WBJEE 2024 Math Paper

To find the values of a that satisfy the inequality

1

√a

‌

a

∫

1

(

3

2

√x+1−

1

√x

)‌dx<4, we need to first evaluate the integral and then solve the inequality.
Consider the integral

a

∫

1

(

3

2

√x+1−

1

√x

)‌dx
We will solve this integral term by term:
For the term ‌

3

2

√x :
For the term 1 :
∫1dx=x
For the term −‌

1

√x

:
∫−‌

1

√x

dx=−∫x−

1

2

dx=−2x1∕2
Now, combining these results, we have:
Evaluating this at the bounds x=1 and x=a
(x

3

2

+x−2x

1

2

)|1a=(a

3

2

+a−2a

1

2

)−(1

3

2

+1−2.1

1

2

)
‌=a3∕2+a−2a1∕2−(1+1−2)
‌=a3∕2+a−2a1∕2−0
‌=a3∕2+a−2a1∕2
Now, plugging this back into the inequality:
‌

1

√a

(a3∕2+a−2a1∕2)<4
Simplify the expression inside the inequality:

Now, let √a=t. Hence, a=t2. Substituting t into the inequality:
t2+t−2<4
Rearranging the inequality:
t2+t−6<0
We solve the quadratic equation t2+t−6=0 by factoring it:
(t+3)(t−2)=0
The roots are t=−3 and t=2. The quadratic inequality t2+t−6<0 is satisfied between the roots:

−3<t<2
Since √a=t, this translates to:
−3<√a<2
However, since √a represents a real, non-negative quantity, the inequality reduces to:
0≤√a<2
Squaring both sides to return to a-terms:
0≤a<4
The appropriate interval for a is:
Therefore, the values of a that satisfy the original inequality lie in the interval [0,4). Since a cannot be exactly 0 (as it would make the original integral undefined), the interval we need is:
Option C: (0,4)