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NCERT Exemplar · Class 12 Mathematics Application of Derivatives

64 questions · 64 still being checked

EXERCISE 6.3 51–60 (part 6 of 7)

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

    Exercise 51

    The function f(x)=tan⁡x−x\displaystyle f(x)=\tan x-x
    (A)
    always increases
    (B)
    always decreases
    (C)
    never increases
    (D)
    sometimes increases and sometimes decreases.

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    NCERT’s answer
    A
    (A) always increases.\[f'(x)=\sec^2x-1=\tan^2x\ge0 \]\[f'(x)=0 \text{ only at } x=n\pi \]Zeros of \(\displaystyle f'\) are isolated, so \(\displaystyle f\) is increasing on every interval of its domain.

  2. Exercise 52

    If x\displaystyle x is real, the minimum value of x2−8x+17\displaystyle x^2-8 x+17 is
    (A)
    -1\displaystyle 1 (B) 0\displaystyle 0 (C) 1\displaystyle 1 (D) 2\displaystyle 2

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    NCERT’s answer
    C
    (C) \(\displaystyle 1\).\[x^2-8x+17=(x-4)^2+1 \]\[(x-4)^2\ge0 \Rightarrow x^2-8x+17\ge1 \]\[\text{equality at } x=4 \]

  3. Exercise 53

    The smallest value of the polynomial x3−18x2+96x\displaystyle x^3-18 x^2+96 x in [0\displaystyle 0, 9\displaystyle 9] is
    (A)
    126\displaystyle 126
    (B)
    0\displaystyle 0 (C) 135\displaystyle 135
    (D)
    160\displaystyle 160

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    NCERT’s answer
    B
    (B) \(\displaystyle 0\).\[g'(x)=3x^2-36x+96=3(x-4)(x-8) \]Critical points \(\displaystyle 4,\,8\) lie in \(\displaystyle [0,9]\). Compare with the end points:\[g(0)=0 \]\[g(4)=64-288+384=160 \]\[g(8)=512-1152+768=128 \]\[g(9)=729-1458+864=135 \]The smallest value is \(\displaystyle 0\).

  4. Exercise 54

    The function f(x)=2x3−3x2−12x+4\displaystyle f(x)=2 x^3-3 x^2-12 x+4, has
    (A)
    two points of local maximum
    (B)
    two points of local minimum
    (C)
    one maxima and one minima
    (D)
    no maxima or minima

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    NCERT’s answer
    C
    (C) one maxima and one minima.\[f'(x)=6x^2-6x-12=6(x+1)(x-2) \]\[f'(x)=0 \Rightarrow x=-1,\ 2 \]\[f''(x)=12x-6 \]\[f''(-1)=-18<0 \Rightarrow \text{local maximum at } x=-1 \]\[f''(2)=18>0 \Rightarrow \text{local minimum at } x=2 \]

  5. Exercise 55

    The maximum value of sin⁡x.cos⁡x\displaystyle \sin x . \cos x is
    (A)
    14\displaystyle \frac{1}{4}
    (B)
    12\displaystyle \frac{1}{2}
    (C)
    2\displaystyle \sqrt{2}
    (D)
    22\displaystyle 2 \sqrt{2}

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    NCERT’s answer
    B
    (B) \(\displaystyle \frac{1}{2}\).\[\sin x\cos x=\tfrac12\sin 2x \]\[\sin 2x\le1 \Rightarrow \sin x\cos x\le\tfrac12 \]\[\text{equality at } x=\tfrac{\pi}{4} \]

  6. Exercise 56

    At x=5π6,f(x)=2sin⁡3x+3cos⁡3x\displaystyle x=\frac{5 \pi}{6}, f(x)=2 \sin 3 x+3 \cos 3 x is:
    (A)
    maximum
    (B)
    minimum
    (C)
    zero
    (D)
    neither maximum nor minimum.

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    (D) neither maximum nor minimum.\[f'(x)=6\cos 3x-9\sin 3x \]\[f'\!\left(\tfrac{5\pi}{6}\right)=6\cos\tfrac{5\pi}{2}-9\sin\tfrac{5\pi}{2}=6(0)-9(1)=-9 \]\[f'\!\left(\tfrac{5\pi}{6}\right)\neq0 \]So \(\displaystyle x=\frac{5\pi}{6}\) is not a critical point.

  7. Exercise 57

    Maximum slope of the curve y=−x3+3x2+9x−27\displaystyle y=-x^3+3 x^2+9 x-27 is:
    (A)
    0\displaystyle 0 (B) 12\displaystyle 12
    (C)
    16\displaystyle 16
    (D)
    32\displaystyle 32

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    NCERT’s answer
    B
    (B) \(\displaystyle 12\) — the slope is a downward parabola in \(\displaystyle x\), so its maximum is at the stationary point. \[m = \frac{dy}{dx} = -3x^2 + 6x + 9 \] \[\frac{dm}{dx} = -6x + 6 = 0 \Rightarrow x = 1 \] \[\frac{d^2m}{dx^2} = -6 < 0 \] \[m_{\max} = -3(1)^2 + 6(1) + 9 = 12 \]

  8. Exercise 58

    f(x)=xx\displaystyle f(x)=x^x has a stationary point at
    (A)
    x=e\displaystyle x=e
    (B)
    x=1e\displaystyle x=\frac{1}{e}
    (C)
    x=1\displaystyle x=1
    (D)
    x=e\displaystyle x=\sqrt{e}

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    NCERT’s answer
    B
    (B) \(\displaystyle x = \frac{1}{e}\) — a stationary point is where \(\displaystyle f'(x) = 0\). \[\ln f = x \ln x \] \[\frac{f'(x)}{f(x)} = \ln x + 1 \] \[f'(x) = x^x(\ln x + 1) \] \[x^x > 0 \Rightarrow \ln x + 1 = 0 \Rightarrow x = e^{-1} = \frac{1}{e} \]

  9. Exercise 59

    The maximum value of (1x)x\displaystyle \left(\frac{1}{x}\right)^x is:
    (A)
    e\displaystyle e (B) ee\displaystyle e^e
    (C)
    e1e\displaystyle e^{\frac{1}{e}}
    (D)
    (1e)1e\displaystyle \left(\frac{1}{e}\right)^{\frac{1}{e}}

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    NCERT’s answer
    C
    (C) \(\displaystyle e^{\frac{1}{e}}\) — maximise the logarithm of the function. \[y = \left(\frac{1}{x}\right)^x \Rightarrow \ln y = -x \ln x \] \[\frac{d}{dx}(\ln y) = -(\ln x + 1) = 0 \Rightarrow x = \frac{1}{e} \] \[\frac{d^2}{dx^2}(\ln y) = -\frac{1}{x} < 0 \] \[\ln y_{\max} = -\frac{1}{e}\ln\frac{1}{e} = \frac{1}{e} \] \[y_{\max} = e^{\frac{1}{e}} \]
  10. Fill in the blanks in each of the following Exercises $\displaystyle 60$ to $\displaystyle 64$:

    Exercise 60

    The curves y=4x2+2x−8\displaystyle y=4 x^2+2 x-8 and y=x3−x+13\displaystyle y=x^3-x+13 touch each other at the point ____.

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

    Touching needs a common point with a common slope. \[y_1 = 4x^2 + 2x - 8 \Rightarrow \frac{dy_1}{dx} = 8x + 2 \] \[y_2 = x^3 - x + 13 \Rightarrow \frac{dy_2}{dx} = 3x^2 - 1 \] \[8x + 2 = 3x^2 - 1 \Rightarrow 3x^2 - 8x - 3 = 0 \] \[(3x + 1)(x - 3) = 0 \Rightarrow x = 3 \ \text{or} \ x = -\frac{1}{3} \] \[x = 3: \quad y_1 = 36 + 6 - 8 = 34, \quad y_2 = 27 - 3 + 13 = 37 \] \[x = -\frac{1}{3}: \quad y_1 = -\frac{74}{9}, \quad y_2 = \frac{359}{27} \] At both slope-matching points the \(\displaystyle y\)-values differ, so as printed the curves do not touch anywhere.