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Q.Prove that 13x+1+1x+1−1(3x+1)(x+1)\dfrac{1}{3x + 1} + \dfrac{1}{x + 1} - \dfrac{1}{(3x + 1)(x + 1)} does not lie between 11 and 44, if xx is real.

Telangana TsbieTelangana Board of Intermediate Education 2023Subjective· 4mImportance★★★★★
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The expression equals y=4x+13x2+4x+1y = \dfrac{4x+1}{3x^2+4x+1}; requiring real xx forces (y−1)(y−4)≥0(y-1)(y-4)\ge 0, so yy cannot lie strictly between 11 and 44.

First simplify. The common denominator is (3x+1)(x+1)(3x+1)(x+1):

13x+1+1x+1−1(3x+1)(x+1)=(x+1)+(3x+1)−1(3x+1)(x+1)=4x+1(3x+1)(x+1)\dfrac{1}{3x+1} + \dfrac{1}{x+1} - \dfrac{1}{(3x+1)(x+1)} = \dfrac{(x+1) + (3x+1) - 1}{(3x+1)(x+1)} = \dfrac{4x+1}{(3x+1)(x+1)}.

Since (3x+1)(x+1)=3x2+4x+1(3x+1)(x+1) = 3x^2 + 4x + 1, put

y=4x+13x2+4x+1y = \dfrac{4x+1}{3x^2 + 4x + 1}.

Cross-multiplying: y(3x2+4x+1)=4x+1y(3x^2 + 4x + 1) = 4x + 1, i.e.

3y x2+(4y−4)x+(y−1)=03y\,x^2 + (4y - 4)x + (y - 1) = 0.

For xx to be real, the discriminant must be ≥0\ge 0:

(4y−4)2−4(3y)(y−1)≥0(4y-4)^2 - 4(3y)(y-1) \ge 0

16(y−1)2−12y(y−1)≥016(y-1)^2 - 12y(y-1) \ge 0

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