AP EAMCET Engineering 21 Sep 2020 Shift 2 Paper

(a) The equation of given curve is
y=ax3+bx2+cx+5‌‌......(i)
∵ Curve (i) passes through point P(−2,0), so
−8a+4b−2c+5=0‌‌(ii)
and curve (i) cuts the Y-axis at a point Q with gradient 3, so ‌

dy

dx

|Q=3
⇒‌‌3ax2+2bx+c|x=0⇒c=3‌‌.....(iii)
∵ Point Q have coordinates (0,5), so curve (i) passes through Q(0,5).
From Eqs. (ii) and (iii), we have 8a−4b+1=0.
Now, from the options
c=3,a=−‌

1

2

‌ and ‌b=−‌

3

4