A value of α such that ∫αα+1dx/((x+α)(x+α+1))=log_e9/8 is

Correct answer: (C) -2. I=∫αα+1dx/((x+α)(x+α+1))=∫αα+11/(x+α )–1/(x+α+1 )dx Using partial fractions =lnx+α-lnx+α+1αα+1 =ln(x+α)/(x+α+1) αα+1 =ln(2α+1)/(2α+2)-ln(2α)/(2α+1) =ln2α+1^22α+1^2-1=ln⁡9/8 ⇒(2α+1)^2=9 ⇒2α+1=±3 ⇒α=1 or -2.

Q57 / 90 · JEE Main 2019 (12 Apr Shift 2) · Mathematics

A value of $α$ such that $\int_{α}^{α + 1} \frac{d x}{(x + α) (x + α + 1)} = l o g_{e} (\frac{9}{8})$ is

A
$- \frac{1}{2}$
B
$\frac{1}{2}$
Correct answer
C
$- 2$
D
$2$

Solution

I = \int_{α}^{α + 1} \frac{d x}{(x + α) (x + α + 1)} = \int_{α}^{α + 1} (\frac{1}{x + α} - \frac{1}{x + α + 1}) d x

Using partial fractions

= {[l n |x + α| - l n |x + α + 1|]}_{α}^{α + 1}

= {[l n |\frac{x + α}{x + α + 1}|]}_{α}^{α + 1}

= l n |\frac{2 α + 1}{2 α + 2}| - l n |\frac{2 α}{2 α + 1}|

= l n |\frac{{(2 α + 1)}^{2}}{{(2 α + 1)}^{2} - 1}| = \ln \frac{9}{8}

\Rightarrow {(2 α + 1)}^{2} = 9

\Rightarrow 2 α + 1 = \pm 3

\Rightarrow α = 1

- 2

Hence, the correct answer is option: 'C'.

A value of α such that ∫αα+1dx(x+α)(x+α+1)=loge98 is

Solution

Relevant Questions

A value of $α$ such that $\int_{α}^{α + 1} \frac{d x}{(x + α) (x + α + 1)} = l o g_{e} (\frac{9}{8})$ is