JEE Advanced 2020 Full Test 4 Paper 2

(for Q.47−48)
a,b,c are three roots of the quadratic equation
((4f(−1)−3)x2+(4f(1)−3))x+f(2)=0
⇒4f(−1)=3,4f(1)=3 and f(2)=0
Let f(x)=(x−2)(Ax+B)
Now, 3=4f(−1)=4(−3)(−A+B)⇒A−B=‌

1

4

3=4f(−1)=4(−1)(A+B)⇒A+B=‌

−3

4

∴A=‌

−1

4

,B=‌

−2

4

Thus, f(x)=‌

1

4

(4−x2)
The graph of y=f(x) is given in Figure.
(I). x coordinates of points of intersection of y=f(x) when the x− axis are ±2 .
(II). Area =‌

3

2

‌

2

∫

−2

‌

1

4

(4−x2)dx=‌

3

8

(2)(4x−‌

x2

3

)]02=4
(III). Maximum value of f(x) is 1
(IV). Length of intercept on the x - axis is 4