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# Solution for Let F(X) = X3 + Ax2 + Bx + 5 Sin2x Be an Increasing Function on the Set R. Then, a and B Satisfy (A) A2 − 3b − 15 > 0 (B) A2 − 3b + 15 > 0 (C) A2 − 3b + 15 < 0 (D) a > 0 and B > 0 - CBSE (Commerce) Class 12 - Mathematics

ConceptIncreasing and Decreasing Functions

#### Question

Let f(x) = x3 + ax2 + bx + 5 sin2x be an increasing function on the set R. Then, a and bsatisfy
(a) a2 − 3b − 15 > 0
(b) a2 − 3b + 15 > 0
(c) a2 − 3b + 15 < 0
(d) a > 0 and b > 0

#### Solution

(c) a2 − 3b + 15 < 0

$f\left( x \right) = x^3 + a x^2 + bx + 5 \sin^2 x$

$f'\left( x \right) = 3 x^2 + 2ax + \left( b + 5 \sin 2x \right)$

$\text { Given }:f\left( x \right)\text { is increasing on R }.$

$\Rightarrow f'\left( x \right) > 0, \forall x \in R$

$\Rightarrow 3 x^2 + 2ax + \left( b + 5 \sin 2x \right) > 0, \forall x \in R$

$\text { Since this quadratic function is >0, its discriminant is } <0.$

$\Rightarrow \left( 2a \right)^2 - 4\left( 3 \right)\left( b + 5 \sin 2x \right) < 0$

$\Rightarrow 4 a^2 - 12b - 60 \sin 2x < 0$

$\Rightarrow a^2 - 3b - 15 \sin 2x < 0$

$\text { We know that the minimum value of sin 2x is -1 }.$

$\therefore a^2 - 3b - 15 < 0$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [3]

Solution Let F(X) = X3 + Ax2 + Bx + 5 Sin2x Be an Increasing Function on the Set R. Then, a and B Satisfy (A) A2 − 3b − 15 > 0 (B) A2 − 3b + 15 > 0 (C) A2 − 3b + 15 < 0 (D) a > 0 and B > 0 Concept: Increasing and Decreasing Functions.
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