If fx-x3-3 x2-4 x-b sin x-c cos xx - R is one-one- Then maximum value of b2-c2

The correct option is C Is 1f^'(x)=3x^2 + 6x+ 4+b cos x c sin x ≥ 0 f^'(x) ≥ 0 (it can't be decreasing) 3x^2+6x+4 ≥ c sin x b cos x This should be greater t ...

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Standard XII

Physics

Definite Integrals

If fx=x3+3 x2...

Question

If $f (x) = x^{3} + 3 x^{2} + 4 x + b s i n x + c c o s x (x ϵ R)$ is one-one. Then maximum value of $b^{2} + c^{2}$

Is 3

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Is 2

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Is 1

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Can’t be determined

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Solution

The correct option is C Is 1
$f^{^{'}} (x) = 3 x^{2} + 6 x + 4 + b c o s x - c s i n x \geq 0$
$f^{^{'}} (x) \geq 0$ (it can't be decreasing)
$3 x^{2} + 6 x + 4 \geq c s i n x - b c o s x$
This should be greater than max value of right hand side
$3 x^{2} + 6 x + 4 \geq \sqrt{b^{2} + c^{2}}$
$3 (x + 1)^{2} + 1 \geq \sqrt{b^{2} + c^{2}}$
$\sqrt{b^{2} + c^{2}} \leq 1$

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If f(x)=x3+3x2+4x+b sin x+c cos x(xϵR) is one-one. Then maximum value of b2+c2

The correct option is C Is 1f′(x)=3x2+6x+4+b cos x−c sin x≥0 f′(x)≥0 (it can't be decreasing) 3x2+6x+4≥c sin x−b cos x This should be greater than max value of right hand side 3x2+6x+4≥√b2+c2 3(x+1)2+1≥√b2+c2 √b2+c2≤1

If $f (x) = x^{3} + 3 x^{2} + 4 x + b s i n x + c c o s x (x ϵ R)$ is one-one. Then maximum value of $b^{2} + c^{2}$