You visited us 1 times! Enjoying our articles? Unlock Full Access!

Convexity

Suppose a,b...

Question

Suppose $a, b, c$ are such that the curve $y = a x^{2} + b x + c$ is tangent to $y = 3 x - 3$ at $(1, 0)$ and is also tangent to $y = x + 1$ at $(3, 4)$ then the value of $(2 a - b - 4 c)$ equals

7

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $9$
$y = a x^{2} + b x + c$
Since $(1, 0)$ lies on the curve
$\Rightarrow a + b + c = 0$ ...... (i)
Slope of curve $= \frac{d y}{d x} = 2 a x + b$
Slope of curve at (1,0) $= 2 a + b$
Slope of curve at (3,4) $= 6 a + b$
Slope of tangent $y = 3 x - 3$ is 3
Slope of tangent $y = x + 1$ is 1
$(\frac{d y}{d x})_{x = 1} = 3$ and $(\frac{d y}{d x})_{x = 3} = 1$
$2 a + b = 3$ .... (ii)
$6 a + b = 1$ .... (iii)
on solving we get a $= - \frac{1}{2}, b = 4, c = - \frac{7}{2}$
$∴ 2 a - b - 4 c = - 1 - 4 + 14 = 9$

Suggest Corrections

Similar questions

a, b, c

y = a x^{2} + b x + c

y = 3 x - 3

(1, 0)

y = x + 1

(3, 4)

(2 a - b - 4 c)

y = a x^{2} + b x

(- 1, 0)

y = x

x = 1

(a, b)

x = 1, y = 2 x

y = a x^{2} + b x + c

a, b, c

y = a x^{2} + b x + c, x \in R,

(1, 2)

y = x,

a, b, c

y = a x^{2} + b x + c

Join BYJU'S Learning Program