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Bunuel
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The equation of the curve shown in the figure is given by af(x) = ax2 19 Jul 2019, 00:52
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38% (02:31) correct 62% (02:52) wrong based on 77 sessions

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The equation of the curve shown in the figure is given by \(af(x) = ax^2 + bx – 3b\). If (1, t) is a point on the curve, then in which quadrant does the point lie?

(1) (2, 3) is a point on the curve.
(2) (3, 9) is a point on the curve.


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2019-07-19_1149.png
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A
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The equation of the curve shown in the figure is given by af(x) = ax2 19 Jul 2019, 01:22
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Equation of Curve: af(x)=ax^2+bx–3b

1. (2,3) lies on the curve:
a*f(2) = 4a +2b -3b
a*3 = 4a -b
=> a=b
Thus, if (1,t) lies on the curve then:
a*f(1) = a +b -3b
a*t = a -2*b
t = \(\frac{a-2b}{a}\)
using a=b:
t = \(\frac{a-2a}{a}\) = -1
Thus (1,-1) lies in 4th quadrant.
Sufficient.

2. (3,9) lies on the curve:
a*f(3) = 9a +3b -3b
a*9 = 9a
1=1
Thus we do not know about the relation between a and b, we can`t find the position of (1,t).
Insufficient.

IMO the answer is A.

Please hit kudos if you like the solution.
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The equation of the curve shown in the figure is given by af(x) = ax2 19 Jul 2019, 01:37
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af(x)= ax^2+bx-3b.

Statement1: (2,3) is a point on the curve.

if we put the values of point (2,3) and (1,t) to the Quadratic function:
3a=4a+2b-3b ---> a=b
at=a+b-3b=a-2b ---> at/a=a/a-2b/a --> t=1-2b/a

a=b
t=1-2b/a --> t=1-2=-1
point (1,-1) is in quadrant IV.
Sufficient.


Statement2: (3,9) is a point on the curve.

if we put the value of this point to the Quadratic function:
9a=a*3^2+3b-3b
9a=9a --> 0=0. ---> we can't find the value of t with one point alone. It could be in quadrant I or IV.
Insufficient.

The answer choice is A
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The equation of the curve shown in the figure is given by af(x) = ax2 15 Nov 2022, 00:37
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Bunuel

The equation of the curve shown in the figure is given by \(af(x) = ax^2 + bx – 3b\). If (1, t) is a point on the curve, then in which quadrant does the point lie?

(1) (2, 3) is a point on the curve.
(2) (3, 9) is a point on the curve.


Show Spoiler
Attachment:
2019-07-19_1149.png
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Solution:
Pre Analysis:
  • We are given equation \(af(x) = ax^2 + bx – 3b\) which can be written as \(ay = ax^2 + bx – 3b\)
  • \((1, t)\) lies on the curve which means plsugging \(x=1\) and \(y=t\) should satisfy \(ay = ax^2 + bx – 3b\)
    \(⇒at=a.1^2+b.1-3b\)
    \(⇒at=a+b-3b\)
    \(⇒at=a-2b\)
    \(⇒t=1-2\frac{b}{a}\)
  • To get the quadrant of \((1, t)\), we need the value of 't' for which we either need individual values of a and b or some relation between them

Statement 1: (2, 3) is a point on the curve
  • According to this statement (2, 3) should satisfy \(ay = ax^2 + bx – 3b\)
  • So, upon plugging, we get:
    \(⇒3a=4a+2b-3b\)
    \(⇒-a=-b\)
    \(⇒a=b\)
  • We know \(t=1-2\frac{b}{a}=1-2=-1\)
  • Thus, statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2: (3, 9) is a point on the curve
  • According to this statement (3, 9) should satisfy \(ay = ax^2 + bx – 3b\)
  • So, upon plugging, we get:
    \(⇒9a=9a+3b-3b\)
    \(⇒0=0\)
  • Thus, statement 2 alone is not sufficient

Hence the right answer is Option A
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The equation of the curve shown in the figure is given by af(x) = ax2 11 Feb 2023, 08:55
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SaquibHGMATWhiz
Bunuel

The equation of the curve shown in the figure is given by \(af(x) = ax^2 + bx – 3b\). If (1, t) is a point on the curve, then in which quadrant does the point lie?

(1) (2, 3) is a point on the curve.
(2) (3, 9) is a point on the curve.


Show Spoiler
Attachment:
2019-07-19_1149.png

Solution:
Pre Analysis:
  • We are given equation \(af(x) = ax^2 + bx – 3b\) which can be written as \(ay = ax^2 + bx – 3b\)
  • \((1, t)\) lies on the curve which means plsugging \(x=1\) and \(y=t\) should satisfy \(ay = ax^2 + bx – 3b\)
    \(⇒at=a.1^2+b.1-3b\)
    \(⇒at=a+b-3b\)
    \(⇒at=a-2b\)
    \(⇒t=1-2\frac{b}{a}\)
  • To get the quadrant of \((1, t)\), we need the value of 't' for which we either need individual values of a and b or some relation between them

Statement 1: (2, 3) is a point on the curve
  • According to this statement (2, 3) should satisfy \(ay = ax^2 + bx – 3b\)
  • So, upon plugging, we get:
    \(⇒3a=4a+2b-3b\)
    \(⇒-a=-b\)
    \(⇒a=b\)
  • We know \(t=1-2\frac{b}{a}=1-2=-1\)
  • Thus, statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2: (3, 9) is a point on the curve
  • According to this statement (3, 9) should satisfy \(ay = ax^2 + bx – 3b\)
  • So, upon plugging, we get:
    \(⇒9a=9a+3b-3b\)
    \(⇒0=0\)
  • Thus, statement 2 alone is not sufficient

Hence the right answer is Option A
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Hi Saquib, can you please elaborate why we can't plug in the given (2,3) and (3,9) to t=1−2(b/a) and why it has to be in the original equation af(x)=ax^2+bx-3y? Thanks
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The equation of the curve shown in the figure is given by af(x) = ax2 17 Jun 2025, 15:18
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