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QJEEM22Ma9ea5: JEE Main 2022 Mathematics - Class 11 JEEM22Ma9ea5 Solution: Let be two non-zero real numbers. If and are the roots of the…

Let a,ba, b be two non-zero real numbers. If pp and rr are the roots of the equation x2−8ax+2a=0x^{2}-8 \mathrm{a} x+2 \mathrm{a}=0 and q\mathrm{q} and s are the roots of the equation x2+12 bx+6 b=0x^{2}+12 \mathrm{~b} x+6 \mathrm{~b}=0 , such that 1p,1q,1r,1 s\frac{1}{\mathrm{p}}, \frac{1}{\mathrm{q}}, \frac{1}{\mathrm{r}}, \frac{1}{\mathrm{~s}} are in A.P., then a−1−b−1\mathrm{a}^{-1}-\mathrm{b}^{-1} is equal to _____________.

Step-by-Step Verified Solution

∵\because Roots of 2ax2−8ax+1=02 a x^{2}-8 a x+1=0 are 1p\frac{1}{p} and 1r\frac{1}{r} and roots of 6bx2+12bx+1=06 b x^{2}+12 b x+1=0 are 1q\frac{1}{q} and 1s\frac{1}{s} .

Let 1p,1q,1r,1s\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s} as α−3β,α−β,α+β,α+3β\alpha-3 \beta, \alpha-\beta, \alpha+\beta, \alpha+3 \beta

So sum of roots 2α−2β=42 \alpha-2 \beta=4 and 2α+2β=−22 \alpha+2 \beta=-2

Clearly α=12\alpha=\frac{1}{2} and β=−32\beta=-\frac{3}{2}

Now product of roots, 1p⋅1r=12a=−5⇒1a=−10\frac{1}{p} \cdot \frac{1}{r}=\frac{1}{2 a}=-5 \Rightarrow \frac{1}{a}=-10

and 1q⋅1x=16b=−8⇒1b=−48\frac{1}{q} \cdot \frac{1}{x}=\frac{1}{6 b}=-8 \Rightarrow \frac{1}{b}=-48

So, 1a−1b=38\frac{1}{a}-\frac{1}{b}=38

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