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NCERT Exemplar · Class 11 Mathematics Limits and Derivatives

80 questions · 80 still being checked

EXERCISE 13.3 61–70 (part 7 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 61

    lim⁡x→π4sec⁡2x−2tan⁡x−1\displaystyle \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sec ^2 x-2}{\tan x-1} is
    (A)
    3\displaystyle 3 (B) 1\displaystyle 1 (C) 0\displaystyle 0 (D) 2\displaystyle \sqrt{2}

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

    None of the options: the limit is \(\displaystyle 2\), and the printed options ($\displaystyle 3$, $\displaystyle 1$, $\displaystyle 0$, \(\displaystyle \sqrt{2}\)) do not include 2.\[\sec^2x-2=(1+\tan^2x)-2=\tan^2x-1=(\tan x-1)(\tan x+1) \] \[\frac{\sec^2x-2}{\tan x-1}=\tan x+1 \quad \left(x\ne\tfrac{\pi}{4}\right) \] \[\lim_{x\to\frac{\pi}{4}}\frac{\sec^2x-2}{\tan x-1}=\tan\tfrac{\pi}{4}+1=1+1=2 \]NCERT prints: (D) \(\displaystyle \sqrt{2}\) -- the quotient equals \(\displaystyle \tan x+1\) near \(\displaystyle \tfrac{\pi}{4}\), which tends to $\displaystyle 2$, not \(\displaystyle \sqrt{2}\).

  2. Exercise 62

    lim⁡x→1(x−1)(2x−3)2x2+x−3\displaystyle \lim _{x \rightarrow 1} \frac{(\sqrt{x}-1)(2 x-3)}{2 x^2+x-3} is
    (A)
    110\displaystyle \frac{1}{10}
    (B)
    −110\displaystyle \frac{-1}{10}
    (C)
    1\displaystyle 1 (D) None of these

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    NCERT’s answer
    B
    (B) \(\displaystyle -\dfrac{1}{10}\) Factor and rationalise. \[2x^2+x-3=(2x+3)(x-1),\qquad \sqrt x-1=\frac{x-1}{\sqrt x+1} \] \[\frac{(\sqrt x-1)(2x-3)}{2x^2+x-3}=\frac{2x-3}{(\sqrt x+1)(2x+3)} \] \[\to\frac{2-3}{(1+1)(2+3)}=-\frac{1}{10} \]

  3. Exercise 63

    If f(x)={sin⁡[x][x],[x]≠00,[x]=0\displaystyle f(x)=\left\{\begin{array}{l}\frac{\sin [x]}{[x]},[x] \neq 0 \\ 0 \quad,[x]=0\end{array}\right., where [.] denotes the greatest integer function, then lim⁡x→0f(x)\displaystyle \lim _{x \rightarrow 0} f(x) is equal to
    (A)
    1\displaystyle 1 (B) 0\displaystyle 0 (C) -1\displaystyle 1 (D) None of these

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    NCERT’s answer
    D
    (D) None of these: the limit does not exist. For \(\displaystyle -1<x<0\), \(\displaystyle [x]=-1\); for \(\displaystyle 0<x<1\), \(\displaystyle [x]=0\). \[\lim_{x\to0^-}f(x)=\frac{\sin(-1)}{-1}=\sin 1 \] \[\lim_{x\to0^+}f(x)=0 \] \[\sin1\neq0 \]

  4. Exercise 64

    lim⁡x→0∣sin⁡x∣x\displaystyle \lim _{x \rightarrow 0} \frac{|\sin x|}{x} is
    (A)
    1\displaystyle 1 (B) -1\displaystyle 1 (C) does not exist
    (D)
    None of these

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    NCERT’s answer
    C
    (C) does not exist. \[\lim_{x\to0^+}\frac{|\sin x|}{x}=\lim_{x\to0^+}\frac{\sin x}{x}=1 \] \[\lim_{x\to0^-}\frac{|\sin x|}{x}=\lim_{x\to0^-}\frac{-\sin x}{x}=-1 \] \[1\neq-1 \]

  5. Exercise 65

    Let f(x)={x2−1,0<x<22x+3,2≤x<3\displaystyle f(x)=\left\{\begin{array}{l}x^2-1,0<x<2 \\ 2 x+3,2 \leq x<3\end{array}\right., the quadratic equation whose roots are lim⁡x→2−f(x)\displaystyle \lim _{x \rightarrow 2^{-}} f(x) and lim⁡x→2+f(x)\displaystyle \lim _{x \rightarrow 2^{+}} f(x) is
    (A)
    x2−6x+9=0\displaystyle x^2-6 x+9=0
    (B)
    x2−7x+8=0\displaystyle x^2-7 x+8=0
    (C)
    x2−14x+49=0\displaystyle x^2-14 x+49=0
    (D)
    x2−10x+21=0\displaystyle x^2-10 x+21=0

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    NCERT’s answer
    D
    (D) \(\displaystyle x^2-10x+21=0\)\[\lim_{x\to2^-}f(x)=2^2-1=3 \]\[\lim_{x\to2^+}f(x)=2(2)+3=7 \]\[x^2-(3+7)x+(3)(7)=0 \]\[x^2-10x+21=0 \]

  6. Exercise 66

    lim⁡x→0tan⁡2x−x3x−sin⁡x\displaystyle \lim _{x \rightarrow 0} \frac{\tan 2 x-x}{3 x-\sin x} is
    (A)
    2\displaystyle 2 (B) 12\displaystyle \frac{1}{2}
    (C)
    −12\displaystyle \frac{-1}{2}
    (D)
    14\displaystyle \frac{1}{4}

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    NCERT’s answer
    B
    (B) \(\displaystyle \dfrac12\)Divide numerator and denominator by \(\displaystyle x\).\[\lim_{x\to0}\frac{\tan 2x-x}{3x-\sin x}=\lim_{x\to0}\frac{2\cdot\dfrac{\tan 2x}{2x}-1}{3-\dfrac{\sin x}{x}} \]\[=\frac{2(1)-1}{3-1}=\frac12 \]

  7. Exercise 67

    Let f(x)=x−[x];∈R\displaystyle f(x)=x-[x] ; \in \mathbf{R}, then f′(12)\displaystyle f^{\prime}\left(\frac{1}{2}\right) is
    (A)
    32\displaystyle \frac{3}{2}
    (B)
    1\displaystyle 1 (C) 0\displaystyle 0 (D) -1\displaystyle 1

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    NCERT’s answer
    B
    (B) \(\displaystyle 1\)Near \(\displaystyle x=\tfrac12\) we have \(\displaystyle 0<x<1\), so \(\displaystyle [x]=0\).\[f(x)=x-0=x \]\[f'(x)=1 \;\Rightarrow\; f'\!\left(\tfrac12\right)=1 \]

  8. Exercise 68

    If y=x+1x\displaystyle y=\sqrt{x}+\frac{1}{\sqrt{x}}, then dydx\displaystyle \frac{d y}{d x} at x=1\displaystyle x=1 is
    (A)
    1\displaystyle 1 (B) 12\displaystyle \frac{1}{2}
    (C)
    12\displaystyle \frac{1}{\sqrt{2}}

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    NCERT’s answer
    D
    (D) \(\displaystyle 0\)\[y=x^{1/2}+x^{-1/2} \]\[\frac{dy}{dx}=\frac{1}{2}x^{-1/2}-\frac{1}{2}x^{-3/2} \]\[\left.\frac{dy}{dx}\right|_{x=1}=\frac12-\frac12=0 \]

  9. Exercise 69

    If f(x)=x−42x\displaystyle f(x)=\frac{x-4}{2 \sqrt{x}}, then f′(1)\displaystyle f^{\prime}(1) is
    (A)
    54\displaystyle \frac{5}{4}
    (B)
    45\displaystyle \frac{4}{5}
    (C)
    1\displaystyle 1 (D) 0\displaystyle 0

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac54\)\[f(x)=\frac{x-4}{2\sqrt x}=\frac12x^{1/2}-2x^{-1/2} \]\[f'(x)=\frac14x^{-1/2}+x^{-3/2} \]\[f'(1)=\frac14+1=\frac54 \]

  10. Exercise 70

    If y=1+1x21−1x2\displaystyle y=\frac{1+\dfrac{1}{x^2}}{1-\dfrac{1}{x^2}}, then dydx\displaystyle \frac{d y}{d x} is
    (A)
    −4x(x2−1)2\displaystyle \frac{-4 x}{\left(x^2-1\right)^2}
    (B)
    −4xx2−1\displaystyle \frac{-4 x}{x^2-1}
    (C)
    1−x24x\displaystyle \frac{1-x^2}{4 x}
    (D)
    4xx2−1\displaystyle \frac{4 x}{x^2-1}

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{-4x}{(x^2-1)^2}\)\[y=\frac{1+\frac1{x^2}}{1-\frac1{x^2}}=\frac{x^2+1}{x^2-1} \]\[\frac{dy}{dx}=\frac{2x(x^2-1)-2x(x^2+1)}{(x^2-1)^2} \]\[=\frac{-4x}{(x^2-1)^2} \]