If fx-1-x1-x21-x3- then f-1-A- 28B- 32C- 24D- 20

The correct option is C 24 f(x) = (1 + x) (1 + x^2)(1 + x^3) f’(x) = (1 + x) (1 + x^2) 3x^2+(1 + x)(1 + x^3).2x+ (1 + x^2) (1 + x^3) f’(1) = 12 + 8 + 4 = 24 ...

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

Byju's Answer

Standard XII

Mathematics

Sum of Coefficients of All Terms

If fx=1+x1+x2...

Question

If f(x) = (1 + x) $(1 + x^{2}) (1 + x^{3})$ , then f'(1) =

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

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

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

Open in App

Solution

The correct option is B
24

$f (x) = (1 + x) (1 + x^{2}) (1 + x^{3})$

$f^{'} (x) = (1 + x) (1 + x^{2}) 3 x^{2} + (1 + x) (1 + x^{3}) .2 x + (1 + x^{2}) (1 + x^{3})$

$f^{'} (1) = 12 + 8 + 4 = 24$

Suggest Corrections

Similar questions

Q. If

f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) . . . \infty

then

f (\frac{1}{2}) =

Q. If

x_{1}, x_{2}, x_{3}

be the roots of the equation

x^{3} - x + 1 = 0

, then the value of

(\frac{1 + x_{1}}{1 - x_{1}}) (\frac{1 + x_{2}}{1 - x_{2}}) + (\frac{1 + x_{1}}{1 - x_{1}}) (\frac{1 + x_{3}}{1 - x_{3}}) + (\frac{1 + x_{2}}{1 - x_{2}}) (\frac{1 + x_{3}}{1 - x_{3}})

Q. Find the derivative of the function f (x) given by

f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8})

and hence find f' (1)

Q. Coefficient of

x^{9}

\to (1 + x) ((1 + x^{2}) (1 + x^{3}) . . . . . . (1 + x^{100})) ?

Q. Let

f (x) = \frac{x (1 + a cos x) - b sin x}{x^{3}}, x \neq 0 a n d f (0) = 1

, then values if 'a' and 'b' so that 'f' is continuous are

Join BYJU'S Learning Program

If f(x) = (1 + x) (1+x2)(1+x3), then f'(1) =

The correct option is B 24 f(x)=(1+x)(1+x2)(1+x3) f′(x)=(1+x)(1+x2)3x2+(1+x)(1+x3).2x+(1+x2)(1+x3) f′(1)=12+8+4=24

If f(x) = (1 + x) $(1 + x^{2}) (1 + x^{3})$ , then f'(1) =

The correct option is B
24

$f (x) = (1 + x) (1 + x^{2}) (1 + x^{3})$

$f^{'} (x) = (1 + x) (1 + x^{2}) 3 x^{2} + (1 + x) (1 + x^{3}) .2 x + (1 + x^{2}) (1 + x^{3})$

$f^{'} (1) = 12 + 8 + 4 = 24$