SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 12 Mathematics Continuity and Differentiability

106 questions · 106 still being checked

EXERCISE 5.3 11–20 (part 2 of 11)

  1. Find the value of \(\displaystyle k\) in each of the Exercises $\displaystyle 11$ to $\displaystyle 14$ so that the function \(\displaystyle f\) is continuous at the indicated point:

    Exercise 11

    f(x)={3x−8, if x≤52k, if x>5\displaystyle f(x)=\left\{\begin{array}{ll}3 x-8, & \text { if } x \leq 5 \\ 2 k, & \text { if } x>5\end{array}\right. at x=5\displaystyle x=5

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle k=\frac{7}{2}\)
    \[f(5)=3(5)-8=7 \] \[\lim_{x\to5^-}f(x)=3(5)-8=7 \] \[\lim_{x\to5^+}f(x)=2k \] Continuity at \(\displaystyle x=5\) needs the limits to equal \(\displaystyle f(5)\). \[2k=7 \] Answer: \(\displaystyle k=\dfrac72\)

  2. Exercise 12

    f(x)={2x+2−164x−16, if x≠2k, if x=2\displaystyle f(x)=\left\{\begin{array}{ll}\frac{2^{x+2}-16}{4^x-16}, & \text { if } x \neq 2 \\ k, & \text { if } x=2\end{array}\right. at x=2\displaystyle x=2

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle k=\frac{1}{2}\)
    Put \(\displaystyle t=2^x\); then \(\displaystyle 2^{x+2}=4t\) and \(\displaystyle 4^x=t^2\). \[\frac{2^{x+2}-16}{4^x-16}=\frac{4t-16}{t^2-16}=\frac{4(t-4)}{(t-4)(t+4)}=\frac{4}{t+4} \quad (t\ne 4) \] \[x\to2 \;\Rightarrow\; t\to4 \] \[\lim_{x\to2}f(x)=\frac{4}{4+4}=\frac12 \] Continuity needs \(\displaystyle k=\lim_{x\to2}f(x)\). Answer: \(\displaystyle k=\dfrac12\)

  3. Exercise 13

    f(x)={1+kx−1−kxx, if −1≤x<02x+1x−1, if 0≤x≤1\displaystyle f(x)=\left\{\begin{array}{ll}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} & , \text { if }-1 \leq x<0 \\ \frac{2 x+1}{x-1} & , \text { if } 0 \leq x \leq 1\end{array}\right. at x=0\displaystyle x=0

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle k=-1\)
    Left limit, rationalising the numerator: \[\lim_{x\to0^-}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\cdot\frac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}} \] \[=\lim_{x\to0^-}\frac{(1+kx)-(1-kx)}{x\left(\sqrt{1+kx}+\sqrt{1-kx}\right)} \] \[=\lim_{x\to0^-}\frac{2k}{\sqrt{1+kx}+\sqrt{1-kx}}=\frac{2k}{2}=k \] Value and right limit: \[f(0)=\lim_{x\to0^+}f(x)=\frac{2(0)+1}{0-1}=-1 \] Continuity needs the left limit to equal \(\displaystyle f(0)\). \[k=-1 \] Answer: \(\displaystyle k=-1\)

  4. Exercise 14

    f(x)={1−cos⁡kxxsin⁡x, if x≠012, if x=0\displaystyle f(x)=\left\{\begin{array}{ll}\frac{1-\cos k x}{x \sin x}, & \text { if } x \neq 0 \\ \frac{1}{2} & , \text { if } x=0\end{array}\right. at x=0\displaystyle x=0

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle k=\pm 1\)
    \[1-\cos kx=2\sin^2\frac{kx}{2} \] \[\frac{1-\cos kx}{x\sin x}=\frac{2\sin^2\dfrac{kx}{2}}{x\sin x}=\frac{k^2}{2}\left(\frac{\sin\dfrac{kx}{2}}{\dfrac{kx}{2}}\right)^2\cdot\frac{x}{\sin x} \] \[\lim_{x\to0}f(x)=\frac{k^2}{2}\,(1)^2\,(1)=\frac{k^2}{2} \] Continuity needs this to equal \(\displaystyle f(0)\). \[\frac{k^2}{2}=\frac12 \;\Rightarrow\; k^2=1 \] Answer: \(\displaystyle k=\pm 1\)

  5. Exercise 15

    Prove that the function f\displaystyle f defined by f(x)={x∣x∣+2x2,x≠0k,x=0f(x)= \begin{cases}\frac{x}{|x|+2 x^2}, & x \neq 0 \\ k, & x=0\end{cases} remains discontinuous at x=0\displaystyle x=0, regardless the choice of k\displaystyle k.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Right of \(\displaystyle 0\): \(\displaystyle |x|=x\). \[\lim_{x\to0^+}\frac{x}{x+2x^2}=\lim_{x\to0^+}\frac{1}{1+2x}=1 \] Left of \(\displaystyle 0\): \(\displaystyle |x|=-x\). \[\lim_{x\to0^-}\frac{x}{-x+2x^2}=\lim_{x\to0^-}\frac{1}{-1+2x}=-1 \] \[\lim_{x\to0^-}f(x)=-1\ne 1=\lim_{x\to0^+}f(x) \] So \(\displaystyle \lim_{x\to0}f(x)\) does not exist, and no value \(\displaystyle f(0)=k\) can equal it. Answer: \(\displaystyle f\) is discontinuous at \(\displaystyle x=0\) for every \(\displaystyle k\).

  6. Exercise 16

    Find the values of a\displaystyle a and b\displaystyle b such that the function f\displaystyle f defined by f(x)={x−4∣x−4∣+a, if x<4a+b, if x=4x−4∣x−4∣+b, if x>4f(x)= \begin{cases}\frac{x-4}{|x-4|}+a & , \text { if } x<4 \\ a+b & , \text { if } x=4 \\ \frac{x-4}{|x-4|}+b & , \text { if } x>4\end{cases} is a continuous function at x=4\displaystyle x=4.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle a=1, b=-1\)
    \[x<4:\quad |x-4|=4-x \;\Rightarrow\; \frac{x-4}{|x-4|}=-1 \] \[x>4:\quad |x-4|=x-4 \;\Rightarrow\; \frac{x-4}{|x-4|}=1 \] \[\lim_{x\to4^-}f(x)=a-1 \] \[\lim_{x\to4^+}f(x)=b+1 \] \[f(4)=a+b \] Continuity at \(\displaystyle x=4\): \[a-1=a+b=b+1 \] \[a-1=a+b \;\Rightarrow\; b=-1 \] \[a+b=b+1 \;\Rightarrow\; a=1 \] Answer: \(\displaystyle a=1,\; b=-1\)

  7. Exercise 17

    Given the function f(x)=1x+2\displaystyle f(x)=\frac{1}{x+2}. Find the points of discontinuity of the composite function y=f(f(x))\displaystyle y=f(f(x)).

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Discontinuous at \(\displaystyle x=-2\) and \(\displaystyle x=\frac{-5}{2}\)
    The inner \(\displaystyle f\) needs \(\displaystyle x \neq -2\); the outer \(\displaystyle f\) needs \(\displaystyle f(x) \neq -2\). \[f(f(x)) = \frac{1}{f(x)+2} = \frac{1}{\dfrac{1}{x+2}+2} = \frac{x+2}{2x+5} \] \[x + 2 = 0 \;\Rightarrow\; x = -2 \] \[f(x) + 2 = 0 \;\Rightarrow\; \frac{1}{x+2} = -2 \;\Rightarrow\; x = -\frac{5}{2} \] Elsewhere the composite is a quotient of continuous functions with non-zero denominator.Answer: \(\displaystyle f(f(x))\) is discontinuous at \(\displaystyle x = -2\) and \(\displaystyle x = -\dfrac{5}{2}\).

  8. Exercise 18

    Find all points of discontinuity of the function f(t)=1t2+t−2\displaystyle f(t)=\frac{1}{t^2+t-2}, where t=1x−1\displaystyle t=\frac{1}{x-1}.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Discontinuous at \(\displaystyle x=1, \frac{1}{2}\) and $\displaystyle 2$
    \[t^2 + t - 2 = (t+2)(t-1) \] The composite \(\displaystyle f(t(x))\) exists only if \(\displaystyle t = \frac{1}{x-1}\) exists and the denominator is non-zero. \[x - 1 = 0 \;\Rightarrow\; x = 1 \] \[t = 1 \;\Rightarrow\; \frac{1}{x-1} = 1 \;\Rightarrow\; x = 2 \] \[t = -2 \;\Rightarrow\; \frac{1}{x-1} = -2 \;\Rightarrow\; x = \frac{1}{2} \] \[f(t(x)) = -\frac{(x-1)^2}{(2x-1)(x-2)}, \quad x \neq 1 \] Elsewhere it is a rational function, hence continuous.Answer: discontinuous at \(\displaystyle x = 1,\ \dfrac{1}{2},\ 2\).

  9. Exercise 19

    Show that the function f(x)=∣sin⁡x+cos⁡x∣\displaystyle f(x)=|\sin x+\cos x| is continuous at x=π\displaystyle x=\pi.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[f(\pi) = |\sin\pi + \cos\pi| = |0 - 1| = 1 \] \[\lim_{x\to\pi}(\sin x + \cos x) = 0 - 1 = -1 \] Since \(\displaystyle |\cdot|\) is continuous, the limit passes inside it: \[\lim_{x\to\pi} f(x) = \left|\lim_{x\to\pi}(\sin x + \cos x)\right| = |-1| = 1 \] \[\lim_{x\to\pi} f(x) = f(\pi) \]Answer: \(\displaystyle f\) is continuous at \(\displaystyle x = \pi\).
  10. Examine the differentiability of \(\displaystyle f\), where \(\displaystyle f\) is defined by

    Exercise 20

    f(x)={x[x],, if 0≤x<2(x−1)x, if 2≤x<3\displaystyle f(x)= \begin{cases}x[x], & , \text { if } 0 \leq x<2 \\ (x-1) x, & \text { if } 2 \leq x<3\end{cases} at x=2\displaystyle x=2.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Not differentiable at \(\displaystyle x=2\)
    For \(\displaystyle 1 \le x < 2\), \(\displaystyle [x] = 1\), so \(\displaystyle f(x) = x\) just left of \(\displaystyle 2\). \[f(2) = (2-1)(2) = 2 \] \[Lf'(2) = \lim_{h\to0^-}\frac{f(2+h) - f(2)}{h} = \lim_{h\to0^-}\frac{(2+h) - 2}{h} = 1 \] \[Rf'(2) = \lim_{h\to0^+}\frac{(1+h)(2+h) - 2}{h} = \lim_{h\to0^+}\frac{3h + h^2}{h} = \lim_{h\to0^+}(3 + h) = 3 \] \[Lf'(2) = 1 \neq 3 = Rf'(2) \]Answer: \(\displaystyle f\) is not differentiable at \(\displaystyle x = 2\).