CAT 2022 Slot 3 Paper

Let

x2−6x+10

3−x

=p
x2−6x+10=3p−px
x2−(6−p)x+10−3p=0
Since the equation will have real roots,
(6−p)2−4×(10−3p)≥0
p2−12p+12p+36−40≥0
p2≥4
p≥2,p≤−2
Now, when p=−2,x=4. Since it is given that x<3, thus this value will be discarded.
Now,

1

2

and −

1

2

do not come in the mentioned range.
when p=2,x=2
Thus, the minimum possible value of p will be 2 .
Thus, the correct option is B.

Alternate explanation:
Since x<3
3−x>0
Let 3−x=y. So, y>0.
Now,

x2−6x+10

3−x

=

x2−6x+9+1

3−x

⇒

(3−x)2+1

3−x

Since 3−x=y, the equation will transform to

y2+1

y

or y+

1

y

The minimum value of the expression y+‌

1

y

for y>0 will at y=1 i.e., Minimum value =1+1=2
Thus, the correct option is B.