JEE Main 2022 — Mathematics Calculus

Given \displaystyle 12\times {\int }_{3}^{b}\frac{1dx}{({x}^{2}-4)({x}^{2}-1)}=\mathrm{ln}[\frac{49}{40}]&4b>3

$\Rightarrow$⇒ By Using partial fraction

$$\frac{1}{({x}^{2}-4)({x}^{2}-1)}=\frac{1}{3}(\frac{1}{{x}^{2}-4}-\frac{1}{{x}^{2}-1})$$(x2−4)(x2−1)1​=31​(x2−41​−x2−11​)

$$\Rightarrow \frac{12}{3}{\int }_{3}^{b}[\frac{1}{{x}^{2}-4}-\frac{1}{{x}^{2}-1}]=\mathrm{ln}(\frac{49}{40})$$⇒312​∫3b​[x2−41​−x2−11​]=ln(4049​)

$$\Rightarrow 4{[\frac{1}{4}\mathrm{log}(\frac{x-2}{x+2})-\frac{1}{2}\mathrm{log}(\frac{x-1}{x+1})]}_{3}^{b}=\mathrm{ln}(\frac{49}{40})$$⇒4[41​log(x+2x−2​)−21​log(x+1x−1​)]3b​=ln(4049​)

$$\Rightarrow \mathrm{log}(\frac{(b-2){(b+1)}^{2}}{(b+2){(b-1)}^{2}}\times \frac{5}{4})=\mathrm{log}\frac{49}{40}$$⇒log((b+2)(b−1)2(b−2)(b+1)2​×45​)=log4049​

$$\Rightarrow \frac{(b-2){(b+1)}^{2}}{(b+2){(b-1)}^{2}}\times \frac{5}{4}=\frac{49}{40}$$⇒(b+2)(b−1)2(b−2)(b+1)2​×45​=4049​

$$\Rightarrow \frac{(b-2){(b+1)}^{2}}{(b+2){(b-1)}^{2}}=\frac{49}{50}...(i)$$⇒(b+2)(b−1)2(b−2)(b+1)2​=5049​...(i)

Comparing by hit and trial and taking ${(b+1)}^{2}=49\Rightarrow b+1=7\Rightarrow b=6$(b+1)2=49⇒b+1=7⇒b=6

Now putting $b=6$b=6 in equation (i) to corsscheck,

we get $\frac{4\times {7}^{2}}{8\times {5}^{2}}=\frac{49}{50}$8×524×72​=5049​ which is true

So $b=6$b=6 is correct option,

Hence $b=6$b=6