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

Tangent of a Curve y =f(x)

If at x =1, y...

Question

If at x = 1, y = 2x is tangent to the parabola $y = a x^{2} + b x + c$ , then respective values of a, b, c are

\frac{1}{2}, 1, \frac{1}{2}

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

1, \frac{1}{2}, \frac{1}{2}

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

\frac{1}{2}, \frac{1}{2}, 1

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

1, 1, \frac{1}{2}

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

Open in App

Solution

The correct option is A $\frac{1}{2}, 1, \frac{1}{2}$
For x = 1, y = a + b + c
Tangent at (1, a + b + c) is $\frac{1}{2}$ (y + a + b + c) = ax + $\frac{1}{2}$ (x + 1) + c
Comparing with y = 2x , c = a, b = 2(1-a)

Suggest Corrections

Similar questions

x = 1, y = 2 x

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

a, b, c

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

y = x

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

(1, 1)

(- 1, 0)

x = 1, y = 2 x

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

\int R (x, \sqrt{a x^{2} + b x + c}) d x

I. \sqrt{a x^{2} + b x + c} = t \pm x \sqrt{a}

a > 0;

II. \sqrt{a x^{2} + b x + c} = t x \pm \sqrt{c}

c > 0;

III. \sqrt{a x^{2} + b x + c} = (x - a) t \pm x

a x^{2} + b x + c = a (x - α) (x - β)

α

a x^{2} + b x + c = 0.

\frac{1}{1 + \sqrt{x^{2} + 2 x + 2}}

t

x ?

Join BYJU'S Learning Program