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NCERT Exemplar · Class 11 Mathematics Limits and Derivatives

80 questions · 80 still being checked

EXERCISE 13.3 71–80 (part 8 of 8)

  1. Choose the correct answer out of $\displaystyle 4$ options given against each Exercise $\displaystyle 54$ to $\displaystyle 76$ (M.C.Q).

    Exercise 71

    If y=sin⁡x+cos⁡xsin⁡x−cos⁡x\displaystyle y=\frac{\sin x+\cos x}{\sin x-\cos x}, then dydx\displaystyle \frac{d y}{d x} at x=0\displaystyle x=0 is
    (A)
    -2\displaystyle 2 (B) 0\displaystyle 0 (C) 12\displaystyle \frac{1}{2}
    (D)
    does not exist

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    NCERT’s answer
    A
    (A) \(\displaystyle -2\)Quotient rule:\[\frac{dy}{dx}=\frac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^2} \]\[=\frac{-(\sin x-\cos x)^2-(\sin x+\cos x)^2}{(\sin x-\cos x)^2}=\frac{-2}{(\sin x-\cos x)^2} \]\[\left.\frac{dy}{dx}\right|_{x=0}=\frac{-2}{(0-1)^2}=-2 \]

  2. Exercise 72

    If y=sin⁡(x+9)cos⁡x\displaystyle y=\frac{\sin (x+9)}{\cos x} then dydx\displaystyle \frac{d y}{d x} at x=0\displaystyle x=0 is
    (A)
    cos⁡9\displaystyle \cos 9
    (B)
    sin⁡9\displaystyle \sin 9
    (C)
    0\displaystyle 0 (D) 1\displaystyle 1

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    NCERT’s answer
    A
    (A) \(\displaystyle \cos 9\)Quotient rule:\[\frac{dy}{dx}=\frac{\cos(x+9)\cos x+\sin(x+9)\sin x}{\cos^2x} \]\[=\frac{\cos\big((x+9)-x\big)}{\cos^2x}=\frac{\cos 9}{\cos^2x} \]\[\left.\frac{dy}{dx}\right|_{x=0}=\cos 9 \]

  3. Exercise 73

    If f(x)=1+x+x22+…+x100100\displaystyle f(x)=1+x+\frac{x^2}{2}+\ldots+\frac{x^{100}}{100}, then f′(1)\displaystyle f^{\prime}(1) is equal to
    (A)
    1100\displaystyle \frac{1}{100}
    (B)
    100\displaystyle 100
    (C)
    does not exist

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    NCERT’s answer
    B
    (B) \(\displaystyle 100\). Differentiate term by term.\[f'(x) = 1 + x + x^2 + \ldots + x^{99} \]\[f'(1) = 1 + 1 + \ldots + 1 \;\;(100 \text{ terms}) = 100 \]

  4. Exercise 74

    If f(x)=xn−anx−a\displaystyle f(x)=\frac{x^n-a^n}{x-a} for some constant ‘a\displaystyle a’, then f′(a)\displaystyle f^{\prime}(a) is
    (A)
    1\displaystyle 1 (B) 0\displaystyle 0 (C) does not exist
    (D)
    12\displaystyle \frac{1}{2}

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    NCERT’s answer
    C
    (C) does not exist. The denominator vanishes at \(\displaystyle x = a\), so \(\displaystyle f(a)\) is not defined.\[f(x) = \frac{x^n - a^n}{x - a}, \qquad x - a = 0 \text{ at } x = a \]\[f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]This needs \(\displaystyle f(a)\), so \(\displaystyle f'(a)\) does not exist.

  5. Exercise 75

    If f(x)=x100+x99+…+x+1\displaystyle f(x)=x^{100}+x^{99}+\ldots+x+1, then f′(1)\displaystyle f^{\prime}(1) is equal to
    (A)
    5050\displaystyle 5050
    (B)
    5049\displaystyle 5049
    (C)
    5051\displaystyle 5051
    (D)
    50051\displaystyle 50051

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    NCERT’s answer
    A
    (A) \(\displaystyle 5050\). Differentiate term by term.\[f'(x) = 100x^{99} + 99x^{98} + \ldots + 2x + 1 \]\[f'(1) = 100 + 99 + \ldots + 2 + 1 = \frac{100 \cdot 101}{2} = 5050 \]

  6. Exercise 76

    If f(x)=1−x+x2−x3…−x99+x100\displaystyle f(x)=1-x+x^2-x^3 \ldots-x^{99}+x^{100}, then f′(1)\displaystyle f^{\prime}(1) is euqal to
    (A)
    150\displaystyle 150
    (B)
    -50\displaystyle 50
    (C)
    -150\displaystyle 150
    (D)
    50\displaystyle 50

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    NCERT’s answer
    D
    (D) \(\displaystyle 50\). Differentiate term by term, then pair the terms.\[f'(x) = -1 + 2x - 3x^2 + \ldots - 99x^{98} + 100x^{99} \]\[f'(1) = (-1 + 2) + (-3 + 4) + \ldots + (-99 + 100) \]\[f'(1) = \underbrace{1 + 1 + \ldots + 1}_{50 \text{ pairs}} = 50 \]
  7. Fill in the blanks in Exercises $\displaystyle 77$ to 80.

    Exercise 77

    If f(x)=tan⁡xx−π\displaystyle f(x)=\frac{\tan x}{x-\pi}, then lim⁡x→πf(x)=\displaystyle \lim _{x \rightarrow \pi} f(x)= ____\displaystyle \_\_\_\_

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    NCERT’s answer
    $\displaystyle 1$
    \(\displaystyle 1\). Put \(\displaystyle h = x - \pi\), so \(\displaystyle h \to 0\) as \(\displaystyle x \to \pi\).\[\tan x = \tan(\pi + h) = \tan h \]\[\lim_{x \to \pi} \frac{\tan x}{x - \pi} = \lim_{h \to 0} \frac{\tan h}{h} = 1 \]Answer: \(\displaystyle 1\)

  8. Exercise 78

    lim⁡x→0(sin⁡mxcot⁡x3)=2\displaystyle \lim _{x \rightarrow 0}\left(\sin m x \cot \frac{x}{\sqrt{3}}\right)=2, then m=\displaystyle m= ____\displaystyle \_\_\_\_

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    NCERT’s answer
    \(\displaystyle m=\frac{2 \sqrt{3}}{3}\)
    \(\displaystyle \dfrac{2}{\sqrt{3}}\). Write \(\displaystyle \cot\) as \(\displaystyle \cos/\sin\) and reduce to \(\displaystyle \dfrac{\sin\theta}{\theta}\) forms.\[\lim_{x \to 0} \frac{\sin mx}{\sin \dfrac{x}{\sqrt3}}\cos\frac{x}{\sqrt3} = \lim_{x \to 0} \frac{\sin mx}{mx} \cdot \frac{\dfrac{x}{\sqrt3}}{\sin\dfrac{x}{\sqrt3}} \cdot \frac{mx}{x/\sqrt3} \cdot \cos\frac{x}{\sqrt3} \]\[= 1 \cdot 1 \cdot m\sqrt{3} \cdot 1 = m\sqrt{3} \]\[m\sqrt{3} = 2 \Rightarrow m = \frac{2}{\sqrt{3}} \]Answer: \(\displaystyle m = \dfrac{2}{\sqrt{3}}\)

  9. Exercise 79

    if y=1+x1!+x22!+x33!+…\displaystyle y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots, then dydx=\displaystyle \frac{d y}{d x}= ____\displaystyle \_\_\_\_

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    NCERT’s answer
    \(\displaystyle y\)
    \(\displaystyle y\) (that is, \(\displaystyle e^x\)). Differentiate term by term.\[\frac{dy}{dx} = 0 + 1 + \frac{2x}{2!} + \frac{3x^2}{3!} + \frac{4x^3}{4!} + \ldots \]\[\frac{dy}{dx} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots = y \]Answer: \(\displaystyle y = e^x\)

  10. Exercise 80

    lim⁡x→3+x[x]=\displaystyle \lim _{x \rightarrow 3^{+}} \frac{x}{[x]}= ____\displaystyle \_\_\_\_

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    NCERT’s answer
    $\displaystyle 1$
    \(\displaystyle 1\). For \(\displaystyle x \to 3^{+}\), \(\displaystyle 3 < x < 4\), so \(\displaystyle [x] = 3\).\[\lim_{x \to 3^{+}} \frac{x}{[x]} = \lim_{x \to 3^{+}} \frac{x}{3} = \frac{3}{3} = 1 \]Answer: \(\displaystyle 1\)