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NCERT Exemplar · Class 12 Mathematics Application of Derivatives

64 questions · 64 still being checked

EXERCISE 6.3 11–20 (part 2 of 7)

  1. Exercise 11

    x\displaystyle x and y\displaystyle y are the sides of two squares such that y=x−x2\displaystyle y=x-x^2. Find the rate of change of the area of second square with respect to the area of first square.

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    \[A_1 = x^2 \;\Rightarrow\; \frac{dA_1}{dx} = 2x \]\[A_2 = y^2 = (x-x^2)^2 \;\Rightarrow\; \frac{dA_2}{dx} = 2(x-x^2)(1-2x) \]\[\frac{dA_2}{dA_1} = \frac{dA_2/dx}{dA_1/dx} = \frac{2x(1-x)(1-2x)}{2x} \]\[\frac{dA_2}{dA_1} = (1-x)(1-2x) \]Answer: \(\displaystyle (1-x)(1-2x) = 2x^2-3x+1\)

  2. Exercise 12

    Find the condition that the curves 2x=y2\displaystyle 2 x=y^2 and 2xy=k\displaystyle 2 x y=k intersect orthogonally.

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    NCERT’s answer
    \(\displaystyle k^2=8\)
    Slopes at a common point \(\displaystyle (x,y)\):\[2x = y^2 \;\Rightarrow\; 2 = 2y\frac{dy}{dx} \;\Rightarrow\; m_1 = \frac{1}{y} \]\[2xy = k \;\Rightarrow\; 2y + 2x\frac{dy}{dx} = 0 \;\Rightarrow\; m_2 = -\frac{y}{x} \]Orthogonal means \(\displaystyle m_1m_2=-1\):\[\frac{1}{y}\cdot\left(-\frac{y}{x}\right) = -1 \;\Rightarrow\; x = 1 \]\[y^2 = 2x = 2, \qquad k = 2xy = 2y \]\[k^2 = 4y^2 = 8 \]NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q12Answer: \(\displaystyle k^2 = 8\), i.e. \(\displaystyle k=\pm 2\sqrt2\)

  3. Exercise 13

    Prove that the curves xy=4\displaystyle x y=4 and x2+y2=8\displaystyle x^2+y^2=8 touch each other.

    Not cross-checked

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

    Common points:\[(x-y)^2 = (x^2+y^2) - 2xy = 8 - 8 = 0 \;\Rightarrow\; y = x \]\[xy = 4 \;\Rightarrow\; x^2 = 4 \;\Rightarrow\; (2,2),\ (-2,-2) \]Slopes:\[xy = 4 \;\Rightarrow\; y + x\frac{dy}{dx} = 0 \;\Rightarrow\; m_1 = -\frac{y}{x} \]\[x^2+y^2 = 8 \;\Rightarrow\; 2x + 2y\frac{dy}{dx} = 0 \;\Rightarrow\; m_2 = -\frac{x}{y} \]\[\text{at } (2,2) \text{ and } (-2,-2): \quad m_1 = m_2 = -1 \]NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q13Answer: the curves meet at \(\displaystyle (2,2)\) and \(\displaystyle (-2,-2)\) with the same tangent slope \(\displaystyle -1\), so they touch each other.

  4. Exercise 14

    Find the co-ordinates of the point on the curve x+y=4\displaystyle \sqrt{x}+\sqrt{y}=4 at which tangent is equally inclined to the axes.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle (4, 4)$
    \[\sqrt x + \sqrt y = 4 \]\[\frac{1}{2\sqrt x} + \frac{1}{2\sqrt y}\frac{dy}{dx} = 0 \;\Rightarrow\; \frac{dy}{dx} = -\frac{\sqrt y}{\sqrt x} \]Equally inclined to the axes means slope \(\displaystyle \pm1\); \(\displaystyle -\sqrt y/\sqrt x\) is never positive, so slope \(\displaystyle =-1\).\[\frac{\sqrt y}{\sqrt x} = 1 \;\Rightarrow\; y = x \]\[\sqrt x + \sqrt x = 4 \;\Rightarrow\; x = 4, \quad y = 4 \]NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q14Answer: \(\displaystyle (4,\,4)\)

  5. Exercise 15

    Find the angle of intersection of the curves y=4−x2\displaystyle y=4-x^2 and y=x2\displaystyle y=x^2.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)\)
    Intersection:\[4 - x^2 = x^2 \;\Rightarrow\; x = \pm\sqrt2, \quad y = 2 \]Slopes at \(\displaystyle x=\sqrt2\):\[m_1 = \frac{d}{dx}(4-x^2) = -2x = -2\sqrt2 \]\[m_2 = \frac{d}{dx}(x^2) = 2x = 2\sqrt2 \]\[\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right| = \left|\frac{-4\sqrt2}{1-8}\right| = \frac{4\sqrt2}{7} \]At \(\displaystyle x=-\sqrt2\) the slopes only swap signs, so \(\displaystyle \tan\theta\) is the same.NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q15Answer: \(\displaystyle \theta = \tan^{-1}\dfrac{4\sqrt2}{7}\)

  6. Exercise 16

    Prove that the curves y2=4x\displaystyle y^2=4 x and x2+y2−6x+1=0\displaystyle x^2+y^2-6 x+1=0 touch each other at the point (1,2)\displaystyle (1, 2).

    Not cross-checked

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

    The point \(\displaystyle (1,2)\) lies on both curves:\[2^2 = 4\cdot 1 \quad \text{(on } y^2=4x\text{)} \]\[1^2 + 2^2 - 6(1) + 1 = 0 \quad \text{(on the circle)} \]Slopes at \(\displaystyle (1,2)\):\[y^2 = 4x \;\Rightarrow\; 2y\frac{dy}{dx} = 4 \;\Rightarrow\; m_1 = \frac{2}{y} = 1 \]\[2x + 2y\frac{dy}{dx} - 6 = 0 \;\Rightarrow\; m_2 = \frac{3-x}{y} = \frac{2}{2} = 1 \]NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q16Answer: \(\displaystyle (1,2)\) is common and the slopes are equal \(\displaystyle (m_1=m_2=1)\), so the curves touch at \(\displaystyle (1,2)\).

  7. Exercise 17

    Find the equation of the normal lines to the curve 3x2−y2=8\displaystyle 3 x^2-y^2=8 which are parallel to the line x+3y=4\displaystyle x+3 y=4.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle x+3 y= \pm 8\)
    A normal parallel to \(\displaystyle x+3y=4\) has slope \(\displaystyle -\tfrac13\), so the tangent there has slope \(\displaystyle 3\).\[3x^2-y^2=8 \;\Rightarrow\; 6x-2y\frac{dy}{dx}=0 \;\Rightarrow\; \frac{dy}{dx}=\frac{3x}{y} \]\[\frac{3x}{y}=3 \;\Rightarrow\; y=x \]\[3x^2-x^2=8 \;\Rightarrow\; x=\pm2 \;\Rightarrow\; (2,2),\ (-2,-2) \]Normals of slope \(\displaystyle -\tfrac13\) through these points:\[y-2=-\tfrac13(x-2) \;\Rightarrow\; x+3y=8 \]\[y+2=-\tfrac13(x+2) \;\Rightarrow\; x+3y=-8 \]NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q17Answer: \(\displaystyle x+3y=8\) and \(\displaystyle x+3y=-8\).

  8. Exercise 18

    At what points on the curve x2+y2−2x−4y+1=0\displaystyle x^2+y^2-2 x-4 y+1=0, the tangents are parallel to the y\displaystyle y-axis?

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle (3, 2)$, $\displaystyle (-1, 2)$
    A tangent parallel to the \(\displaystyle y\)-axis is vertical, so \(\displaystyle \dfrac{dy}{dx}\) is not defined.\[2x+2y\frac{dy}{dx}-2-4\frac{dy}{dx}=0 \;\Rightarrow\; \frac{dy}{dx}=\frac{1-x}{y-2} \]\[y-2=0 \;\Rightarrow\; y=2 \]\[x^2+4-2x-8+1=0 \;\Rightarrow\; x^2-2x-3=0 \;\Rightarrow\; (x-3)(x+1)=0 \]\[x=3 \ \text{or}\ x=-1 \]At both points the numerator is non-zero:\[1-x=-2 \ \text{or}\ 2 \neq 0 \]NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q18Answer: \(\displaystyle (3,2)\) and \(\displaystyle (-1,2)\).

  9. Exercise 19

    Show that the line xa+yb=1\displaystyle \frac{x}{a}+\frac{y}{b}=1, touches the curve y=b⋅e−xa\displaystyle y=b \cdot e^{\frac{-x}{a}} at the point where the curve intersects the axis of y\displaystyle y.

    Not cross-checked

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

    The curve meets the \(\displaystyle y\)-axis where \(\displaystyle x=0\).\[y=b\,e^{0}=b \;\Rightarrow\; \text{point } (0,b) \]The line passes through this point:\[\frac{0}{a}+\frac{b}{b}=1 \]Slope of the curve at \(\displaystyle (0,b)\):\[\frac{dy}{dx}=-\frac{b}{a}\,e^{-x/a} \;\Rightarrow\; \left.\frac{dy}{dx}\right|_{(0,b)}=-\frac{b}{a} \]Slope of the line:\[\frac{x}{a}+\frac{y}{b}=1 \;\Rightarrow\; y=b-\frac{b}{a}\,x \;\Rightarrow\; \text{slope}=-\frac{b}{a} \]NCERT_Solution_Class12_Maths_Exemplar_Ch6_Ex6-3_Q19Answer: The line and the curve share the point \(\displaystyle (0,b)\) and the slope \(\displaystyle -\dfrac{b}{a}\), so the line touches the curve at \(\displaystyle (0,b)\).

  10. Exercise 20

    Show that f(x)=2x+cot⁡−1x+log⁡(1+x2−x)\displaystyle f(x)=2 x+\cot ^{-1} x+\log \left(\sqrt{1+x^2}-x\right) is increasing in R\displaystyle \mathbf{R}.

    Not cross-checked

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

    The logarithm is defined for every real \(\displaystyle x\), since \(\displaystyle \sqrt{1+x^2}>|x|\ge x\).\[\frac{d}{dx}\log\left(\sqrt{1+x^2}-x\right)=\frac{\dfrac{x}{\sqrt{1+x^2}}-1}{\sqrt{1+x^2}-x}=\frac{-\left(\sqrt{1+x^2}-x\right)}{\sqrt{1+x^2}\left(\sqrt{1+x^2}-x\right)}=-\frac{1}{\sqrt{1+x^2}} \]\[f'(x)=2-\frac{1}{1+x^2}-\frac{1}{\sqrt{1+x^2}} \]Put \(\displaystyle t=\dfrac{1}{\sqrt{1+x^2}}\), so \(\displaystyle 0<t\le1\):\[f'(x)=2-t^2-t=(1-t)(2+t)\ge0 \]\[f'(x)=0 \iff t=1 \iff x=0 \]Answer: \(\displaystyle f'(x)\ge0\) for all \(\displaystyle x\in\mathbf R\), zero only at \(\displaystyle x=0\), so \(\displaystyle f\) is increasing in \(\displaystyle \mathbf R\).