Test Index

Quadratic Equation and Inequalities

Section: Mathematics

© examsnet.com

Question : 2 of 24

Marks: +1, -0

Let

\mathbb{R}

denote the set of all real numbers. Let

a_i, b_i \in \mathbb{R}

i \in \{1,2,3\}

Define the functions

f: \mathbb{R} \longrightarrow \mathbb{R}, g: \mathbb{R} \longrightarrow \mathbb{R}

h: \mathbb{R} \longrightarrow \mathbb{R}

f(x)=a_1+10x+a_2 x^2+a_3 x^3+x^4

g(x)=b_1+3x+b_2 x^2+b_3 x^3+x^4

h(x)=f(x+1)-g(x+2)

f(x) \neq g(x)

x \in \mathbb{R}

, then the coefficient of

x^3

h(x)

[JEE Adv 2025 P1]

8
2
-4
-6

Solution:

f(x+1)=a_1+10(x+1)+a_2(x+1)^2+a_3(x+1)^3+(x+1)^4

Coefficient of

x^3

in

f(x+1)=a_3+4

g(x+2)=b_1+3(x+2)+b_2(x+2)^2+b_3(x+2)^3+(x+2)^4

Coefficient of

x^3

in

g(x+2)=b_3+8

\Rightarrow

Coefficient of

x^3

in

h(x)=f(x+1)-g(x+2)

is

a_3+4-b_3-8=a_3-b_3-4

But

f(x) \neq g(x) \forall x

\Rightarrow f(x)-g(x) \neq 0

\Rightarrow f(x)-g(x)=0

have no real roots

(a_1-b_1) +7x+ (a_2-b_2) x^2+ (a_3-b_3) x^3=0

have no real roots

\Rightarrow a_3-b_3=0

\Rightarrow

Coefficient of

x^3

in

h(x)=-4

Option (C) is correct.

© examsnet.com

Go to Question:

Prev Question

Next Question