
If $f(x) = a{x^2} + bx + c$, $g(x) = p{x^2} + qx + r$ such that $f(1) = g(1)$, $f(2) = g(2)$ and $f(3) - f(3) = 2$. Then $f(4) - g(4)$ is
A $4$
B $5$
C $6$
D $7$
Answer
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Hint: First we put $x = 1$ in $f(x) = a{x^2} + bx + c$ and $g(x) = p{x^2} + qx + r$ then find an expression using $f(1) = g(1)$. Then put $x = 2$ in $f(x) = a{x^2} + bx + c$ and $g(x) = p{x^2} + qx + r$ then find an expression using $f(2) = g(2)$. Then using $f(3) - f(3) = 2$ and all other resultant expressions we will find the value of $a - p$, $b - q$ and $c - r$. Then put these values in $f(4) - g(4)$to get the required answer.
Complete step by step solution: Given, $f(x) = a{x^2} + bx + c$----(1)
Put $x = 1$
$f(1) = a + b + c$
$g(x) = p{x^2} + qx + r$ ----(2)
Put $x = 1$
$g(1) = p + q + r$
Given, $f(1) = g(1)$
$a + b + c = p + q + r$---- (3)
Put $x = 2$ in equation (1)
$f(2) = 4a + 2b + c$
Put $x = 2$ in equation (2)
$g(2) = 4p + 2q + r$
Given, $f(2) = g(2)$
$4a + 2b + c = 4p + 2q + r$ ----(4)
Subtracting equation (3) from equation (4), we get
$3a + b = 3p + q$
Put $x = 3$ in equation (1)
$f(3) = 9a + b + c$
Put $x = 3$ in equation (2)
$g(3) = 9p + 3q + r$
Given, $f(3) - f(3) = 2$
$9a + 3b + c - 9p - 3q - r = 2$
$3(3a + b) + c - 3(3p + q) - r = 2$
$3(3p + q) + c - 3(3p + q) - r = 2$
$c - r = 2$
From equation (3)
$(a - p) + (b - q) + (c - r) = 0$
$(a - p) + (b - q) + 2 = 0$----(5)
From equation (4)
$4(a - p) + 2(b - q) + (c - r) = 0$
$4(a - p) + 2(b - q) + 2 = 0$
Dividing both sides by 2
$2(a - p) + (b - q) + 1 = 0$ ----(6)
Subtracting equation (5) from equation (6)
$(a - p) - 1 = 0$
$a - p = 1$
Putting in equation (5)
$1 + (b - q) + 2 = 0$
$b - q = - 3$
$f(4) - g(4) = (16a + 4b + c) - (16p + 4q + r)$
$ = 16(a - p) + 4(b - q) + (c - r)$
$ = 16(1) + 4( - 3) + 2$
$ = 16 - 12 + 2$
$ = 6$
Hence, the value of $f(4) - g(4)$ is $6$
Therefore, option C is correct.
Note: Students should do calculations carefully and do not skip any step as the solution is lengthy. If they skip any step they can make mistakes. And should use the given information correctly to get the required solution.
Complete step by step solution: Given, $f(x) = a{x^2} + bx + c$----(1)
Put $x = 1$
$f(1) = a + b + c$
$g(x) = p{x^2} + qx + r$ ----(2)
Put $x = 1$
$g(1) = p + q + r$
Given, $f(1) = g(1)$
$a + b + c = p + q + r$---- (3)
Put $x = 2$ in equation (1)
$f(2) = 4a + 2b + c$
Put $x = 2$ in equation (2)
$g(2) = 4p + 2q + r$
Given, $f(2) = g(2)$
$4a + 2b + c = 4p + 2q + r$ ----(4)
Subtracting equation (3) from equation (4), we get
$3a + b = 3p + q$
Put $x = 3$ in equation (1)
$f(3) = 9a + b + c$
Put $x = 3$ in equation (2)
$g(3) = 9p + 3q + r$
Given, $f(3) - f(3) = 2$
$9a + 3b + c - 9p - 3q - r = 2$
$3(3a + b) + c - 3(3p + q) - r = 2$
$3(3p + q) + c - 3(3p + q) - r = 2$
$c - r = 2$
From equation (3)
$(a - p) + (b - q) + (c - r) = 0$
$(a - p) + (b - q) + 2 = 0$----(5)
From equation (4)
$4(a - p) + 2(b - q) + (c - r) = 0$
$4(a - p) + 2(b - q) + 2 = 0$
Dividing both sides by 2
$2(a - p) + (b - q) + 1 = 0$ ----(6)
Subtracting equation (5) from equation (6)
$(a - p) - 1 = 0$
$a - p = 1$
Putting in equation (5)
$1 + (b - q) + 2 = 0$
$b - q = - 3$
$f(4) - g(4) = (16a + 4b + c) - (16p + 4q + r)$
$ = 16(a - p) + 4(b - q) + (c - r)$
$ = 16(1) + 4( - 3) + 2$
$ = 16 - 12 + 2$
$ = 6$
Hence, the value of $f(4) - g(4)$ is $6$
Therefore, option C is correct.
Note: Students should do calculations carefully and do not skip any step as the solution is lengthy. If they skip any step they can make mistakes. And should use the given information correctly to get the required solution.
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