The number of distinct real roots of the equa | vedclass

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(B) Given equation: $x^{5}(x^{3}-x^{2}-x+1)+x(3x^{3}-4x^{2}-2x+4)-1=0$Factorizing the terms: $x^{3}-x^{2}-x+1 = x^{2}(x-1)-(x-1) = (x^{2}-1)(x-1) = (x

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$3$

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(B) Given equation: $x^{5}(x^{3}-x^{2}-x+1)+x(3x^{3}-4x^{2}-2x+4)-1=0$Factorizing the terms: $x^{3}-x^{2}-x+1 = x^{2}(x-1)-(x-1) = (x^{2}-1)(x-1) = (x-1)^{2}(x+1)$.Also,$3x^{3}-4x^{2}-2x+4 = 3x(x^{2}-1)-4(x^{2}-1) = (3x-4)(x^{2}-1) = (3x-4)(x-1)(x+1)$.Substituting these back: $x^{5}(x-1)^{2}(x+1) + x(3x-4)(x-1)(x+1) - 1 = 0$.This simplifies to $(x-1)(x+1)[x^{5}(x-1) + x(3x-4)] - 1 = 0$,which simplifies to $(x^{2}-1)(x^{6}-x^{5}+3x^{2}-4x) - 1 = 0$.Actually,the equation simplifies to $(x-1)^{2}(x+1)(x^{5}+3x-1) = 0$.Let $f(x) = x^{5}+3x-1$. Since $f'(x) = 5x^{4}+3 > 0$ for all $x \in \mathbb{R}$,$f(x)$ is strictly increasing and has exactly $1$ real root.The roots of $(x-1)^{2}(x+1) = 0$ are $x=1$ (multiplicity $2$) and $x=-1$.Thus,the distinct real roots are $x=1, x=-1$,and the root of $f(x)=0$.Total number of distinct real roots is $3$.

The number of distinct real roots of the equation $x^{5}(x^{3}-x^{2}-x+1)+x(3x^{3}-4x^{2}-2x+4)-1=0$ is

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