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Mark the correct alternative in the following:

Question:

Mark the correct alternative in the following:

Let $f(x)=x^{3}+a x^{2}+b x+5 \sin ^{2} x$ be an increasing function on the set $R$. Then, $a$ and $b$ satisfy.

A. $a^{2}-3 b-15>0$

B. $a^{2}-3 b+15>0$

C. $a^{2}-3 b+15<0$

D. $a>0$ and $b>0$

Solution:

Formula:- (i) $a x^{2}+b x+c>0$ for all $x \Rightarrow a>0$ and $b^{2}-4 a c<0$

(ii) $a x^{2}+b x+c<0$ for all $x \Rightarrow a<0$ and $b^{2}-4 a c<0$

(iii) The necessary and sufficient condition for differentiable function defined on $(a, b)$ to be strictly increasing on $(a, b)$ is that $f^{\prime}(x)>0$ for all $x \in(a, b)$

Given:-

$f(x)=x^{3}+a x^{2}+b x+5 \sin ^{2} x$

$\mathrm{d}\left(\frac{\mathrm{f}(\mathrm{x})}{\mathrm{dx}}\right)=3 \mathrm{x}^{2}+2 \mathrm{ax}+\mathrm{b}+5 \sin ^{2} \mathrm{x}=\mathrm{f}(\mathrm{x})$

For increasing function $f^{\prime}(x)>0$

$3 x^{2}+2 a x+b+5 \sin 2 x>0$

Then

$3 x^{2}+2 a x+b-5<0$

And $b^{2}-4 a c<0$

$\Rightarrow 4 a^{2}-12(b-5)<0$

$\Rightarrow a^{2}-3 b+15<0$

$\Rightarrow a^{2}-3 b+15<0$

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