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NCERT Exemplar · Class 12 Mathematics Application of Derivatives

64 questions · 64 still being checked

EXERCISE 6.3 21–30 (part 3 of 7)

  1. Exercise 21

    Show that for a≥1,f(x)=3sin⁡x−cos⁡x−2ax+b\displaystyle a \geq 1, f(x)=\sqrt{3} \sin x-\cos x-2 a x+b is decreasing in R\displaystyle \mathbf{R}.

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    \[f'(x)=\sqrt3\cos x+\sin x-2a \]\[\sin x+\sqrt3\cos x=2\sin\left(x+\frac{\pi}{3}\right) \]\[f'(x)=2\sin\left(x+\frac{\pi}{3}\right)-2a\le2-2a \]\[a\ge1 \;\Rightarrow\; 2-2a\le0 \;\Rightarrow\; f'(x)\le0 \quad\text{for all } x\in\mathbf R \]Equality can hold only at isolated points:\[f'(x)=0 \iff a=1,\ x=\frac{\pi}{6}+2n\pi \]Answer: \(\displaystyle f'(x)\le0\) on \(\displaystyle \mathbf R\), zero only at isolated points, so \(\displaystyle f\) is decreasing in \(\displaystyle \mathbf R\).

  2. Exercise 22

    Show that f(x)=tan⁡−1(sin⁡x+cos⁡x)\displaystyle f(x)=\tan ^{-1}(\sin x+\cos x) is an increasing function in (0,π4)\displaystyle \left(0, \frac{\pi}{4}\right).

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    \[f'(x)=\frac{\cos x-\sin x}{1+(\sin x+\cos x)^2} \]\[1+(\sin x+\cos x)^2=2+\sin2x>0 \]\[0<x<\frac{\pi}{4} \;\Rightarrow\; \sin x<\cos x \;\Rightarrow\; \cos x-\sin x>0 \]\[f'(x)>0 \quad\text{for all } x\in\left(0,\frac{\pi}{4}\right) \]Answer: \(\displaystyle f'(x)>0\) on \(\displaystyle \left(0,\dfrac{\pi}{4}\right)\), so \(\displaystyle f\) is increasing there.

  3. Exercise 23

    At what point, the slope of the curve y=−x3+3x2+9x−27\displaystyle y=-x^3+3 x^2+9 x-27 is maximum? Also find the maximum slope.

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    NCERT’s answer
    $\displaystyle (1, -16)$, max. slope \(\displaystyle =12\)
    \[m=\frac{dy}{dx}=-3x^2+6x+9 \]\[\frac{dm}{dx}=-6x+6=0 \;\Rightarrow\; x=1 \]\[\frac{d^2m}{dx^2}=-6<0 \;\Rightarrow\; m \text{ is maximum at } x=1 \]\[m_{\max}=-3(1)^2+6(1)+9=12 \]\[y(1)=-1+3+9-27=-16 \]Answer: The slope is maximum at \(\displaystyle (1,-16)\), and the maximum slope is \(\displaystyle 12\).

  4. Exercise 24

    Prove that f(x)=sin⁡x+3cos⁡x\displaystyle f(x)=\sin x+\sqrt{3} \cos x has maximum value at x=π6\displaystyle x=\frac{\pi}{6}.

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    \[f'(x)=\cos x-\sqrt3\sin x \]\[f'\left(\frac{\pi}{6}\right)=\frac{\sqrt3}{2}-\sqrt3\cdot\frac12=0 \]\[f''(x)=-\sin x-\sqrt3\cos x \]\[f''\left(\frac{\pi}{6}\right)=-\frac12-\frac32=-2<0 \]So \(\displaystyle x=\dfrac{\pi}{6}\) is a point of maximum, with value:\[f\left(\frac{\pi}{6}\right)=\frac12+\sqrt3\cdot\frac{\sqrt3}{2}=2 \]This is the greatest value of \(\displaystyle f\), since:\[f(x)=2\sin\left(x+\frac{\pi}{3}\right)\le2 \]Answer: \(\displaystyle f\) has its maximum value \(\displaystyle 2\) at \(\displaystyle x=\dfrac{\pi}{6}\).

  5. Exercise 25

    If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is π3\displaystyle \frac{\pi}{3}.

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    Right angle at \(\displaystyle B\), \(\displaystyle AB=x\), \(\displaystyle AC=h\), \(\displaystyle \theta=\angle A\), and \(\displaystyle h+x=k\) (constant). NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q25 \[BC=\sqrt{(h-x)(h+x)}=\sqrt{k(k-2x)} \quad (h-x=k-2x) \] \[S=\tfrac12\,x\cdot BC \;\Rightarrow\; S^2=\frac{k}{4}\left(kx^2-2x^3\right) \] \[\frac{d(S^2)}{dx}=\frac{k}{2}\,x\,(k-3x)=0 \;\Rightarrow\; x=\frac{k}{3} \] \[\frac{d^2(S^2)}{dx^2}=\frac{k}{4}(2k-12x)=-\frac{k^2}{2}<0 \quad\text{(maximum)} \] \[h=k-x=\frac{2k}{3} \] \[\cos\theta=\frac{x}{h}=\frac12 \;\Rightarrow\; \theta=\frac{\pi}{3} \] Answer: the area is maximum when \(\displaystyle \theta=\frac{\pi}{3}\).

  6. Exercise 26

    Find the points of local maxima, local minima and the points of inflection of the function f(x)=x5−5x4+5x3−1\displaystyle f(x)=x^5-5 x^4+5 x^3-1. Also find the corresponding local maximum and local minimum values.

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    \[f'(x)=5x^4-20x^3+15x^2=5x^2(x-1)(x-3) \] \[f'(x)=0 \;\Rightarrow\; x=0,\;1,\;3 \] \[f''(x)=20x^3-60x^2+30x=10x\,(2x^2-6x+3) \] \[f''(1)=-10<0 \;\Rightarrow\; \text{local maximum at } x=1 \] \[f''(3)=90>0 \;\Rightarrow\; \text{local minimum at } x=3 \] \[f(1)=1-5+5-1=0 \] \[f(3)=243-405+135-1=-28 \] \[f''(x)=0 \;\Rightarrow\; x=0,\;\frac{3\pm\sqrt3}{2} \] \[f'''(x)=60x^2-120x+30 \] \[f'''(0)=30\neq0,\qquad f'''\!\left(\frac{3\pm\sqrt3}{2}\right)=30\,(1\pm\sqrt3)\neq0 \] \(\displaystyle f''\) changes sign at each of these three roots, so each is a point of inflection (\(\displaystyle x=0\) is not an extremum). Answer: local maximum at \(\displaystyle x=1\), value \(\displaystyle 0\); local minimum at \(\displaystyle x=3\), value \(\displaystyle -28\); points of inflection at \(\displaystyle x=0,\ \frac{3\pm\sqrt3}{2}\).

  7. Exercise 27

    A telephone company in a town has 500\displaystyle 500 subscribers on its list and collects fixed charges of Rs 300\displaystyle 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Re 1\displaystyle 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?

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    NCERT’s answer
    Rs $\displaystyle 100$
    Let the annual charge be raised by Rs \(\displaystyle x\). No costs are given, so profit is the income. \[\text{subscribers}=500-x,\qquad \text{charge per subscriber}=300+x \] \[R(x)=(500-x)(300+x)=150000+200x-x^2 \] \[R'(x)=200-2x=0 \;\Rightarrow\; x=100 \] \[R''(x)=-2<0 \quad\text{(maximum)} \] \[R(100)=400\times400=160000 \] Answer: an increase of Rs \(\displaystyle 100\).

  8. Exercise 28

    If the straight line xcos⁡α+ysin⁡α=p\displaystyle x \cos \alpha+y \sin \alpha=p touches the curve x2a2+y2b2=1\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, then prove that a2cos⁡2α+b2sin⁡2α=p2\displaystyle a^2 \cos ^2 \alpha+b^2 \sin ^2 \alpha=p^2.

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    Let the line touch the ellipse at \(\displaystyle (x_1,y_1)\). \[\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0 \;\Rightarrow\; \frac{dy}{dx}=-\frac{b^2x_1}{a^2y_1} \] \[\text{slope of the line}=-\frac{\cos\alpha}{\sin\alpha} \] \[b^2x_1\sin\alpha=a^2y_1\cos\alpha \] \[x_1=\lambda a^2\cos\alpha,\qquad y_1=\lambda b^2\sin\alpha \] Put \(\displaystyle D=a^2\cos^2\alpha+b^2\sin^2\alpha\). \[\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1 \;\Rightarrow\; \lambda^2D=1 \quad\text{(point on the ellipse)} \] \[x_1\cos\alpha+y_1\sin\alpha=p \;\Rightarrow\; \lambda D=p \quad\text{(point on the line)} \] \[p^2=\lambda^2D\cdot D=D \] \[a^2\cos^2\alpha+b^2\sin^2\alpha=p^2 \]

  9. Exercise 29

    An open box with square base is to be made of a given quantity of card board of area c2\displaystyle c^2. Show that the maximum volume of the box is c363\displaystyle \frac{c^3}{6 \sqrt{3}} cubic units.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Base side \(\displaystyle x\), height \(\displaystyle h\).\[x^2+4xh=c^2 \;\Rightarrow\; h=\frac{c^2-x^2}{4x} \] \[V=x^2h=\frac{c^2x-x^3}{4} \] \[\frac{dV}{dx}=\frac{c^2-3x^2}{4}=0 \;\Rightarrow\; x=\frac{c}{\sqrt3} \] \[\frac{d^2V}{dx^2}=-\frac{3x}{2}<0 \quad\text{(maximum)} \] \[V_{\max}=\frac14\left(\frac{c^3}{\sqrt3}-\frac{c^3}{3\sqrt3}\right)=\frac14\cdot\frac{2c^3}{3\sqrt3}=\frac{c^3}{6\sqrt3} \] Answer: the maximum volume is \(\displaystyle \dfrac{c^3}{6\sqrt3}\) cubic units.

  10. Exercise 30

    Find the dimensions of the rectangle of perimeter 36\displaystyle 36 cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume.

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    NCERT’s answer
    6cm, $\displaystyle 12$ cm, \(\displaystyle 864\pi\mathrm{cm}^3\)
    Sides \(\displaystyle x\) and \(\displaystyle y\); revolving about the side \(\displaystyle x\) sweeps a cylinder of radius \(\displaystyle y\) and height \(\displaystyle x\).\[2(x+y)=36 \;\Rightarrow\; x=18-y \] \[V=\pi y^2x=\pi\left(18y^2-y^3\right) \] \[\frac{dV}{dy}=\pi(36y-3y^2)=3\pi y(12-y)=0 \;\Rightarrow\; y=12 \] \[\frac{d^2V}{dy^2}=\pi(36-6y)=-36\pi<0 \quad\text{(maximum)} \] \[x=18-12=6 \] \[V_{\max}=\pi(12)^2(6)=864\pi \ \text{cm}^3 \] Answer: sides \(\displaystyle 6\) cm and \(\displaystyle 12\) cm, revolved about the \(\displaystyle 6\) cm side; maximum volume \(\displaystyle 864\pi\) cm\(\displaystyle ^3\).