If frac{e^x + 2}{(e^x - 1)(2e^x - 3)} = -frac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $u = e^x$. The equation becomes $\frac{u + 2}{(u - 1)(2u - 3)} = -\frac{3}{u - 1} + \frac{B}{2u - 3}$.Combining the terms on the right side:$-

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(D) Let $u = e^x$. The equation becomes $\frac{u + 2}{(u - 1)(2u - 3)} = -\frac{3}{u - 1} + \frac{B}{2u - 3}$.Combining the terms on the right side:$-\frac{3}{u - 1} + \frac{B}{2u - 3} = \frac{-3(2u - 3) + B(u - 1)}{(u - 1)(2u - 3)} = \frac{-6u + 9 + Bu - B}{(u - 1)(2u - 3)} = \frac{(B - 6)u + (9 - B)}{(u - 1)(2u - 3)}$.Comparing the numerators of the original expression and the simplified expression:$u + 2 = (B - 6)u + (9 - B)$.Equating the coefficients of $u$:$B - 6 = 1 \implies B = 7$.Checking the constant term:$9 - B = 9 - 7 = 2$,which matches the constant term in the original numerator.Thus,$B = 7$.

If $\frac{e^x + 2}{(e^x - 1)(2e^x - 3)} = -\frac{3}{e^x - 1} + \frac{B}{2e^x - 3}$,then $B = $

Similar Questions

Which type of cartilage is found at the ends of long bones?

The midpoint of the chord $4x - 3y = 5$ of the hyperbola $2x^2 - 3y^2 = 12$ is

If the circle $x^2+y^2+6x-2y+k=0$ bisects the circumference of the circle $x^2+y^2+2x-6y-15=0$,then $k=$

$A$ coordination complex contains $Co^{3+}$,$Cl^{-}$,and $NH_3$. When dissolved in water,one mole of this complex gives a total of $3$ moles of ions. The complex is

If the amplitude of $z - 2 - 3i$ is $\pi / 4$,then the locus of $z = x + iy$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module