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NCERT Solutions · Class 12 Mathematics Application of Derivatives

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EXERCISE 6.3 1–10 (part 5 of 9)

  1. Exercise 1

    Find the maximum and minimum values, if any, of the following functions given by
    (i)
    f(x)=(2x−1)2+3\displaystyle f(x)=(2 x-1)^{2}+3
    (ii)
    f(x)=9x2+12x+2\displaystyle f(x)=9 x^{2}+12 x+2
    (iii)
    f(x)=−(x−1)2+10\displaystyle f(x)=-(x-1)^{2}+10
    (iv)
    g(x)=x3+1\displaystyle g(x)=x^{3}+1

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    NCERT’s answer
    (i)
    Minimum Value \(\displaystyle =3\) (ii) Minimum Value \(\displaystyle =-2\) (iii) Maximum Value \(\displaystyle =10\) (iv) Neither minimum nor maximum value
    Each part is a polynomial: complete the square (or use the sign of the derivative) and read off the range.
    (i)
    \(\displaystyle f(x)=(2x-1)^{2}+3\). A square is never negative, so \(\displaystyle (2x-1)^{2}\ge 0\) and hence \(\displaystyle f(x)\ge 3\), with equality only when \(\displaystyle 2x-1=0\), i.e. \(\displaystyle x=\frac{1}{2}\). Also \(\displaystyle f(x)\to\infty\) as \(\displaystyle x\to\infty\), so \(\displaystyle f\) is not bounded above.
    Minimum value \(\displaystyle 3\), attained at \(\displaystyle x=\frac{1}{2}\); no maximum value.
    (ii)
    \(\displaystyle f(x)=9x^{2}+12x+2=9\left(x^{2}+\frac{4}{3}x\right)+2=9\left(x+\frac{2}{3}\right)^{2}-4+2=9\left(x+\frac{2}{3}\right)^{2}-2\).
    So \(\displaystyle f(x)\ge -2\), with equality at \(\displaystyle x=-\frac{2}{3}\), and \(\displaystyle f(x)\to\infty\) as \(\displaystyle x\to\infty\).
    Minimum value \(\displaystyle -2\), attained at \(\displaystyle x=-\frac{2}{3}\); no maximum value.
    (iii)
    \(\displaystyle f(x)=-(x-1)^{2}+10\). Since \(\displaystyle -(x-1)^{2}\le 0\), \(\displaystyle f(x)\le 10\), with equality at \(\displaystyle x=1\), and \(\displaystyle f(x)\to-\infty\) as \(\displaystyle x\to\infty\).
    Maximum value \(\displaystyle 10\), attained at \(\displaystyle x=1\); no minimum value.
    (iv)
    \(\displaystyle g(x)=x^{3}+1\). Here \(\displaystyle g'(x)=3x^{2}\ge 0\), so \(\displaystyle g\) is increasing, and \(\displaystyle g(x)\to\infty\) as \(\displaystyle x\to\infty\) while \(\displaystyle g(x)\to-\infty\) as \(\displaystyle x\to-\infty\); the range is the whole of \(\displaystyle \mathbb{R}\).
    Neither a maximum nor a minimum value exists.

  2. Exercise 2

    Find the maximum and minimum values, if any, of the following functions given by
    (i)
    f(x)=∣x+2∣−1\displaystyle f(x)=|x+2|-1
    (ii)
    g(x)=−∣x+1∣+3\displaystyle g(x)=-|x+1|+3
    (iii)
    h(x)=sin⁡(2x)+5\displaystyle h(x)=\sin (2 x)+5
    (iv)
    f(x)=∣sin⁡4x+3∣\displaystyle f(x)=|\sin 4 x+3|
    (v)
    h(x)=x+1,x∈(−1,1)\displaystyle h(x)=x+1, x \in(-1,1)

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    NCERT’s answer
    (i)
    Minimum Value \(\displaystyle =-1\); No maximum value (ii) Maximum Value = $\displaystyle 3$; No minimum value (iii) Minimum Value \(\displaystyle =4\); Maximum Value = $\displaystyle 6$ (iv) Minimum Value = $\displaystyle 2$; Maximum Value = $\displaystyle 4$ (v) Neither minimum nor Maximum Value
    Use the bounds \(\displaystyle |t|\ge 0\) and \(\displaystyle -1\le\sin\theta\le 1\); on an open interval, check whether a bound is actually attained.
    (i)
    \(\displaystyle f(x)=|x+2|-1\). Since \(\displaystyle |x+2|\ge 0\), \(\displaystyle f(x)\ge -1\), with equality when \(\displaystyle x=-2\); and \(\displaystyle f(x)\to\infty\) as \(\displaystyle x\to\infty\).
    Minimum value \(\displaystyle -1\) at \(\displaystyle x=-2\); no maximum value.
    (ii)
    \(\displaystyle g(x)=-|x+1|+3\). Since \(\displaystyle -|x+1|\le 0\), \(\displaystyle g(x)\le 3\), with equality when \(\displaystyle x=-1\); and \(\displaystyle g(x)\to-\infty\) as \(\displaystyle x\to\infty\).
    Maximum value \(\displaystyle 3\) at \(\displaystyle x=-1\); no minimum value.
    (iii)
    \(\displaystyle h(x)=\sin 2x+5\). Since \(\displaystyle -1\le\sin 2x\le 1\), \(\displaystyle 4\le h(x)\le 6\), and both bounds are attained (at \(\displaystyle x=\frac{3\pi}{4}\) and \(\displaystyle x=\frac{\pi}{4}\)).
    Maximum value \(\displaystyle 6\), minimum value \(\displaystyle 4\).
    (iv)
    \(\displaystyle f(x)=|\sin 4x+3|\). This is the step to watch: since \(\displaystyle \sin 4x\ge -1\), the quantity inside the modulus satisfies \(\displaystyle \sin 4x+3\ge 2>0\), so the modulus can simply be dropped and \(\displaystyle f(x)=\sin 4x+3\). Then \(\displaystyle 2\le f(x)\le 4\), both values being attained.
    Maximum value \(\displaystyle 4\), minimum value \(\displaystyle 2\).
    (v)
    \(\displaystyle h(x)=x+1\) on the open interval \(\displaystyle (-1,1)\). Here \(\displaystyle 0<h(x)<2\). The bounds \(\displaystyle 0\) and \(\displaystyle 2\) would need \(\displaystyle x=-1\) and \(\displaystyle x=1\), which are not in the domain; and given any \(\displaystyle x_{0}\in(-1,1)\) there is always a point of \(\displaystyle (-1,1)\) nearer to \(\displaystyle 1\) (so \(\displaystyle h(x_{0})\) is not the largest value) and one nearer to \(\displaystyle -1\) (so it is not the smallest).
    Neither a maximum nor a minimum value exists.

  3. Exercise 3

    Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:
    (i)
    f(x)=x2\displaystyle f(x)=x^{2}
    (ii)
    g(x)=x3−3x\displaystyle g(x)=x^{3}-3 x
    (iii)
    h(x)=sin⁡x+cos⁡x,0<x<π2\displaystyle h(x)=\sin x+\cos x, 0<x<\frac{\pi}{2}
    (iv)
    f(x)=sin⁡x−cos⁡x,0<x<2π\displaystyle f(x)=\sin x-\cos x, 0<x<2 \pi
    (v)
    f(x)=x3−6x2+9x+15\displaystyle f(x)=x^{3}-6 x^{2}+9 x+15
    (vi)
    g(x)=x2+2x,x>0\displaystyle g(x)=\frac{x}{2}+\frac{2}{x}, \quad x>0
    (vii)
    g(x)=1x2+2\displaystyle g(x)=\frac{1}{x^{2}+2}
    (viii)
    f(x)=x1−x,0<x<1\displaystyle f(x)=x \sqrt{1-x}, \quad 0<x<1

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    NCERT’s answer
    (i)
    local minimum at \(\displaystyle x=0\), local minimum value \(\displaystyle =0\) (ii) local minimum at \(\displaystyle x=1, \quad\) local minimum value \(\displaystyle =-2\) local maximum at \(\displaystyle x=-1\), local maximum value \(\displaystyle =2\) (iii) local maximum at \(\displaystyle x=\frac{\pi}{4}\), local maximum value \(\displaystyle =\sqrt{2}\) (iv) local maximum at \(\displaystyle x=\frac{3 \pi}{4}\), local maximum value \(\displaystyle =\sqrt{2}\) local minimum at \(\displaystyle x=\frac{7 \pi}{4}\), local minimum value \(\displaystyle =-\sqrt{2}\) (v) local maximum at \(\displaystyle x=1, \quad\) local maximum value \(\displaystyle =19\) local minimum at \(\displaystyle x=3, \quad\) local minimum value \(\displaystyle =15\) (vi) local minimum at \(\displaystyle x=2, \quad\) local minimum value \(\displaystyle =2\) (vii) local maximum at \(\displaystyle x=0, \quad\) local maximum value \(\displaystyle =\frac{1}{2}\) (viii) local maximum at \(\displaystyle x=\frac{2}{3}\), local maximum value \(\displaystyle =\frac{2 \sqrt{3}}{9}\)
    In each part solve \(\displaystyle f'(x)=0\) for the critical points, then classify them by the second derivative test (or by the sign change of \(\displaystyle f'\)).
    (i)
    \(\displaystyle f(x)=x^{2}\): \(\displaystyle f'(x)=2x=0\Rightarrow x=0\); \(\displaystyle f''(x)=2>0\).
    Local minimum at \(\displaystyle x=0\), local minimum value \(\displaystyle f(0)=0\); no local maximum.
    (ii)
    \(\displaystyle g(x)=x^{3}-3x\): \(\displaystyle g'(x)=3x^{2}-3=3(x-1)(x+1)=0\Rightarrow x=\pm1\); \(\displaystyle g''(x)=6x\).
    At \(\displaystyle x=1\): \(\displaystyle g''(1)=6>0\), so a local minimum, value \(\displaystyle g(1)=1-3=-2\).
    At \(\displaystyle x=-1\): \(\displaystyle g''(-1)=-6<0\), so a local maximum, value \(\displaystyle g(-1)=-1+3=2\).
    (iii)
    \(\displaystyle h(x)=\sin x+\cos x\), \(\displaystyle 0<x<\frac{\pi}{2}\): \(\displaystyle h'(x)=\cos x-\sin x=0\Rightarrow\tan x=1\Rightarrow x=\frac{\pi}{4}\), the only such point in the interval. \(\displaystyle h''(x)=-\sin x-\cos x\), so \(\displaystyle h''\left(\frac{\pi}{4}\right)=-\sqrt{2}<0\).
    Local maximum at \(\displaystyle x=\frac{\pi}{4}\), local maximum value \(\displaystyle \sqrt{2}\); no local minimum in the interval.
    (iv)
    \(\displaystyle f(x)=\sin x-\cos x\), \(\displaystyle 0<x<2\pi\): \(\displaystyle f'(x)=\cos x+\sin x=0\Rightarrow\tan x=-1\Rightarrow x=\frac{3\pi}{4}\) or \(\displaystyle x=\frac{7\pi}{4}\). \(\displaystyle f''(x)=-\sin x+\cos x\).
    At \(\displaystyle x=\frac{3\pi}{4}\): \(\displaystyle f''=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\sqrt{2}<0\), a local maximum, value \(\displaystyle \frac{1}{\sqrt2}+\frac{1}{\sqrt2}=\sqrt{2}\).
    At \(\displaystyle x=\frac{7\pi}{4}\): \(\displaystyle f''=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}=\sqrt{2}>0\), a local minimum, value \(\displaystyle -\frac{1}{\sqrt2}-\frac{1}{\sqrt2}=-\sqrt{2}\).
    (v)
    \(\displaystyle f(x)=x^{3}-6x^{2}+9x+15\): \(\displaystyle f'(x)=3x^{2}-12x+9=3(x-1)(x-3)=0\Rightarrow x=1,\,3\); \(\displaystyle f''(x)=6x-12\).
    At \(\displaystyle x=1\): \(\displaystyle f''=-6<0\), local maximum, value \(\displaystyle f(1)=1-6+9+15=19\).
    At \(\displaystyle x=3\): \(\displaystyle f''=6>0\), local minimum, value \(\displaystyle f(3)=27-54+27+15=15\).
    (vi)
    \(\displaystyle g(x)=\frac{x}{2}+\frac{2}{x}\), \(\displaystyle x>0\): \(\displaystyle g'(x)=\frac{1}{2}-\frac{2}{x^{2}}=0\Rightarrow x^{2}=4\Rightarrow x=2\) (the root \(\displaystyle x=-2\) is rejected, it is outside the domain \(\displaystyle x>0\)). \(\displaystyle g''(x)=\frac{4}{x^{3}}\), so \(\displaystyle g''(2)=\frac{1}{2}>0\).
    Local minimum at \(\displaystyle x=2\), local minimum value \(\displaystyle 1+1=2\); no local maximum.
    (vii)
    \(\displaystyle g(x)=\frac{1}{x^{2}+2}\): \(\displaystyle g'(x)=\frac{-2x}{\left(x^{2}+2\right)^{2}}=0\Rightarrow x=0\). The denominator is always positive, so \(\displaystyle g'>0\) for \(\displaystyle x<0\) and \(\displaystyle g'<0\) for \(\displaystyle x>0\): the sign changes from \(\displaystyle +\) to \(\displaystyle -\).
    Local maximum at \(\displaystyle x=0\), local maximum value \(\displaystyle \frac{1}{2}\); no local minimum.
    (viii)
    \(\displaystyle f(x)=x\sqrt{1-x}\), \(\displaystyle 0<x<1\): by the product rule,
    \[f'(x)=\sqrt{1-x}+x\cdot\frac{-1}{2\sqrt{1-x}}=\frac{2(1-x)-x}{2\sqrt{1-x}}=\frac{2-3x}{2\sqrt{1-x}}. \]
    On \(\displaystyle 0<x<1\) the denominator is positive, so the sign of \(\displaystyle f'\) is the sign of \(\displaystyle 2-3x\): positive for \(\displaystyle x<\frac{2}{3}\) and negative for \(\displaystyle x>\frac{2}{3}\).
    Local maximum at \(\displaystyle x=\frac{2}{3}\), local maximum value \(\displaystyle \frac{2}{3}\sqrt{1-\frac{2}{3}}=\frac{2}{3\sqrt{3}}=\frac{2\sqrt{3}}{9}\); no local minimum.

  4. Exercise 4

    Prove that the following functions do not have maxima or minima:
    (i)
    f(x)=ex\displaystyle f(x)=e^{x}
    (ii)
    g(x)=log⁡x\displaystyle g(x)=\log x
    (iii)
    h(x)=x3+x2+x+1\displaystyle h(x)=x^{3}+x^{2}+x+1

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    For a differentiable function, an interior maximum or minimum can occur only at a point where \(\displaystyle f'(x)=0\). So it is enough to show the derivative never vanishes -- equivalently that the function is strictly monotonic on its whole domain.
    (i)
    \(\displaystyle f(x)=e^{x}\), \(\displaystyle x\in\mathbb{R}\): \(\displaystyle f'(x)=e^{x}>0\) for every \(\displaystyle x\), so \(\displaystyle f'(x)=0\) has no solution and \(\displaystyle f\) is strictly increasing on \(\displaystyle \mathbb{R}\). A function strictly increasing on all of \(\displaystyle \mathbb{R}\) has no greatest and no least value (given any \(\displaystyle x_{0}\), \(\displaystyle f(x_{0}+1)>f(x_{0})>f(x_{0}-1)\)). Hence \(\displaystyle e^{x}\) has neither maxima nor minima.
    (ii)
    \(\displaystyle g(x)=\log x\), domain \(\displaystyle x>0\): \(\displaystyle g'(x)=\frac{1}{x}>0\) for all \(\displaystyle x>0\) and is never zero, so \(\displaystyle g\) is strictly increasing on \(\displaystyle (0,\infty)\) and by the same argument has no maxima or minima.
    (iii)
    \(\displaystyle h(x)=x^{3}+x^{2}+x+1\): \(\displaystyle h'(x)=3x^{2}+2x+1\). Its discriminant is \(\displaystyle 2^{2}-4\cdot3\cdot1=-8<0\) with leading coefficient \(\displaystyle 3>0\), so \(\displaystyle h'(x)>0\) for every real \(\displaystyle x\); explicitly \(\displaystyle h'(x)=3\left(x+\frac{1}{3}\right)^{2}+\frac{2}{3}>0\). Thus \(\displaystyle h'\) never vanishes, \(\displaystyle h\) is strictly increasing on \(\displaystyle \mathbb{R}\), and \(\displaystyle h\) has no maxima or minima.

  5. Exercise 5

    Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
    (i)
    f(x)=x3,x∈[−2,2]\displaystyle f(x)=x^{3}, x \in[-2,2]
    (ii)
    f(x)=sin⁡x+cos⁡x,x∈[0,π]\displaystyle f(x)=\sin x+\cos x, x \in[0, \pi]
    (iii)
    f(x)=4x−12x2,x∈[−2,92]\displaystyle f(x)=4 x-\frac{1}{2} x^{2}, x \in\left[-2, \frac{9}{2}\right]
    (iv)
    f(x)=(x−1)2+3,x∈−3,1]\displaystyle \left.f(x)=(x-1)^{2}+3, x \in-3,1\right]

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    NCERT’s answer
    (i)
    Absolute minimum value \(\displaystyle =-8\), absolute maximum value \(\displaystyle =8\) (ii) Absolute minimum value \(\displaystyle =-1\), absolute maximum value \(\displaystyle =\sqrt{2}\) (iii) Absolute minimum value = -$\displaystyle 10$, absolute maximum value = $\displaystyle 8$ (iv) Absolute minimum value \(\displaystyle =19\), absolute maximum value \(\displaystyle =3\)
    A continuous function on a closed interval attains its absolute extrema either at a critical point inside the interval or at an endpoint, so compare the values of \(\displaystyle f\) at the roots of \(\displaystyle f'(x)=0\) lying in the interval with the two endpoint values.
    (i)
    \(\displaystyle f(x)=x^{3}\) on \(\displaystyle [-2,2]\): \(\displaystyle f'(x)=3x^{2}=0\Rightarrow x=0\).
    \(\displaystyle f(-2)=-8\), \(\displaystyle f(0)=0\), \(\displaystyle f(2)=8\).
    Absolute maximum \(\displaystyle 8\) at \(\displaystyle x=2\); absolute minimum \(\displaystyle -8\) at \(\displaystyle x=-2\).
    (ii)
    \(\displaystyle f(x)=\sin x+\cos x\) on \(\displaystyle [0,\pi]\): \(\displaystyle f'(x)=\cos x-\sin x=0\Rightarrow\tan x=1\Rightarrow x=\frac{\pi}{4}\).
    \(\displaystyle f(0)=1\), \(\displaystyle f\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}=\sqrt{2}\), \(\displaystyle f(\pi)=0-1=-1\).
    Absolute maximum \(\displaystyle \sqrt{2}\) at \(\displaystyle x=\frac{\pi}{4}\); absolute minimum \(\displaystyle -1\) at \(\displaystyle x=\pi\).
    (iii)
    \(\displaystyle f(x)=4x-\frac{1}{2}x^{2}\) on \(\displaystyle \left[-2,\frac{9}{2}\right]\): \(\displaystyle f'(x)=4-x=0\Rightarrow x=4\), which lies in the interval.
    \(\displaystyle f(-2)=-8-2=-10\), \(\displaystyle f(4)=16-8=8\), \(\displaystyle f\left(\frac{9}{2}\right)=18-\frac{81}{8}=\frac{63}{8}=7.875\).
    Absolute maximum \(\displaystyle 8\) at \(\displaystyle x=4\); absolute minimum \(\displaystyle -10\) at \(\displaystyle x=-2\).
    (iv)
    \(\displaystyle f(x)=(x-1)^{2}+3\) on \(\displaystyle [-3,1]\): \(\displaystyle f'(x)=2(x-1)=0\Rightarrow x=1\), which is the right endpoint, so only the endpoints need comparing.
    \(\displaystyle f(-3)=(-4)^{2}+3=19\), \(\displaystyle f(1)=0+3=3\).
    Absolute maximum \(\displaystyle 19\) at \(\displaystyle x=-3\); absolute minimum \(\displaystyle 3\) at \(\displaystyle x=1\).

  6. Exercise 6

    Find the maximum profit that a company can make, if the profit function is given by p(x)=41−72x−18x2p(x)=41-72 x-18 x^{2}

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    NCERT’s answer
    Maximum profit = $\displaystyle 113$ unit.
    Apply the second derivative test to the profit function. \[p(x)=41-72x-18x^{2},\qquad p'(x)=-72-36x,\qquad p''(x)=-36. \] Critical point: \(\displaystyle p'(x)=0\Rightarrow -72-36x=0\Rightarrow x=-2\). Since \(\displaystyle p''(-2)=-36<0\), \(\displaystyle x=-2\) gives a maximum; and as \(\displaystyle p\) is a downward parabola this is the absolute maximum of \(\displaystyle p\). \[p(-2)=41-72(-2)-18(-2)^{2}=41+144-72=113. \] The maximum profit the company can make is \(\displaystyle 113\) units.

  7. Exercise 7

    Find both the maximum value and the minimum value of 3x4−8x3+12x2−48x+25\displaystyle 3 x^{4}-8 x^{3}+12 x^{2}-48 x+25 on the interval [0\displaystyle 0, 3\displaystyle 3].

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    NCERT’s answer
    Minima at \(\displaystyle x=2\), minimum value \(\displaystyle =-39\), Maxima at \(\displaystyle x=0\), maximum value \(\displaystyle =25\).
    The interval \(\displaystyle [0,3]\) is closed and \(\displaystyle f\) is a polynomial, hence continuous: compare the critical values with the endpoint values. Let \(\displaystyle f(x)=3x^{4}-8x^{3}+12x^{2}-48x+25\). Then \[f'(x)=12x^{3}-24x^{2}+24x-48=12\left(x^{3}-2x^{2}+2x-4\right)=12\left[x^{2}(x-2)+2(x-2)\right]=12(x-2)\left(x^{2}+2\right). \] Since \(\displaystyle x^{2}+2>0\) for every \(\displaystyle x\), the only critical point is \(\displaystyle x=2\), and it lies in \(\displaystyle [0,3]\). \(\displaystyle f(0)=25\); \(\displaystyle f(2)=48-64+48-96+25=-39\); \(\displaystyle f(3)=243-216+108-144+25=16\). Maximum value \(\displaystyle 25\) (at \(\displaystyle x=0\)); minimum value \(\displaystyle -39\) (at \(\displaystyle x=2\)).

  8. Exercise 8

    At what points in the interval [0,2π]\displaystyle [0,2 \pi], does the function sin⁡2x\displaystyle \sin 2 x attain its maximum value?

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    NCERT’s answer
    At \(\displaystyle x=\frac{\pi}{4}\) and \(\displaystyle \frac{5 \pi}{4}\)
    Let \(\displaystyle f(x)=\sin 2x\) on \(\displaystyle [0,2\pi]\). Then \(\displaystyle f'(x)=2\cos 2x=0\Rightarrow \cos 2x=0\), and since \(\displaystyle 2x\in[0,4\pi]\), \[2x=\frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{5\pi}{2},\ \frac{7\pi}{2}\qquad\Longrightarrow\qquad x=\frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4},\ \frac{7\pi}{4}. \] Values at these points and at the endpoints: \(\displaystyle f\left(\frac{\pi}{4}\right)=\sin\frac{\pi}{2}=1\), \(\displaystyle f\left(\frac{3\pi}{4}\right)=\sin\frac{3\pi}{2}=-1\), \(\displaystyle f\left(\frac{5\pi}{4}\right)=\sin\frac{5\pi}{2}=1\), \(\displaystyle f\left(\frac{7\pi}{4}\right)=\sin\frac{7\pi}{2}=-1\), \(\displaystyle f(0)=0\), \(\displaystyle f(2\pi)=\sin 4\pi=0\). The greatest of these is \(\displaystyle 1\) (which is also the largest value \(\displaystyle \sin\) can ever take). The function \(\displaystyle \sin 2x\) attains its maximum value \(\displaystyle 1\) at \(\displaystyle x=\frac{\pi}{4}\) and \(\displaystyle x=\frac{5\pi}{4}\).

  9. Exercise 9

    What is the maximum value of the function sin⁡x+cos⁡x\displaystyle \sin x+\cos x ?

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    NCERT’s answer
    Maximum value \(\displaystyle =\sqrt{2}\)
    Write the expression as a single sine, using \(\displaystyle a\sin x+b\cos x=R\sin(x+\alpha)\) with \(\displaystyle R=\sqrt{a^{2}+b^{2}}\): \[\sin x+\cos x=\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x\right)=\sqrt{2}\,\sin\left(x+\frac{\pi}{4}\right). \] Since \(\displaystyle \sin\theta\le 1\), the expression is at most \(\displaystyle \sqrt{2}\), and the value \(\displaystyle \sqrt{2}\) is actually reached, at \(\displaystyle x=\frac{\pi}{4}\). By calculus the same conclusion: with \(\displaystyle f(x)=\sin x+\cos x\), \(\displaystyle f'(x)=\cos x-\sin x=0\Rightarrow\tan x=1\Rightarrow x=\frac{\pi}{4}\) (taking the value in \(\displaystyle [0,2\pi)\)), and \(\displaystyle f''(x)=-\sin x-\cos x\) gives \(\displaystyle f''\left(\frac{\pi}{4}\right)=-\sqrt{2}<0\), a maximum. The maximum value is \(\displaystyle \sqrt{2}\).

  10. Exercise 10

    Find the maximum value of 2x3−24x+107\displaystyle 2 x^{3}-24 x+107 in the interval [1\displaystyle 1, 3\displaystyle 3]. Find the maximum value of the same function in [-3\displaystyle 3, -1\displaystyle 1].

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    NCERT’s answer
    Maximum at \(\displaystyle x=3\), maximum value $\displaystyle 89$ ; maximum at \(\displaystyle x=-2\), maximum value \(\displaystyle =139\)
    Let \(\displaystyle f(x)=2x^{3}-24x+107\). Then \[f'(x)=6x^{2}-24=6(x-2)(x+2), \] so the critical points are \(\displaystyle x=2\) and \(\displaystyle x=-2\). On each closed interval compare the critical value inside it with the two endpoint values.On \(\displaystyle [1,3]\): the critical point \(\displaystyle x=2\) lies inside (\(\displaystyle x=-2\) does not). \(\displaystyle f(1)=2-24+107=85\), \(\displaystyle f(2)=16-48+107=75\), \(\displaystyle f(3)=54-72+107=89\). Maximum value on \(\displaystyle [1,3]\) is \(\displaystyle 89\), at \(\displaystyle x=3\).On \(\displaystyle [-3,-1]\): the critical point \(\displaystyle x=-2\) lies inside. \(\displaystyle f(-3)=-54+72+107=125\), \(\displaystyle f(-2)=-16+48+107=139\), \(\displaystyle f(-1)=-2+24+107=129\). Maximum value on \(\displaystyle [-3,-1]\) is \(\displaystyle 139\), at \(\displaystyle x=-2\).