SolveItClass 12 · NCERT

NCERT Solutions · Class 12 Mathematics Application of Derivatives

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EXERCISE 6.2 11–19 (part 4 of 9)

  1. Exercise 11

    Prove that the function f\displaystyle f given by f(x)=x2−x+1\displaystyle f(x)=x^{2}-x+1 is neither strictly increasing nor decreasing on ( −1,1\displaystyle -1,1 ).

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    \[f(x)=x^{2}-x+1 \quad\Rightarrow\quad f'(x)=2x-1. \]The zero of \(\displaystyle f'\) is \(\displaystyle x=\frac{1}{2}\), and this point lies inside the given interval \(\displaystyle (-1,1)\). That is what settles the question: the derivative changes sign inside the interval.For \(\displaystyle -1<x<\frac12\): \(\displaystyle 2x<1\Rightarrow f'(x)<0\), so \(\displaystyle f\) is decreasing there.For \(\displaystyle \frac12<x<1\): \(\displaystyle 2x>1\Rightarrow f'(x)>0\), so \(\displaystyle f\) is increasing there.A numerical check confirms it: \(\displaystyle f(0)=1\), \(\displaystyle f\!\left(\tfrac12\right)=\tfrac34\), \(\displaystyle f\!\left(\tfrac34\right)=\tfrac{13}{16}\). So \(\displaystyle f(0)>f\!\left(\tfrac12\right)\) but \(\displaystyle f\!\left(\tfrac12\right)<f\!\left(\tfrac34\right)\).Hence on \(\displaystyle (-1,1)\) the function \(\displaystyle f\) is neither strictly increasing nor strictly decreasing.

  2. Exercise 12

    Which of the following functions are decreasing on 0,π2\displaystyle 0, \frac{\pi}{2} ? (A) cos⁡x\displaystyle \cos x (B) cos⁡2x\displaystyle \cos 2 x (C) cos⁡3x\displaystyle \cos 3 x (D) tan⁡x\displaystyle \tan x

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    NCERT’s answer
    A, B
    Test each option by the sign of its derivative on the whole of \(\displaystyle \left(0,\frac{\pi}{2}\right)\).
    (A)
    \(\displaystyle y=\cos x\Rightarrow y'=-\sin x\). For \(\displaystyle 0<x<\frac{\pi}{2}\), \(\displaystyle \sin x>0\), so \(\displaystyle y'<0\): decreasing.
    (B)
    \(\displaystyle y=\cos 2x\Rightarrow y'=-2\sin 2x\). As \(\displaystyle x\) runs over \(\displaystyle \left(0,\frac{\pi}{2}\right)\), \(\displaystyle 2x\) runs over \(\displaystyle (0,\pi)\), where \(\displaystyle \sin 2x>0\); so \(\displaystyle y'<0\): decreasing.
    (C)
    \(\displaystyle y=\cos 3x\Rightarrow y'=-3\sin 3x\). Here \(\displaystyle 3x\) runs over \(\displaystyle \left(0,\frac{3\pi}{2}\right)\). For \(\displaystyle \frac{\pi}{3}<x<\frac{\pi}{2}\) we get \(\displaystyle \pi<3x<\frac{3\pi}{2}\), where \(\displaystyle \sin 3x<0\) and hence \(\displaystyle y'>0\). So \(\displaystyle \cos 3x\) is not decreasing throughout the interval.
    (D)
    \(\displaystyle y=\tan x\Rightarrow y'=\sec^{2}x>0\) on \(\displaystyle \left(0,\frac{\pi}{2}\right)\): increasing, not decreasing.
    Hence the decreasing functions on \(\displaystyle \left(0,\frac{\pi}{2}\right)\) are (A) \(\displaystyle \cos x\) and (B) \(\displaystyle \cos 2x\).

  3. Exercise 13

    On which of the following intervals is the function f\displaystyle f given by f(x)=x100+sin⁡x−1\displaystyle f(x)=x^{100}+\sin x-1 decreasing ? (A) (0,1)\displaystyle (0,1) (B) π2,π\displaystyle \frac{\pi}{2}, \pi (C) 0,π2\displaystyle 0, \frac{\pi}{2} (D) None of these

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    NCERT’s answer
    D
    \[f(x)=x^{100}+\sin x-1 \quad\Rightarrow\quad f'(x)=100x^{99}+\cos x. \]
    For \(\displaystyle f\) to be decreasing on an interval we need \(\displaystyle f'(x)<0\) there. Check each option.
    (A)
    On \(\displaystyle (0,1)\): \(\displaystyle x>0\Rightarrow 100x^{99}>0\); and since \(\displaystyle 1<\frac{\pi}{2}\), every \(\displaystyle x\in(0,1)\) is in the first quadrant, so \(\displaystyle \cos x>0\). Hence \(\displaystyle f'(x)>0\): increasing.
    (B)
    On \(\displaystyle \left(\frac{\pi}{2},\pi\right)\): here \(\displaystyle \cos x\) is negative but \(\displaystyle \cos x\ge -1\), while \(\displaystyle x>\frac{\pi}{2}>1\) gives \(\displaystyle 100x^{99}>100\). So
    \[f'(x)=100x^{99}+\cos x>100-1=99>0: \text{ increasing}. \]
    (C)
    On \(\displaystyle \left(0,\frac{\pi}{2}\right)\): both \(\displaystyle 100x^{99}>0\) and \(\displaystyle \cos x>0\), so \(\displaystyle f'(x)>0\): increasing.
    The function is increasing on all three intervals, so it is decreasing on none of them.
    Hence the correct option is (D) None of these.

  4. Exercise 14

    For what values of a\displaystyle a the function f\displaystyle f given by f(x)=x2+ax+1\displaystyle f(x)=x^{2}+a x+1 is increasing on [1\displaystyle 1, 2\displaystyle 2]?

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    NCERT’s answer
    \(\displaystyle a>-2\)
    \[f(x)=x^{2}+ax+1 \quad\Rightarrow\quad f'(x)=2x+a. \]For \(\displaystyle f\) to be increasing on \(\displaystyle [1,2]\) we need \[f'(x)=2x+a\ \ge\ 0 \quad\text{for every } x\in[1,2]. \]Since \(\displaystyle 2x+a\) is itself an increasing function of \(\displaystyle x\), its least value on \(\displaystyle [1,2]\) occurs at the left endpoint \(\displaystyle x=1\) -- this is the step to get right: it is enough to impose the condition at \(\displaystyle x=1\).\[f'(1)=2(1)+a=2+a\ \ge\ 0 \quad\Rightarrow\quad a\ \ge\ -2. \](Then for all \(\displaystyle x\in[1,2]\), \(\displaystyle 2x+a\ge 2+a\ge 0\), so the condition holds on the whole interval.)Hence \(\displaystyle f\) is increasing on \(\displaystyle [1,2]\) for all \(\displaystyle a\ge -2\).

  5. Exercise 15

    Let I be any interval disjoint from [−1,1]\displaystyle [-1,1]. Prove that the function f\displaystyle f given by f(x)=x+1x\displaystyle f(x)=x+\frac{1}{x} is increasing on I.

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    \[f(x)=x+\frac{1}{x} \quad\Rightarrow\quad f'(x)=1-\frac{1}{x^{2}}=\frac{x^{2}-1}{x^{2}}. \]Let \(\displaystyle \mathrm{I}\) be an interval disjoint from \(\displaystyle [-1,1]\), and let \(\displaystyle x\in \mathrm{I}\). Then \(\displaystyle x\notin[-1,1]\), which means \[x<-1 \quad\text{or}\quad x>1, \qquad\text{i.e.}\qquad |x|>1 . \](In particular \(\displaystyle x\ne 0\), so \(\displaystyle f\) and \(\displaystyle f'\) are defined on \(\displaystyle \mathrm{I}\).)Squaring the inequality \(\displaystyle |x|>1\) gives \(\displaystyle x^{2}>1\), hence \(\displaystyle x^{2}-1>0\), while \(\displaystyle x^{2}>0\). Therefore \[f'(x)=\frac{x^{2}-1}{x^{2}}>0 \quad\text{for every } x\in \mathrm{I}. \]Hence \(\displaystyle f(x)=x+\dfrac{1}{x}\) is increasing on every interval \(\displaystyle \mathrm{I}\) disjoint from \(\displaystyle [-1,1]\).

  6. Exercise 16

    Prove that the function f\displaystyle f given by f(x)=log⁡sin⁡x\displaystyle f(x)=\log \sin x is increasing on 0,π2\displaystyle 0, \frac{\pi}{2} and decreasing on π2,π\displaystyle \frac{\pi}{2}, \pi.

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    \[f(x)=\log\sin x. \]By the chain rule (valid wherever \(\displaystyle \sin x>0\), which holds on \(\displaystyle (0,\pi)\), so \(\displaystyle f\) is defined there), \[f'(x)=\frac{1}{\sin x}\cdot\cos x=\cot x. \]On \(\displaystyle \left(0,\frac{\pi}{2}\right)\): both \(\displaystyle \sin x>0\) and \(\displaystyle \cos x>0\), so \(\displaystyle \cot x>0\), i.e. \(\displaystyle f'(x)>0\).On \(\displaystyle \left(\frac{\pi}{2},\pi\right)\): \(\displaystyle \sin x>0\) but \(\displaystyle \cos x<0\), so \(\displaystyle \cot x<0\), i.e. \(\displaystyle f'(x)<0\).Hence \(\displaystyle f(x)=\log\sin x\) is increasing on \(\displaystyle \left(0,\frac{\pi}{2}\right)\) and decreasing on \(\displaystyle \left(\frac{\pi}{2},\pi\right)\).

  7. Exercise 17

    Prove that the function f\displaystyle f given by f(x)=log⁡∣cos⁡x∣\displaystyle f(x)=\log |\cos x| is decreasing on (0,π2)\displaystyle \left(0, \frac{\pi}{2}\right) and increasing on (3π2,2π)\displaystyle \left(\frac{3 \pi}{2}, 2 \pi\right).

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    \[f(x)=\log|\cos x|. \]Using \(\displaystyle \dfrac{d}{dx}\log|u|=\dfrac{u'}{u}\) with \(\displaystyle u=\cos x\) (the absolute value is exactly what lets us differentiate this way even where \(\displaystyle \cos x<0\); we only need \(\displaystyle \cos x\ne 0\)), \[f'(x)=\frac{-\sin x}{\cos x}=-\tan x. \]On \(\displaystyle \left(0,\frac{\pi}{2}\right)\): \(\displaystyle x\) is in the first quadrant, where \(\displaystyle \tan x>0\). Hence \[f'(x)=-\tan x<0, \] so \(\displaystyle f\) is decreasing on \(\displaystyle \left(0,\frac{\pi}{2}\right)\).On \(\displaystyle \left(\frac{3\pi}{2},2\pi\right)\): \(\displaystyle x\) is in the fourth quadrant, where \(\displaystyle \sin x<0\) and \(\displaystyle \cos x>0\), so \(\displaystyle \tan x<0\). Hence \[f'(x)=-\tan x>0, \] so \(\displaystyle f\) is increasing on \(\displaystyle \left(\frac{3\pi}{2},2\pi\right)\).Hence \(\displaystyle f(x)=\log|\cos x|\) is decreasing on \(\displaystyle \left(0,\frac{\pi}{2}\right)\) and increasing on \(\displaystyle \left(\frac{3\pi}{2},2\pi\right)\).

  8. Exercise 18

    Prove that the function given by f(x)=x3−3x2+3x−100\displaystyle f(x)=x^{3}-3 x^{2}+3 x-100 is increasing in R\displaystyle \mathbf{R}.

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    \[f(x)=x^{3}-3x^{2}+3x-100 \quad\Rightarrow\quad f'(x)=3x^{2}-6x+3. \]Factor the derivative -- this is the whole point of the question: \[f'(x)=3\left(x^{2}-2x+1\right)=3(x-1)^{2}. \]A square is never negative, so \[f'(x)=3(x-1)^{2}\ \ge\ 0 \quad\text{for all } x\in\mathbf{R}, \] with equality only at the single point \(\displaystyle x=1\).Since \(\displaystyle f'\) is non-negative everywhere and vanishes only at one isolated point (never on a whole interval), \(\displaystyle f\) is increasing on \(\displaystyle \mathbf{R}\).

  9. Exercise 19

    The interval in which y=x2e−x\displaystyle y=x^{2} e^{-x} is increasing is (A) (−∞,∞)\displaystyle (-\infty, \infty) (B) (−2,0)\displaystyle (-2, 0) (C) (2,∞)\displaystyle (2, \infty) (D) (0,2)\displaystyle (0,2)

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    NCERT’s answer
    D
    \[y=x^{2}e^{-x}. \]By the product rule, \[\frac{dy}{dx}=2x\,e^{-x}+x^{2}\left(-e^{-x}\right)=x\,e^{-x}(2-x). \]Since \(\displaystyle e^{-x}>0\) for every real \(\displaystyle x\), the sign of \(\displaystyle \dfrac{dy}{dx}\) is the sign of \(\displaystyle x(2-x)\). Thus \[\frac{dy}{dx}>0 \iff x(2-x)>0 \iff 0<x<2 . \](For \(\displaystyle x<0\) and for \(\displaystyle x>2\) the product \(\displaystyle x(2-x)\) is negative, so \(\displaystyle y\) is decreasing there; this rules out (A), (B) and (C).)Hence \(\displaystyle y=x^{2}e^{-x}\) is increasing on \(\displaystyle (0,2)\), which is option (D).