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NCERT Exemplar · Class 12 Mathematics Continuity and Differentiability

106 questions · 106 still being checked

EXERCISE 5.3 1–10 (part 1 of 11)

  1. Exercise 1

    Examine the continuity of the function f(x)=x3+2x2−1 at x=1f(x)=x^3+2 x^2-1 \text { at } x=1

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    NCERT’s answer
    Continuous at \(\displaystyle x=1\)
    Compare the limit with the value at \(\displaystyle x=1\).\[f(1)=1^3+2(1)^2-1=2 \]\[\lim_{x\to1}\left(x^3+2x^2-1\right)=1+2-1=2 \]\[\lim_{x\to1}f(x)=f(1) \]Answer: \(\displaystyle f\) is continuous at \(\displaystyle x=1\).
  2. Find which of the functions in Exercises $\displaystyle 2$ to $\displaystyle 10$ is continuous or discontinuous at the indicated points:

    Exercise 2

    f(x)={3x+5, if x≥2x2, if x<2\displaystyle f(x)=\left\{\begin{array}{l}3 x+5, \text { if } x \geq 2 \\ x^2, \text { if } x<2\end{array}\right. at x=2\displaystyle x=2

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    NCERT’s answer
    Discontinuous
    Check the left and right limits at \(\displaystyle x=2\).\[f(2)=3(2)+5=11 \]\[\lim_{x\to2^+}f(x)=\lim_{x\to2^+}(3x+5)=11 \]\[\lim_{x\to2^-}f(x)=\lim_{x\to2^-}x^2=4 \]\[4\neq 11 \]Answer: The two limits differ, so \(\displaystyle f\) is discontinuous at \(\displaystyle x=2\).

  3. Exercise 3

    f(x)={1−cos⁡2xx2, if x≠05, if x=0\displaystyle f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2}, & \text { if } x \neq 0 \\ 5, & \text { if } x=0\end{cases} at x=0\displaystyle x=0

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    NCERT’s answer
    Discontinuous
    Use the double-angle identity.\[1-\cos 2x=2\sin^2 x \]\[\lim_{x\to0}\frac{1-\cos 2x}{x^2}=2\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2=2(1)^2=2 \]\[f(0)=5\neq 2 \]Answer: \(\displaystyle f\) is discontinuous at \(\displaystyle x=0\).

  4. Exercise 4

    f(x)={2x2−3x−2x−2, if x≠25, if x=2\displaystyle f(x)= \begin{cases}\frac{2 x^2-3 x-2}{x-2}, & \text { if } x \neq 2 \\ 5, & \text { if } x=2\end{cases} at x=2\displaystyle x=2

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    NCERT’s answer
    Continuous
    Factorise the numerator.\[2x^2-3x-2=(2x+1)(x-2) \]\[\lim_{x\to2}\frac{(2x+1)(x-2)}{x-2}=\lim_{x\to2}(2x+1)=5 \quad (x\neq 2) \]\[f(2)=5 \]\[\lim_{x\to2}f(x)=f(2) \]Answer: \(\displaystyle f\) is continuous at \(\displaystyle x=2\).

  5. Exercise 5

    f(x)={∣x−4∣2(x−4), if x≠40, if x=4\displaystyle f(x)= \begin{cases}\frac{|x-4|}{2(x-4)}, & \text { if } x \neq 4 \\ 0, & \text { if } x=4\end{cases} at x=4\displaystyle x=4

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    NCERT’s answer
    Discontinuous
    Remove the modulus on each side of \(\displaystyle x=4\).\[x>4:\quad |x-4|=x-4 \]\[x<4:\quad |x-4|=-(x-4) \]\[\lim_{x\to4^+}\frac{x-4}{2(x-4)}=\frac12 \]\[\lim_{x\to4^-}\frac{-(x-4)}{2(x-4)}=-\frac12 \]\[\frac12\neq-\frac12 \]Answer: \(\displaystyle \lim_{x\to4}f(x)\) does not exist, so \(\displaystyle f\) is discontinuous at \(\displaystyle x=4\).

  6. Exercise 6

    f(x)={∣x∣cos⁡1x, if x≠00, if x=0\displaystyle f(x)= \begin{cases}|x| \cos \frac{1}{x}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{cases} at x=0\displaystyle x=0

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    NCERT’s answer
    Continuous
    Squeeze \(\displaystyle f\) between \(\displaystyle 0\) and \(\displaystyle |x|\).\[\left|\cos\frac1x\right|\le 1 \]\[0\le\left||x|\cos\frac1x\right|\le |x| \quad (x\neq 0) \]\[\lim_{x\to0}|x|=0 \]\[\lim_{x\to0}|x|\cos\frac1x=0=f(0) \]Answer: \(\displaystyle f\) is continuous at \(\displaystyle x=0\).

  7. Exercise 7

    f(x)={∣x−a∣sin⁡1x−a, if x≠00, if x=a\displaystyle f(x)=\left\{\begin{array}{l}|x-a| \sin \frac{1}{x-a}, \text { if } x \neq 0 \\ 0, \quad \text { if } x=a\end{array}\right. at x=a\displaystyle x=a

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    NCERT’s answer
    Continuous
    Squeeze \(\displaystyle f\) near \(\displaystyle x=a\), where \(\displaystyle x\neq a\).\[\left|\sin\frac{1}{x-a}\right|\le 1 \]\[-|x-a|\le |x-a|\sin\frac{1}{x-a}\le |x-a| \]\[\lim_{x\to a}|x-a|=0 \]\[\lim_{x\to a}f(x)=0=f(a) \]Answer: \(\displaystyle f\) is continuous at \(\displaystyle x=a\).

  8. Exercise 8

    f(x)={e1x1+e1x, if x≠00, if x=0\displaystyle f(x)=\left\{\begin{array}{l}\frac{e^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, \text { if } x \neq 0 \\ 0, \quad \text { if } x=0\end{array}\right. at x=0\displaystyle x=0

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    NCERT’s answer
    Discontinuous
    Write \(\displaystyle f\) so the exponential tends to \(\displaystyle 0\) on each side.\[\lim_{x\to0^+}\frac{e^{1/x}}{1+e^{1/x}}=\lim_{x\to0^+}\frac{1}{e^{-1/x}+1}=\frac{1}{0+1}=1 \]\[\lim_{x\to0^-}\frac{e^{1/x}}{1+e^{1/x}}=\frac{0}{1+0}=0 \]\[1\neq 0 \]Answer: \(\displaystyle \lim_{x\to0}f(x)\) does not exist, so \(\displaystyle f\) is discontinuous at \(\displaystyle x=0\).

  9. Exercise 9

    f(x)={x22, if 0≤x≤12x2−3x+32, if 1<x≤2\displaystyle f(x)=\left\{\begin{array}{l}\frac{x^2}{2}, \text { if } 0 \leq x \leq 1 \\ 2 x^2-3 x+\frac{3}{2}, \text { if } 1<x \leq 2\end{array}\right. at x=1\displaystyle x=1

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    NCERT’s answer
    Continuous
    \[f(1)=\frac{1^2}{2}=\frac12 \] \[\lim_{x\to1^-}f(x)=\lim_{x\to1^-}\frac{x^2}{2}=\frac12 \] \[\lim_{x\to1^+}f(x)=\lim_{x\to1^+}\left(2x^2-3x+\frac32\right)=2-3+\frac32=\frac12 \] \[\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1) \] Answer: \(\displaystyle f\) is continuous at \(\displaystyle x=1\).

  10. Exercise 10

    f(x)=∣x∣+∣x−1∣\displaystyle f(x)=|x|+|x-1| at x=1\displaystyle x=1

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    NCERT’s answer
    Continuous
    Near \(\displaystyle x=1\) we have \(\displaystyle x>0\), so \(\displaystyle |x|=x\); only \(\displaystyle |x-1|\) changes form. \[x<1:\quad f(x)=x+(1-x)=1 \] \[x\ge 1:\quad f(x)=x+(x-1)=2x-1 \] \[\lim_{x\to1^-}f(x)=1 \] \[\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(2x-1)=1 \] \[f(1)=2(1)-1=1 \] \[\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1) \] Answer: \(\displaystyle f\) is continuous at \(\displaystyle x=1\).