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Miscellaneous 7 I · Q101

Q.lim⁡x→3(1x2−11x+24+1x2−x−6)=\displaystyle\lim_{x\to 3}\left(\frac{1}{x^2-11x+24}+\frac{1}{x^2-x-6}\right)= (A) −225-\dfrac{2}{25} (B) 225\dfrac{2}{25} (C) 725\dfrac{7}{25} (D) −725-\dfrac{7}{25}

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✓ Free question

x2−11x+24=(x−3)(x−8)x^2-11x+24=(x-3)(x-8); x2−x−6=(x−3)(x+2)x^2-x-6=(x-3)(x+2). Sum =1(x−3)(x−8)+1(x−3)(x+2)=1x−3[1x−8+1x+2]=1x−3⋅2x−6(x−8)(x+2)=1x−3⋅2(x−3)(x−8)(x+2)=2(x−8)(x+2)=\dfrac{1}{(x-3)(x-8)}+\dfrac{1}{(x-3)(x+2)}=\dfrac{1}{x-3}\left[\dfrac{1}{x-8}+\dfrac1{x+2}\right]=\dfrac{1}{x-3}\cdot\dfrac{2x-6}{(x-8)(x+2)}=\dfrac{1}{x-3}\cdot\dfrac{2(x-3)}{(x-8)(x+2)}=\dfrac{2}{(x-8)(x+2)} after cancelling (x−3)(x-3). At x=3x=3: 2(−5)(5)=−225\dfrac{2}{(-5)(5)}=-\dfrac{2}{25}.

✓Final answer

−225-\dfrac{2}{25} (option A)

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