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NCERT Exemplar · Class 11 Mathematics Limits and Derivatives

80 questions · 80 still being checked

EXERCISE 13.3 21–30 (part 3 of 8)

  1. Evaluate :

    Exercise 21

    lim⁡x→π4sin⁡x−cos⁡xx−π4\displaystyle \lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\dfrac{\pi}{4}}

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    NCERT’s answer
    \(\displaystyle \sqrt{2}\)
    \[\sin x-\cos x = \sqrt2\left(\sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4}\right) = \sqrt2\,\sin\!\left(x-\frac{\pi}{4}\right) \] Put \(\displaystyle h=x-\frac{\pi}{4}\), so \(\displaystyle h\to0\): \[\lim_{x\to\frac{\pi}{4}}\frac{\sin x-\cos x}{x-\dfrac{\pi}{4}} = \sqrt2\lim_{h\to0}\frac{\sin h}{h} = \sqrt2 \] Answer: \(\displaystyle \sqrt2\)

  2. Exercise 22

    lim⁡x→π63sin⁡x−cos⁡xx−π6\displaystyle \lim _{x \rightarrow \frac{\pi}{6}} \frac{\sqrt{3} \sin x-\cos x}{x-\dfrac{\pi}{6}}

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    NCERT’s answer
    $\displaystyle 2$
    \[\sqrt3\sin x-\cos x = 2\left(\sin x\cos\frac{\pi}{6}-\cos x\sin\frac{\pi}{6}\right) = 2\sin\!\left(x-\frac{\pi}{6}\right) \] Put \(\displaystyle h=x-\frac{\pi}{6}\), so \(\displaystyle h\to0\): \[\lim_{x\to\frac{\pi}{6}}\frac{\sqrt3\sin x-\cos x}{x-\dfrac{\pi}{6}} = 2\lim_{h\to0}\frac{\sin h}{h} = 2 \] Answer: \(\displaystyle 2\)

  3. Exercise 23

    lim⁡x→0sin⁡2x+3x2x+tan⁡3x\displaystyle \lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\tan 3 x}

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    NCERT’s answer
    $\displaystyle 1$
    Divide numerator and denominator by \(\displaystyle x\): \[\frac{\sin 2x+3x}{2x+\tan 3x} = \frac{2\cdot\dfrac{\sin 2x}{2x}+3}{2+3\cdot\dfrac{\tan 3x}{3x}} \] \[\lim_{x\to0}\frac{\sin 2x}{2x}=1, \qquad \lim_{x\to0}\frac{\tan 3x}{3x}=1 \] \[\lim_{x\to0}\frac{\sin 2x+3x}{2x+\tan 3x} = \frac{2+3}{2+3} = 1 \] Answer: \(\displaystyle 1\)

  4. Exercise 24

    lim⁡x→asin⁡x−sin⁡ax−a\displaystyle \lim _{x \rightarrow a} \frac{\sin x-\sin a}{\sqrt{x}-\sqrt{a}}

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    NCERT’s answer
    \(\displaystyle 2\sqrt{a}\cos a\)
    For \(\displaystyle a>0\): \[\sin x-\sin a = 2\cos\frac{x+a}{2}\,\sin\frac{x-a}{2} \] \[x-a = \left(\sqrt x-\sqrt a\right)\left(\sqrt x+\sqrt a\right) \] \[\frac{\sin x-\sin a}{\sqrt x-\sqrt a} = \cos\frac{x+a}{2}\cdot\frac{\sin\dfrac{x-a}{2}}{\dfrac{x-a}{2}}\cdot\left(\sqrt x+\sqrt a\right) \] \[\lim_{x\to a}\frac{\sin x-\sin a}{\sqrt x-\sqrt a} = \cos a\cdot 1\cdot 2\sqrt a \] Answer: \(\displaystyle 2\sqrt a\,\cos a\)

  5. Exercise 25

    lim⁡x→π6cot⁡2x−3cosec⁡x−2\displaystyle \lim _{x \rightarrow \frac{\pi}{6}} \frac{\cot ^2 x-3}{\operatorname{cosec} x-2}

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    NCERT’s answer
    $\displaystyle 4$
    Use \(\displaystyle \cot^2 x=\operatorname{cosec}^2 x-1\).\[\frac{\cot^2 x-3}{\operatorname{cosec} x-2}=\frac{\operatorname{cosec}^2 x-4}{\operatorname{cosec} x-2} \]\[=\frac{(\operatorname{cosec} x-2)(\operatorname{cosec} x+2)}{\operatorname{cosec} x-2}=\operatorname{cosec} x+2 \quad (x\neq \tfrac{\pi}{6}) \]\[\lim_{x\to \pi/6}(\operatorname{cosec} x+2)=\operatorname{cosec}\frac{\pi}{6}+2=2+2 \]Answer: \(\displaystyle 4\)

  6. Exercise 26

    lim⁡x→02−1+cos⁡xsin⁡2x\displaystyle \lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^2 x}

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    NCERT’s answer
    \(\displaystyle \frac{1}{4\sqrt{2}}\)
    Rationalise the numerator.\[\frac{\sqrt2-\sqrt{1+\cos x}}{\sin^2 x}=\frac{2-(1+\cos x)}{\sin^2 x\left(\sqrt2+\sqrt{1+\cos x}\right)} \]\[=\frac{1-\cos x}{(1-\cos x)(1+\cos x)\left(\sqrt2+\sqrt{1+\cos x}\right)} \quad \left(\sin^2 x=1-\cos^2 x\right) \]\[=\frac{1}{(1+\cos x)\left(\sqrt2+\sqrt{1+\cos x}\right)} \quad (x\neq0) \]\[\lim_{x\to0}\frac{\sqrt2-\sqrt{1+\cos x}}{\sin^2 x}=\frac{1}{(1+1)\left(\sqrt2+\sqrt2\right)}=\frac{1}{4\sqrt2} \]Answer: \(\displaystyle \dfrac{1}{4\sqrt2}=\dfrac{\sqrt2}{8}\)

  7. Exercise 27

    lim⁡x→0sin⁡x−2sin⁡3x+sin⁡5xx\displaystyle \lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x}

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    NCERT’s answer
    $\displaystyle 0$
    Split into three limits, each in the form \(\displaystyle \lim_{x\to0}\dfrac{\sin ax}{ax}=1\).\[\frac{\sin x-2\sin 3x+\sin 5x}{x}=\frac{\sin x}{x}-6\cdot\frac{\sin 3x}{3x}+5\cdot\frac{\sin 5x}{5x} \]\[\lim_{x\to0}\frac{\sin x-2\sin 3x+\sin 5x}{x}=1-6\cdot1+5\cdot1=0 \]Answer: \(\displaystyle 0\)

  8. Exercise 28

    If lim⁡x→1x4−1x−1=lim⁡x→kx3−k3x2−k2\displaystyle \lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}, then find the value of k\displaystyle k.

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

    Use \(\displaystyle \lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}\).\[\lim_{x\to1}\frac{x^4-1}{x-1}=4\cdot1^3=4 \]For \(\displaystyle k\neq0\):\[\lim_{x\to k}\frac{x^3-k^3}{x^2-k^2}=\frac{\lim\dfrac{x^3-k^3}{x-k}}{\lim\dfrac{x^2-k^2}{x-k}}=\frac{3k^2}{2k}=\frac{3k}{2} \]\[\frac{3k}{2}=4 \Rightarrow k=\frac{8}{3} \]\(\displaystyle k=0\) gives \(\displaystyle \lim_{x\to0}x=0\neq4\), so it is rejected.Answer: \(\displaystyle k=\dfrac{8}{3}\)
  9. Differentiate each of the functions w. r. to \(\displaystyle x\) in Exercises $\displaystyle 29$ to 42.

    Exercise 29

    x4+x3+x2+1x\displaystyle \frac{x^4+x^3+x^2+1}{x}

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    NCERT’s answer
    \(\displaystyle 3 x^2+2 x+1-\frac{1}{x^2}\)
    Divide term by term.\[y=\frac{x^4+x^3+x^2+1}{x}=x^3+x^2+x+x^{-1} \]\[\frac{dy}{dx}=3x^2+2x+1-x^{-2} \]Answer: \(\displaystyle 3x^2+2x+1-\dfrac{1}{x^2}\)

  10. Exercise 30

    (x+1x)3\displaystyle \left(x+\frac{1}{x}\right)^3

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    NCERT’s answer
    \(\displaystyle 3 x^2-\frac{3}{x^2}-\frac{3}{x^4}+3\)
    Expand \(\displaystyle (a+b)^3\) with \(\displaystyle a=x,\ b=\frac1x\).\[y=\left(x+\frac1x\right)^3=x^3+3x+\frac3x+\frac1{x^3} \]\[y=x^3+3x+3x^{-1}+x^{-3} \]\[\frac{dy}{dx}=3x^2+3-3x^{-2}-3x^{-4} \]Answer: \(\displaystyle 3x^2+3-\dfrac{3}{x^2}-\dfrac{3}{x^4}\)