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Numbers which are not in the range of \( f(x)=\frac{x^{4}-5 x^{2}+4}{x\left(x^{3}+2 x^{2}-x-2\right)} \), are(A) 1 (B) 2(C) 3 (D) \( -1 \) |
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Answer» Correct option are (A) 1 and (B) 2 Given function is f(x) = \(\cfrac{x^4-5x^2+4}{x(x^3+2x^2-x-2}\) = \(\cfrac{(x^2-1)(x^2-4)}{x(x-1)(x^2+3x+2)}\) = \(\cfrac{(x-1)(x+1)(x+2)}{x(x-1)(x+1)(x+2)}\) \(\therefore\) Domain of f(x) = R -{0,1,-1,-2} And f(x) = \(\cfrac{x-2}x\) (By taking f(x) = y = x = f-1(y) = xy = x - 2 = xy - x = -2 = x (y-1) = -2 = x = \(\cfrac{2}{1-y}\) \(\therefore\) f-1(y) = \(\cfrac{2}{1-y}\) (\(\because\) f(x) = y = x = f-1(y)) = f-1(x) = \(\cfrac{2}{1-x}\) ; x ≠ 1 put x = 0 then f-1(0) = \(\cfrac{2}{1-0}\) = 2 x = 1 then f-1(1) is not exist x = -1 then f-1 (-1) = \(\cfrac{2}{1-(-1)}\) = \(\cfrac22\) = 1 x = -2 then f-1 (-2) = \(\cfrac{2}{1-(-2)}\) = \(\cfrac23\) f(x) is not defined at x = 0,1,-1 and -2 \(\therefore\) f(x) never gives the values 2,1 and \(\cfrac23\) \(\therefore\) Range of f(x) is R - {2,1,\(\cfrac23\)} Very Complicated complicated questions jyada dimag Naa lgaye method set h answer paoo |
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