If Rolle-s theorem holds for the function f x -2 x - 3- bx - 2- cx - x lin -1-1- at the point x -1-2- then 2 b - c equals -A- 1B- 2C- -1D- -3

The correct option is C 1 nbsp;By Rolle's theorem f(1)=f( 1) 2 + b + c = – 2 + b – c ⇒ c = – 2 f'(x)=6x^2+ 2bx + c f'(1/2)=3/2+b+c=0 ⇒ b=1/2 So,(2b + c ...

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

Byju's Answer

Standard XII

Mathematics

Range of Quadratic Expression

If Rolle's th...

Question

If Rolle's theorem holds for the function $f (x) = 2 x^{3} + b x^{2} + c x, x \in [- 1, 1],$ at the point $x = \frac{1}{2}$ , then $2 b + c$ equals :

1

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

2

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

- 1

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

- 3

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

Open in App

Solution

The correct option is C $- 1$

By Rolle's theorem
$f (1) = f (- 1) 2 + b + c = - 2 + b - c \Rightarrow c = - 2 f^{'} (x) = 6 x^{2} + 2 b x + c f^{'} (\frac{1}{2}) = \frac{3}{2} + b + c = 0 \Rightarrow b = \frac{1}{2} So, (2 b + c) = - 1$

Suggest Corrections

Similar questions

Q. If Rolle's theorem holds true for the function

f (x) = 2 x^{3} + b x^{2} + c x, x \in [- 1, 1]

at the point

x = \frac{1}{2}

, then

(2 b + c)

is equal to

Q. If Rolle's theorem holds for the function

f (x) = 2 x^{3} + b x^{2} + c x, x \in [- 1, 1],

at the point

x = \frac{1}{2},

then

2 b + c

equals:

Q. If Rolle's theorem holds true for the function

f (x) = 2 x^{3} + b x^{2} + c x, x \in [- 1, 1]

at the point

x = \frac{1}{2}

, then

(2 b + c)

is equal to

Q. If Rolle's theorem holds for the function

f (x) = 2 x^{3} + b x^{2} + c x, x \in [- 1, 1],

at the point

x = \frac{1}{2}

, then

8 b + c

Q. Rolle's theorem hold for the function

f (x) = x^{3} + b x^{2} + c x, 1 \leq x \leq 2

at the point

4 / 3

, the values of b and c are

Join BYJU'S Learning Program

If Rolle's theorem holds for the function f(x)=2x3+bx2+cx,x∈[–1,1],at the point x=12, then 2b+c equals :

The correct option is C −1 By Rolle's theorem f(1)=f(−1)2+b+c=–2+b–c⇒c=–2f′(x)=6x2+2bx+cf′(12)=32+b+c=0⇒b=12So,(2b+c)=−1

If Rolle's theorem holds for the function $f (x) = 2 x^{3} + b x^{2} + c x, x \in [- 1, 1],$ at the point $x = \frac{1}{2}$ , then $2 b + c$ equals :

The correct option is C $- 1$

By Rolle's theorem
$f (1) = f (- 1) 2 + b + c = - 2 + b - c \Rightarrow c = - 2 f^{'} (x) = 6 x^{2} + 2 b x + c f^{'} (\frac{1}{2}) = \frac{3}{2} + b + c = 0 \Rightarrow b = \frac{1}{2} So, (2 b + c) = - 1$