
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
163.8k+ views
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
