JEE Mains 25-July-2022 Shift 1 Solved Paper

f(x)=y=ax3+bx2+cx+5‌‌...‌ (i) ‌
‌

dy

dx

=3ax2+2bx+c‌‌...(ii)
Touches x-axis at P(−2,0)
.⇒y|x=−2=0⇒−8a+4b−2c+5=0‌‌......‌ (iii) ‌
Touches x-axis at P(−2,0) also implies
.‌

dy

dx

|x=−2=0⇒12a−4b+c=0‌‌......(iv)
y=f(x) cuts y-axis at (0,5)
Given, .‌

dy

dx

|x=0=c=3
From (iii), (iv) and (v)
a=−‌

1

2

,b=−‌

3

4

,c=3
⇒f(x)=‌

−x2

2

−‌

3

4

x2+3x+5
f′(x)=‌

−3

2

x2−‌

3

2

x+3
=‌

−3

2

(x+2)(x−1)
f′(x)=0 at x=−2 and x=1
By first derivative test x=1 in point of local maximum Hence local maximum value of f(x) is f(1)
i.e., ‌

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