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

Equation of Tangent at a Point (x,y) in Terms of f'(x)

If y = fx b...

Question

If $y = f (x)$ be the equation of a parabola which is touched by the line $y = x$ at the point where $x = 1$ Then

f^{'} (1) = 1

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

f^{'} (0) = f^{'} (1)

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

2 f (0) = 1 - f^{'} (0)

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

f (0) + f^{'} (0) + f " (0) = 1

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

Open in App

Solution

The correct options are
C $2 f (0) = 1 - f^{'} (0)$
D $f^{'} (1) = 1$
Let, $y = a x^{2} + b x + c$
$∴ y^{'} = 2 a x + b$
Given line $y = x$ touches this parabola at $x = 1$
$\Rightarrow y (1) = 1 \Rightarrow a + b + c = 1 . . (1)$
and $y " (1) = 1 \Rightarrow 2 a + b = 1 . . (2)$
Solving (1) and (2) we get
$2 c = 1 - b \Rightarrow 2 f (0) = 1 - f^{'} (0)$

Suggest Corrections

Similar questions

y = f (x)

y -

y = x

x = 1

y = f (x)

f (0) = 0

f^{''} (x) = f^{'} (x)

f^{'} (0) = 1

y = f (x)

x = 0, x - 1 = 0

y + 1 = 0

y = f (x)

f (x + y) = f (x) \cdot f (y)

x, y \in R

f (1) = 25

S = f (2) + f (1) + f (0) + f (- 1) + . . . \infty

[(f (1) - 1) S]^{1 / 2}

y = f_{1} (x)

y = f_{2} (x)

y d x + d y = - e^{x} y^{2} d y

f_{1} (0) = 1

f_{2} (0) = - 1

f_{1} (x) . f_{2} (x) + x^{2} = 0

f (x) = {cos}^{- 1} (\frac{x^{- 1} - x}{x^{- 1} + x})

f (0) + f^{'} (0) + \frac{f^{''} (0)}{2!} + . . . + \frac{f^{n} (0)}{n!}

Join BYJU'S Learning Program