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NCERT Exemplar · Class 12 Mathematics Application of Derivatives

64 questions · 64 still being checked

EXERCISE 6.3 31–40 (part 4 of 7)

  1. Exercise 31

    If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?

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    NCERT’s answer
    $\displaystyle 1$:$\displaystyle 1$
    Edge of the cube \(\displaystyle x\), radius of the sphere \(\displaystyle r\), total surface area \(\displaystyle S\) constant. \[6x^2+4\pi r^2=S \;\Rightarrow\; 12x+8\pi r\frac{dr}{dx}=0 \;\Rightarrow\; \frac{dr}{dx}=-\frac{3x}{2\pi r} \] \[V=x^3+\frac43\pi r^3 \] \[\frac{dV}{dx}=3x^2+4\pi r^2\frac{dr}{dx}=3x^2-6xr=3x(x-2r) \] \[\frac{dV}{dx}=0 \;\Rightarrow\; x=2r \] \[\frac{d^2V}{dx^2}=6x-6r-6x\frac{dr}{dx}=6x-6r+\frac{9x^2}{\pi r} \] \[\left.\frac{d^2V}{dx^2}\right|_{x=2r}=12r-6r+\frac{36r}{\pi}>0 \quad\text{(minimum)} \] \[\frac{\text{edge}}{\text{diameter}}=\frac{x}{2r}=1 \] Answer: \(\displaystyle 1:1\)

  2. Exercise 32

    AB is a diameter of a circle and C is any point on the circle. Show that the area of ΔABC\displaystyle \Delta \mathrm{ABC} is maximum, when it is isosceles.

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    Let \(\displaystyle r\) be the radius, so \(\displaystyle AB=2r\), and \(\displaystyle \theta=\angle CAB\). NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q32 \[\angle ACB=90^\circ \quad\text{(angle in a semicircle)} \] \[AC=2r\cos\theta,\qquad BC=2r\sin\theta \] \[S=\tfrac12\,AC\cdot BC=2r^2\sin\theta\cos\theta=r^2\sin2\theta \] \[\frac{dS}{d\theta}=2r^2\cos2\theta=0 \;\Rightarrow\; \theta=\frac{\pi}{4} \] \[\frac{d^2S}{d\theta^2}=-4r^2\sin2\theta=-4r^2<0 \quad\text{(maximum)} \] \[AC=2r\cos\frac{\pi}{4}=\sqrt2\,r,\qquad BC=2r\sin\frac{\pi}{4}=\sqrt2\,r \] \[AC=BC \] Answer: the area is maximum when \(\displaystyle AC=BC\), that is, when \(\displaystyle \Delta ABC\) is isosceles.

  3. Exercise 33

    A metal box with a square base and vertical sides is to contain 1024 cm3\displaystyle 1024 \mathrm{~cm}^3. The material for the top and bottom costs Rs 5/cm2\displaystyle 5 / \mathrm{cm}^2 and the material for the sides costs Rs 2.50/cm2\displaystyle 2.50 / \mathrm{cm}^2. Find the least cost of the box.

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    NCERT’s answer
    Rs $\displaystyle 1920$
    Let the base side be \(\displaystyle x\) cm and the height \(\displaystyle h\) cm.\[x^2 h = 1024 \Rightarrow h = \frac{1024}{x^2} \]\[C = 2x^2(5) + 4xh(2.5) = 10x^2 + 10xh \]\[C = 10x^2 + \frac{10240}{x} \]\[\frac{dC}{dx} = 20x - \frac{10240}{x^2} = 0 \Rightarrow x^3 = 512 \Rightarrow x = 8 \]\[\frac{d^2C}{dx^2} = 20 + \frac{20480}{x^3} > 0 \quad \text{(minimum at } x = 8) \]\[h = \frac{1024}{64} = 16 \]\[C_{\min} = 10(64) + \frac{10240}{8} = 640 + 1280 = 1920 \]Answer: Least cost = Rs \(\displaystyle 1920\) (base \(\displaystyle 8\) cm, height \(\displaystyle 16\) cm).

  4. Exercise 34

    The sum of the surface areas of a rectangular parallelopiped with sides x,2x\displaystyle x, 2 x and x3\displaystyle \frac{x}{3} and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x\displaystyle x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.

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    NCERT’s answer
    \(\displaystyle \frac{2}{3}x^3\left(1+\frac{2\pi}{27}\right)\)
    Let the sphere have radius \(\displaystyle r\).\[\text{Cuboid surface} = 2\left(x\cdot 2x + 2x\cdot\frac{x}{3} + \frac{x}{3}\cdot x\right) = 6x^2 \]\[S = 6x^2 + 4\pi r^2 = \text{constant} \]\[V = x\cdot 2x\cdot\frac{x}{3} + \frac{4}{3}\pi r^3 = \frac{2}{3}x^3 + \frac{4}{3}\pi r^3 \]Differentiate \(\displaystyle S\) with respect to \(\displaystyle x\):\[12x + 8\pi r\,\frac{dr}{dx} = 0 \Rightarrow \frac{dr}{dx} = -\frac{3x}{2\pi r} \]\[\frac{dV}{dx} = 2x^2 + 4\pi r^2\,\frac{dr}{dx} = 2x^2 - 6xr = 2x(x - 3r) \]\[\frac{dV}{dx} = 0 \Rightarrow x = 3r \]\[\frac{d^2V}{dx^2} = 4x - 6r - 6x\,\frac{dr}{dx} = 4x - 6r + \frac{9x^2}{\pi r} \]\[\left.\frac{d^2V}{dx^2}\right|_{x=3r} = 6r + \frac{81r}{\pi} > 0 \]So \(\displaystyle V\) is minimum when \(\displaystyle x = 3r\).\[V_{\min} = \frac{2}{3}(3r)^3 + \frac{4}{3}\pi r^3 = 18r^3 + \frac{4}{3}\pi r^3 = \frac{2}{3}(27 + 2\pi)r^3 \]\[S = 54r^2 + 4\pi r^2 \Rightarrow V_{\min} = \frac{rS}{3} \]Answer: \(\displaystyle x = 3r\) gives the minimum, \(\displaystyle V_{\min} = \dfrac{2}{3}(27 + 2\pi)r^3\).
  5. Choose the correct answer from the given four options in each of the following questions $\displaystyle 35$ to $\displaystyle 39$:

    Exercise 35

    The sides of an equilateral triangle are increasing at the rate of 2\displaystyle 2 cm/sec. The rate at which the area increases, when side is 10\displaystyle 10 cm is:
    (A)
    10 cm2/s\displaystyle 10 \mathrm{~cm}^2 / \mathrm{s}
    (B)
    3 cm2/s\displaystyle \sqrt{3} \mathrm{~cm}^2 / \mathrm{s}
    (C)
    103 cm2/s\displaystyle 10 \sqrt{3} \mathrm{~cm}^2 / \mathrm{s}
    (D)
    103 cm2/s\displaystyle \frac{10}{3} \mathrm{~cm}^2 / \mathrm{s}

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    NCERT’s answer
    C
    (C) \(\displaystyle 10\sqrt{3}\ \text{cm}^2/\text{s}\)\[A = \frac{\sqrt{3}}{4}s^2 \]\[\frac{dA}{dt} = \frac{\sqrt{3}}{2}\,s\,\frac{ds}{dt} \]\[\left.\frac{dA}{dt}\right|_{s=10} = \frac{\sqrt{3}}{2}(10)(2) = 10\sqrt{3} \]

  6. Exercise 36

    A ladder, 5\displaystyle 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec\displaystyle 10 \mathrm{~cm} / \mathrm{sec}, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2\displaystyle 2 metres from the wall is:
    (A)
    110radian/sec\displaystyle \frac{1}{10} \mathrm{radian} / \mathrm{sec}
    (B)
    120radian/sec\displaystyle \frac{1}{20} \mathrm{radian} / \mathrm{sec}
    (C)
    20\displaystyle 20 radian/sec
    (D)
    10\displaystyle 10 radian/sec

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    NCERT’s answer
    B
    (B) \(\displaystyle \dfrac{1}{20}\) radian/secLet \(\displaystyle \angle ABO = \theta\) and \(\displaystyle OA = y\) m.NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q36\[y = 5\sin\theta \]\[\frac{dy}{dt} = 5\cos\theta\,\frac{d\theta}{dt} \]\[\frac{dy}{dt} = -10\ \text{cm/s} = -0.1\ \text{m/s}, \qquad \cos\theta = \frac{OB}{AB} = \frac{2}{5} \]\[-0.1 = 5\cdot\frac{2}{5}\,\frac{d\theta}{dt} \Rightarrow \frac{d\theta}{dt} = -\frac{1}{20} \]\(\displaystyle \theta\) decreases at \(\displaystyle \frac{1}{20}\) radian/sec.

  7. Exercise 37

    The curve y=x15\displaystyle y=x^{\frac{1}{5}} has at (0,0)\displaystyle (0,0)
    (A)
    a vertical tangent (parallel to y\displaystyle y-axis)
    (B)
    a horizontal tangent (parallel to x\displaystyle x-axis)
    (C)
    an oblique tangent
    (D)
    no tangent

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    NCERT’s answer
    A
    (A) a vertical tangentNCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q37\[\frac{dy}{dx} = \frac{1}{5}x^{-4/5} \to \infty \quad (x \to 0) \]\[x = y^5 \Rightarrow \frac{dx}{dy} = 5y^4 = 0 \quad \text{at } (0,0) \]The tangent at the origin is the \(\displaystyle y\)-axis.

  8. Exercise 38

    The equation of normal to the curve 3x2−y2=8\displaystyle 3 x^2-y^2=8 which is parallel to the line x+3y=8\displaystyle x+3 y=8 is
    (A)
    3x−y=8\displaystyle 3 x-y=8
    (B)
    3x+y+8=0\displaystyle 3 x+y+8=0
    (C)
    x+3y±8=0\displaystyle x+3 y \pm 8=0
    (D)
    x+3y=0\displaystyle x+3 y=0

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    NCERT’s answer
    C
    (C) \(\displaystyle x + 3y \pm 8 = 0\)\[3x^2 - y^2 = 8 \Rightarrow \frac{dy}{dx} = \frac{3x}{y} \]\[-\frac{y}{3x} = -\frac{1}{3} \quad \text{(slope of } x + 3y = 8) \Rightarrow y = x \]\[3x^2 - x^2 = 8 \Rightarrow x = \pm 2 \Rightarrow (2,2),\ (-2,-2) \]\[y - 2 = -\frac{1}{3}(x - 2) \Rightarrow x + 3y - 8 = 0 \]\[y + 2 = -\frac{1}{3}(x + 2) \Rightarrow x + 3y + 8 = 0 \]

  9. Exercise 39

    If the curve ay+x2=7\displaystyle a y+x^2=7 and x3=y\displaystyle x^3=y, cut orthogonally at (1,1)\displaystyle (1,1), then the value of a\displaystyle a is:
    (A)
    1\displaystyle 1 (B) 0\displaystyle 0 (C) −6\displaystyle -6 (D) .6

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    NCERT’s answer
    D
    (D) \(\displaystyle a = 6\)\[m_1 = \frac{dy}{dx}\bigg|_{(1,1)} = -\frac{2x}{a} = -\frac{2}{a} \]\[m_2 = \frac{d(x^3)}{dx}\bigg|_{(1,1)} = 3x^2 = 3 \]\[m_1 m_2 = -1 \Rightarrow -\frac{6}{a} = -1 \Rightarrow a = 6 \]\[\text{Check: } (1,1) \text{ on } ay + x^2 = 7 \Rightarrow a + 1 = 7 \Rightarrow a = 6 \]

  10. Exercise 40

    If y=x4−10\displaystyle y=x^4-10 and if x\displaystyle x changes from 2\displaystyle 2 to 1.99\displaystyle 1.99, what is the change in y\displaystyle y
    (A)
    .32
    (B)
    .032
    (C)
    5.68\displaystyle 68
    (D)
    5.968\displaystyle 968

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    NCERT’s answer
    A
    (A) \(\displaystyle y\) falls by about \(\displaystyle 0.32\).\[y = x^4 - 10 \Rightarrow dy = 4x^3\,dx \]\[x = 2,\quad dx = 1.99 - 2 = -0.01 \]\[dy = 4(2)^3(-0.01) = -0.32 \]\[\text{Check: } 1.99^4 - 2^4 = 15.6824 - 16 = -0.3176 \]