CAT 2020 Exam Shift 2 Solved Paper

© examsiri.com

Question : 72 of 76

Marks: +1, -0

Let

f(x)=x^{2}+a x+b

and

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

. If

f(x) ≥ 0

for all real

{x}

, and

g(20)=72

, then the smallest possible value of

{b}

is

1
16
0
4

Solution:

f(x)=x^{2}+a x+b

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

=\{(x+1)^{2}+a(x+1)+b\}-\{(x-1)^{2}+a(x-1)+b\}

g(x)=4 x+2 a

g(20)=72

80+2 a=72 ⇒ a=-4

∴ f(x)=x^{2}-4 x+b

f(x)=(x-2)^{2}+b-4

when

b ≥ 4 f(x) ≥ 0

for all

x

∴

The minimum value of

b

is 4

© examsiri.com

Go to Question:

Prev Question

Next Question