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NCERT Exemplar · Class 12 Mathematics Application of Derivatives

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EXERCISE 6.3 41–50 (part 5 of 7)

  1. Choose the correct answer from the given four options in each of the following questions $\displaystyle 35$ to $\displaystyle 39$:

    Exercise 41

    The equation of tangent to the curve y(1+x2)=2−x\displaystyle y\left(1+x^2\right)=2-x, where it crosses x\displaystyle x-axis is:
    (A)
    x+5y=2\displaystyle x+5 y=2
    (B)
    x−5y=2\displaystyle x-5 y=2
    (C)
    5x−y=2\displaystyle 5 x-y=2
    (D)
    5x+y=2\displaystyle 5 x+y=2

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    NCERT’s answer
    A
    (A) \(\displaystyle x+5y=2\). The curve meets the \(\displaystyle x\)-axis where \(\displaystyle y=0\).\[2-x=0 \;\Rightarrow\; x=2 \]\[y'(1+x^2)+2xy=-1 \]\[\text{at } (2,0):\quad 5y'=-1 \;\Rightarrow\; y'=-\tfrac15 \]\[y-0=-\tfrac15\,(x-2) \;\Rightarrow\; x+5y=2 \]

  2. Exercise 42

    The points at which the tangents to the curve y=x3−12x+18\displaystyle y=x^3-12 x+18 are parallel to x\displaystyle x-axis are:
    (A)
    (2,−2),(−2,−34)\displaystyle (2,-2),(-2,-34)
    (B)
    (2,34),(−2,0)\displaystyle (2,34),(-2,0)
    (C)
    (0,34),(−2,0)\displaystyle (0,34),(-2,0)
    (D)
    (2,2),(−2,34)\displaystyle (2,2),(-2,34)

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    NCERT’s answer
    D
    (D) \(\displaystyle (2,2),(-2,34)\). Tangent parallel to the \(\displaystyle x\)-axis means slope \(\displaystyle 0\).\[\frac{dy}{dx}=3x^2-12=0 \;\Rightarrow\; x=\pm 2 \]\[x=2:\quad y=8-24+18=2 \]\[x=-2:\quad y=-8+24+18=34 \]

  3. Exercise 43

    The tangent to the curve y=e2x\displaystyle y=e^{2 x} at the point (0,1)\displaystyle (0,1) meets x\displaystyle x-axis at:
    (A)
    (0,1)\displaystyle (0,1)
    (B)
    (−12,0)\displaystyle \left(-\frac{1}{2}, 0\right)
    (C)
    (2,0)\displaystyle (2,0)
    (D)
    (0,2)\displaystyle (0,2)

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    NCERT’s answer
    B
    (B) \(\displaystyle \left(-\frac12,0\right)\).\[\frac{dy}{dx}=2e^{2x} \;\Rightarrow\; \left.\frac{dy}{dx}\right|_{(0,1)}=2 \]\[y-1=2(x-0) \]\[y=0 \;\Rightarrow\; x=-\tfrac12 \]

  4. Exercise 44

    The slope of tangent to the curve x=t2+3t−8,y=2t2−2t−5\displaystyle x=t^2+3 t-8, y=2 t^2-2 t-5 at the point (2,−1)\displaystyle (2,-1) is:
    (A)
    227\displaystyle \frac{22}{7}
    (B)
    67\displaystyle \frac{6}{7}
    (C)
    −67\displaystyle \frac{-6}{7}

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    NCERT’s answer
    B
    (B) \(\displaystyle \frac67\). First find \(\displaystyle t\) at the point \(\displaystyle (2,-1)\).\[t^2+3t-8=2 \;\Rightarrow\; (t+5)(t-2)=0 \]\[2t^2-2t-5=-1 \;\Rightarrow\; (t-2)(t+1)=0 \]\[\text{common root: } t=2 \]\[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4t-2}{2t+3} \]\[\left.\frac{dy}{dx}\right|_{t=2}=\frac{6}{7} \]

  5. Exercise 45

    The two curves x3−3xy2+2=0\displaystyle x^3-3 x y^2+2=0 and 3x2y−y3−2=0\displaystyle 3 x^2 y-y^3-2=0 intersect at an angle of
    (A)
    π4\displaystyle \frac{\pi}{4}
    (B)
    π3\displaystyle \frac{\pi}{3}
    (C)
    π2\displaystyle \frac{\pi}{2}
    (D)
    π6\displaystyle \frac{\pi}{6}

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    NCERT’s answer
    C
    (C) \(\displaystyle \frac{\pi}{2}\)Differentiate each curve implicitly: \[3x^2-3y^2-6xy\,y'=0 \Rightarrow m_1=\frac{x^2-y^2}{2xy} \] \[6xy+3x^2y'-3y^2y'=0 \Rightarrow m_2=-\frac{2xy}{x^2-y^2} \] \[m_1m_2=-1 \quad (x^2\neq y^2) \] Both curves pass through \(\displaystyle (1,1)\), where \(\displaystyle x^2=y^2\): \[m_1=0,\qquad \frac{dx}{dy}=-\frac{x^2-y^2}{2xy}=0 \] The tangents are perpendicular in every case.

  6. Exercise 46

    The interval on which the function f(x)=2x3+9x2+12x−1\displaystyle f(x)=2 x^3+9 x^2+12 x-1 is decreasing is:
    (A)
    [−1,∞)\displaystyle [-1, \infty)
    (B)
    [-2\displaystyle 2, -1\displaystyle 1]
    (C)
    (−∞,−2]\displaystyle (-\infty,-2]
    (D)
    [−1,1]\displaystyle [-1,1]

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    NCERT’s answer
    B
    (B) \(\displaystyle [-2,-1]\).\[f'(x)=6x^2+18x+12=6(x+1)(x+2) \]\[f'(x)\le 0 \;\Longleftrightarrow\; -2\le x\le -1 \]So \(\displaystyle f\) is decreasing on \(\displaystyle [-2,-1]\).

  7. Exercise 47

    Let the f:R→R\displaystyle f: \mathbf{R} \rightarrow \mathbf{R} be defined by f(x)=2x+cos⁡x\displaystyle f(x)=2 x+\cos x, then f\displaystyle f :
    (A)
    has a minimum at x=π\displaystyle x=\pi
    (B)
    has a maximum, at x=0\displaystyle x=0
    (C)
    is a decreasing function
    (D)
    is an increasing function

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    NCERT’s answer
    D
    (D) \(\displaystyle f\) is an increasing function.\[f'(x)=2-\sin x \]\[-1\le \sin x\le 1 \;\Rightarrow\; f'(x)\ge 1>0 \quad \text{for all } x\in\mathbf{R} \]\(\displaystyle f'\) never vanishes, so \(\displaystyle f\) has no extremum and is increasing throughout.

  8. Exercise 48

    y=x(x−3)2\displaystyle y=x(x-3)^2 decreases for the values of x\displaystyle x given by :
    (A)
    1<x<3\displaystyle 1<x<3
    (B)
    x<0\displaystyle x<0
    (C)
    x>0\displaystyle x>0
    (D)
    0<x<32\displaystyle 0<x<\frac{3}{2}

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    NCERT’s answer
    A
    (A) \(\displaystyle 1<x<3\).\[\frac{dy}{dx}=(x-3)^2+2x(x-3)=3(x-1)(x-3) \]\[\frac{dy}{dx}<0 \;\Longleftrightarrow\; 1<x<3 \]

  9. Exercise 49

    The function f(x)=4sin⁡3x−6sin⁡2x+12sin⁡x+100\displaystyle f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100 is strictly
    (A)
    increasing in (π,3π2)\displaystyle \left(\pi, \frac{3 \pi}{2}\right)
    (B)
    decreasing in (π2,π)\displaystyle \left(\frac{\pi}{2}, \pi\right)
    (C)
    decreasing in [−π2,π2]\displaystyle \left[\frac{-\pi}{2}, \frac{\pi}{2}\right]
    (D)
    decreasing in [0,π2]\displaystyle \left[0, \frac{\pi}{2}\right]

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    NCERT’s answer
    B
    (B) decreasing in \(\displaystyle \left(\frac{\pi}{2},\pi\right)\).\[f'(x)=12\sin^2x\cos x-12\sin x\cos x+12\cos x \]\[f'(x)=12\cos x\,(\sin^2x-\sin x+1) \]\[\sin^2x-\sin x+1=\left(\sin x-\tfrac12\right)^2+\tfrac34>0 \]So \(\displaystyle f'\) has the sign of \(\displaystyle \cos x\), which is negative on \(\displaystyle \left(\frac{\pi}{2},\pi\right)\). Option (A) fails since \(\displaystyle \cos x<0\) on \(\displaystyle \left(\pi,\frac{3\pi}{2}\right)\); (C), (D) fail since \(\displaystyle \cos x\ge0\) there.

  10. Exercise 50

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

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    NCERT’s answer
    C
    (C) \(\displaystyle \cos x\).\[\frac{d}{dx}\cos x=-\sin x<0 \quad \text{for } x\in\left(0,\tfrac{\pi}{2}\right) \]Each other function increases somewhere in the interval:\[\frac{d}{dx}\tan x=\sec^2x>0 \]\[\frac{d}{dx}\sin 2x=2\cos 2x>0 \quad \text{for } x\in\left(0,\tfrac{\pi}{4}\right) \]\[\frac{d}{dx}\cos 3x=-3\sin 3x>0 \quad \text{for } x\in\left(\tfrac{\pi}{3},\tfrac{\pi}{2}\right) \]