y^2 + y$ to find $f(1)$:
$f(1) = -1 + 2(1)^2 + 1 = -1 + 2 + 1 = 2$. This is consistent with the given $f(1)=2$.
This means the function $f(x) = 2x^2+x-1$ is a solution to $f(y) = -1+2y^2+y$.
Let's check if $f(x) = 2x^2+x-1$ satisfies the original functional equation for some $a$.
$f(x+y) = 2(x+y)^2 + (x+y) - 1 = 2(x^2+2xy+y^2) + x+y - 1 = 2x^2+4xy+2y^2+x+y-1$.
We are given $f(x+y) = f(x) + 2y^2 + y + axy$.
Substitute $f(x) = 2x^2+x-1$:
$f(x) + 2y^2 + y + axy = (2x^2+x-1) + 2y^2 + y + axy$.
Comparing the two expressions for $f(x+y)$:
$2x^2+4xy+2y^2+x+y-1 = 2x^2+x-1 + 2y^2+y + axy$.
$4xy = axy$.
This must hold for all $x,y \in R$. So $a=4$.
Now we need to calculate $\sum_{n=1}^{5}(a + f(n))$.
$a=4$.
$f(n) = 2n^2+n-1$.
So we need to calculate $\sum_{n=1}^{5}(4 + 2n^2+n-1) = \sum_{n=1}^{5}(2n^2+n+3)$.
$= 2\sum_{n=1}^{5}n^2 + \sum_{n=1}^{5}n + \sum_{n=1}^{5}3$.
We know $\sum_{n=1}^{N}n = \frac{N(N+1)}{2}$ and $\sum_{n=1}^{N}n^2 = \frac{N(N+1)(2N+1)}{6}$.
For $N=5$:
$\sum_{n=1}^{5}n = \frac{5(6)}{2} = 15$.
$\sum_{n=1}^{5}n^2 = \frac{5(6)(11)}{6} = 55$.
$\sum_{n=1}^{5}3 = 3 	imes 5 = 15$.
So the sum is $2(55) + 15 + 15 = 110 + 30 = 140$.
This matches option (2).

Answer

110

Answer

140

Answer

150

Answer

170

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