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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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On which of the following intervals is the function $f $ given by $f (x) = x^{100} + sin x –1$ strictly decreasing ?

$ (A)\; (0,1) \quad (B) \;\left( \large\frac{\pi}{2}, \pi\right) \quad (C)\; \left(0, \: \large\frac{\pi}{2}\right) \quad (D)\; \text{None of these}$

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1 Answer

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Toolbox:
  • A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
  • If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
  • A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
  • The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
  • The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Let $f(x)=x^{100}+\sin x-1$
Differentiating w.r.t $x$ we get,
$f'(x)=100x^{99}+\cos x$
Consider the interval $(0,1)$
$\cos x > 0$ and $100x^{99} >0$
Therefore $f'(x)> 0$
Hence the function $f$ is strictly increasing in interval $(0,1)$
Step 2:
Consider the interval $(\large\frac{\pi}{2},$$x),\cos x < 0$ and $100x^{99} >0$.
Also $100x^{99} > \cos x$
Therefore $f'(x) >0$ in $(\large\frac{\pi}{2},$$\pi)$
Hence the function is strictly increasing in the interval $(\large\frac{\pi}{2},$$\pi)$
Step 3:
Consider the interval $(0,\large\frac{\pi}{2})$$,\cos x > 0$ and $100x^{99} > 0$
Therefore $100x^{99}+\cos x >0$
Hence $f'(x) >0$ on $(0,\large\frac{\pi}{2})$
Therefore $f$ is strictly increasing in interval $(0,\large\frac{\pi}{2})$
Hence the function $f$ is strictly decreasing in none of the intervals.
The correct answer is $D$
answered Jul 9, 2013 by sreemathi.v
 

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