Question

Number of ordered pair $(a,b)$ from the set $A=\{1,2,3,4,5\}$ so that the function $f\left(x\right)=\frac{x^3}{3}+\frac{a}{2}{x}^2+bx+10$ is an injective mapping $\forall x\in R$:

• A. $13$
• B. $14$
• C. $15$
• D. $16$

Answer 1

The correct option is C $15$

Given

f(x)=x³/3+ax²/2+bx+10

f'(x)=3x²/3+2ax/2+b

f'(x)=x²+ax+b

Since, D

We know that,

D=b²-4ac

There fore,

D=a²-4b

a=1,. b=(1,2,3,4,5) = 5 Pairs

a=2, b=(1,2,3,4,5) = 5 Pairs

a=3, b=(3,4,5). =3 pairs. (remaining doesn't satisfied the equ)

a=4, b=(4,5). =2 pairs

a=5, b=(5). =0 pairs

Hence,

No. Of ordered pair is 15. For injective function.