Let f(x) = sin x + (x^3 - 3x^2 + 4x - 2) cos | vedclass

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(A) Given $f(x) = \sin x + (x^3 - 3x^2 + 4x - 2) \cos x$.Note that $x^3 - 3x^2 + 4x - 2 = (x-1)^3 + (x-1)$.So,$f(x) = \sin x + ((x-1)^3 + (x-1)) \cos

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$I$ and $II$ are true

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(A) Given $f(x) = \sin x + (x^3 - 3x^2 + 4x - 2) \cos x$.Note that $x^3 - 3x^2 + 4x - 2 = (x-1)^3 + (x-1)$.So,$f(x) = \sin x + ((x-1)^3 + (x-1)) \cos x$.Calculating the derivative:$f'(x) = \cos x + [3(x-1)^2 + 1] \cos x - [(x-1)^3 + (x-1)] \sin x$$f'(x) = \cos x [1 + 3(x-1)^2 + 1] - (x-1)((x-1)^2 + 1) \sin x$$f'(x) = \cos x [3(x-1)^2 + 2] + (1-x)((x-1)^2 + 1) \sin x$.For $x \in (0, 1)$,$\cos x > 0$,$\sin x > 0$,$(1-x) > 0$,and the terms in brackets are positive.Thus,$f'(x) > 0$ for all $x \in (0, 1)$,which means $f$ is strictly increasing (monotone).Also,$f(0) = \sin(0) + (-2) \cos(0) = -2  0$.Since $f$ is continuous and changes sign on $(0, 1)$,by the Intermediate Value Theorem,$f$ has a zero in $(0, 1)$.Therefore,both statements $I$ and $II$ are true.

Let $f(x) = \sin x + (x^3 - 3x^2 + 4x - 2) \cos x$ for $x \in (0, 1)$. Consider the following statements:
$I.$ $f$ has a zero in $(0, 1)$.
$II.$ $f$ is monotone in $(0, 1)$.
Then,

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Let $f(x) = \sin x + (x^3 - 3x^2 + 4x - 2) \cos x$ for $x \in (0, 1)$. Consider the following statements:$I.$ $f$ has a zero in $(0, 1)$.$II.$ $f$ is monotone in $(0, 1)$.Then,