3x-4x-12x-1-Ax-1-Bx-1-Cx-12 the A-

The correct option is A 7/4 3x+4(x+1)^2(x 1)=Ax 1+Bx+1+C(x+1)^2 ⇒ 3x+4=A(x+1)^2+B(x 1)(x+1)+C(x 1) keeping the value of x=1 we get 4A=7 ...

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Byju's Answer

Standard XII

Mathematics

Inverse of a Function

3x+4x+12x-1=A...

Question

$\frac{3 x + 4}{(x + 1)^{2} (x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^{2}}$ the $A =$

\frac{7}{4}

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\frac{1}{4}

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\frac{10}{3}

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Solution

The correct option is A $\frac{7}{4}$
$\frac{3 x + 4}{(x + 1)^{2} (x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^{2}}$
$\Rightarrow 3 x + 4 = A (x + 1)^{2} + B (x - 1) (x + 1) + C (x - 1)$
keeping the value of $x = 1$
we get
$4 A = 7$
$A = 7 / 4$

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Similar questions

Q. If

\frac{3 x + 4}{(x + 1)^{2} (x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^{2}}

, then

C =

Q. If

\frac{3 x^{3} - 8 x^{2} + 10}{(x - 1)^{4}} = \frac{A}{x - 1} + \frac{B}{(x - 1)^{2}} + \frac{C}{(x - 1)^{3}} + \frac{D}{(x - 1)^{4}}

, then

A + B + C + D

is equal to

Q. Factorise the expression 4x + 12xy.

Q. Question 13
x + 1 is a factor of the polynomial

A)

x^{3} + x^{2} - x + 1

x^{3} + x^{2} + x + 1

x^{4} + x^{3} + x^{2} + 1

x^{4} + 3 x^{3} + 3 x^{2} + x + 1

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$

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3x+4(x+1)2(x−1)=Ax−1+Bx+1+C(x+1)2 the A=

The correct option is A 743x+4(x+1)2(x−1)=Ax−1+Bx+1+C(x+1)2⇒3x+4=A(x+1)2+B(x−1)(x+1)+C(x−1)keeping the value of x=1we get4A=7A=7/4

$\frac{3 x + 4}{(x + 1)^{2} (x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^{2}}$ the $A =$

The correct option is A $\frac{7}{4}$
$\frac{3 x + 4}{(x + 1)^{2} (x - 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{(x + 1)^{2}}$
$\Rightarrow 3 x + 4 = A (x + 1)^{2} + B (x - 1) (x + 1) + C (x - 1)$
keeping the value of $x = 1$
we get
$4 A = 7$
$A = 7 / 4$