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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2014 · Mathematics · Application of Derivatives

The interval in which the function 2 ^ 3 +15 increases less rapidly...

The interval in which the function 2\mathrmx3+152 \mathrmx^{3}+15 increases less rapidly than the function 9x2−12x,9 x^{2}-12 x, is −-

Options

  1. A(−∞,1)(-\infty, 1)
  2. B(1,2)
  3. C(2,∞)(2, \infty)
  4. DNone of these

Correct answer

B. (1,2)

Step-by-step solution

Let f(x)=2x3+15f(x)=2 x^{3}+15 and g(x)=9x2−12xg(x)=9 x^{2}-12 x then \mathrmf′(x)=6\mathrmx2∀x∈R∴f(x)\mathrmf^{\prime}(\mathrm{x})=6 \mathrmx^{2} \forall \mathrm{x} \in \mathrm{R}\therefore \mathrm{f}(\mathrm{x}) is increasing function ∀x∈R Also, \forall \mathrm{x} \in \mathrm{R}\text{ Also, } g^{\prime}(x)>0 \Rightarrow 18 x-12>0 \Rightarrow x>23\frac{2}{3} Thus, f(x)\mathrm{f}(\mathrm{x}) and g(x)\mathrm{g}(\mathrm{x}) both increases for x>23 Let \mathrm{x} > \frac{2}{3}\text{ Let } F(x)=f(x)-g(x), F^{\prime}(x)<0 (∵f(x)(\because \mathrm{f}(\mathrm{x}) increases less rapidly than the function g(x))⇒6\mathrmx2−18x+12<0⇒1<x\mathrm{g}(\mathrm{x})) \Rightarrow 6 \mathrmx^{2}-18 \mathrm{x}+12<0 \Rightarrow 1<\mathrm{x}<2
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