NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

\(\displaystyle f(x)=x^{3}-3\) for \(\displaystyle x\le 2\) and \(\displaystyle f(x)=x^{2}+1\) for \(\displaystyle x>2\). Both pieces are polynomials, hence continuous on \(\displaystyle (-\infty,2)\) and \(\displaystyle (2,\infty)\) respectively; test the joining point \(\displaystyle x=2\).At \(\displaystyle x=2\): \[\lim_{x\to 2^{-}}\left(x^{3}-3\right)=8-3=5,\qquad \lim_{x\to 2^{+}}\left(x^{2}+1\right)=4+1=5,\qquad f(2)=2^{3}-3=5.\]All three are equal, so \(\displaystyle f\) is continuous at \(\displaystyle x=2\).Answer: \(\displaystyle f\) is continuous on \(\displaystyle \mathbf{R}\); it has no point of discontinuity.

\(\displaystyle f(x)=x^{10}-1\) for \(\displaystyle x\le 1\) and \(\displaystyle f(x)=x^{2}\) for \(\displaystyle x>1\). Both pieces are polynomials, hence continuous away from the joining point; test \(\displaystyle x=1\).At \(\displaystyle x=1\): \[\lim_{x\to 1^{-}}\left(x^{10}-1\right)=1-1=0,\qquad \lim_{x\to 1^{+}}x^{2}=1,\qquad f(1)=1^{10}-1=0.\]Since LHL \(\displaystyle =0\neq 1=\) RHL, \(\displaystyle \lim_{x\to 1}f(x)\) does not exist.Answer: \(\displaystyle x=1\) is the only point of discontinuity of \(\displaystyle f\).

\(\displaystyle f(x)=x+5\) for \(\displaystyle x\le 1\) and \(\displaystyle f(x)=x-5\) for \(\displaystyle x>1\). Each piece is a polynomial, so \(\displaystyle f\) is continuous at every point of \(\displaystyle (-\infty,1)\) and of \(\displaystyle (1,\infty)\). The only doubtful point is \(\displaystyle x=1\).At \(\displaystyle x=1\): \[\lim_{x\to 1^{-}}(x+5)=6,\qquad \lim_{x\to 1^{+}}(x-5)=-4,\qquad f(1)=1+5=6.\]Since LHL \(\displaystyle =6\neq -4=\) RHL, the limit at \(\displaystyle x=1\) does not exist, so \(\displaystyle f\) is discontinuous at \(\displaystyle x=1\).Answer: \(\displaystyle f\) is not a continuous function; it is continuous at every real point except \(\displaystyle x=1\).

The function is defined on the domain \(\displaystyle [0,10]\), and is constant on each of \(\displaystyle [0,1]\), \(\displaystyle (1,3)\) and \(\displaystyle [3,10]\). A constant function is continuous, so \(\displaystyle f\) is continuous at every point of \(\displaystyle (0,1)\), \(\displaystyle (1,3)\) and \(\displaystyle (3,10)\), and one-sidedly continuous at the end points \(\displaystyle x=0\) and \(\displaystyle x=10\). Only \(\displaystyle x=1\) and \(\displaystyle x=3\) need testing.At \(\displaystyle x=1\): \[\lim_{x\to 1^{-}}f(x)=3,\qquad \lim_{x\to 1^{+}}f(x)=4,\qquad f(1)=3.\] LHL \(\displaystyle \neq\) RHL, so \(\displaystyle f\) is discontinuous at \(\displaystyle x=1\).At \(\displaystyle x=3\): \[\lim_{x\to 3^{-}}f(x)=4,\qquad \lim_{x\to 3^{+}}f(x)=5,\qquad f(3)=5.\] LHL \(\displaystyle \neq\) RHL, so \(\displaystyle f\) is discontinuous at \(\displaystyle x=3\).Answer: \(\displaystyle f\) is continuous on \(\displaystyle [0,10]\) except at \(\displaystyle x=1\) and \(\displaystyle x=3\), which are its only points of discontinuity.

\(\displaystyle f(x)=2x\) for \(\displaystyle x<0\), \(\displaystyle f(x)=0\) for \(\displaystyle 0\le x\le 1\), and \(\displaystyle f(x)=4x\) for \(\displaystyle x>1\). Each piece is a polynomial, so continuity can fail only at the joining points \(\displaystyle x=0\) and \(\displaystyle x=1\).At \(\displaystyle x=0\): \[\lim_{x\to 0^{-}}2x=0,\qquad \lim_{x\to 0^{+}}0=0,\qquad f(0)=0.\] All three agree, so \(\displaystyle f\) is continuous at \(\displaystyle x=0\).At \(\displaystyle x=1\): \[\lim_{x\to 1^{-}}f(x)=0,\qquad \lim_{x\to 1^{+}}4x=4,\qquad f(1)=0.\] Since LHL \(\displaystyle =0\neq 4=\) RHL, \(\displaystyle f\) is discontinuous at \(\displaystyle x=1\).Answer: \(\displaystyle f\) is continuous everywhere except at \(\displaystyle x=1\), which is its only point of discontinuity.

\(\displaystyle f(x)=-2\) for \(\displaystyle x\le -1\), \(\displaystyle f(x)=2x\) for \(\displaystyle -1<x\le 1\), and \(\displaystyle f(x)=2\) for \(\displaystyle x>1\). Each piece is a constant or a polynomial, so only the joining points \(\displaystyle x=-1\) and \(\displaystyle x=1\) need testing.At \(\displaystyle x=-1\): \[\lim_{x\to -1^{-}}f(x)=-2,\qquad \lim_{x\to -1^{+}}2x=2(-1)=-2,\qquad f(-1)=-2.\] All three agree, so \(\displaystyle f\) is continuous at \(\displaystyle x=-1\).At \(\displaystyle x=1\): \[\lim_{x\to 1^{-}}2x=2,\qquad \lim_{x\to 1^{+}}f(x)=2,\qquad f(1)=2(1)=2.\] All three agree, so \(\displaystyle f\) is continuous at \(\displaystyle x=1\).Answer: \(\displaystyle f\) is continuous on \(\displaystyle \mathbf{R}\); it has no point of discontinuity.

Continuity at \(\displaystyle x=3\) requires LHL \(\displaystyle =\) RHL \(\displaystyle =f(3)\).Value: \(\displaystyle f(3)=a(3)+1=3a+1\) (the point \(\displaystyle x=3\) belongs to the first branch).Left-hand limit: \(\displaystyle \displaystyle\lim_{x\to 3^{-}}(ax+1)=3a+1\).Right-hand limit: \(\displaystyle \displaystyle\lim_{x\to 3^{+}}(bx+3)=3b+3\).The left-hand limit already equals \(\displaystyle f(3)\), so the single condition is \[3a+1=3b+3\ \Longrightarrow\ 3a-3b=2\ \Longrightarrow\ a-b=\frac{2}{3}.\]Answer: \(\displaystyle f\) is continuous at \(\displaystyle x=3\) precisely when \(\displaystyle a=b+\dfrac{2}{3}\), i.e. \(\displaystyle a-b=\dfrac{2}{3}\).

At \(\displaystyle x=0\), compare the two one-sided limits with \(\displaystyle f(0)\).Value: \(\displaystyle f(0)=\lambda\left(0^{2}-2\cdot 0\right)=0\).Left-hand limit: \(\displaystyle \displaystyle\lim_{x\to 0^{-}}\lambda\left(x^{2}-2x\right)=\lambda(0-0)=0\).Right-hand limit: \(\displaystyle \displaystyle\lim_{x\to 0^{+}}(4x+1)=1\).The right-hand limit is \(\displaystyle 1\) and does not involve \(\displaystyle \lambda\) at all, while the left-hand limit is \(\displaystyle 0\) for every \(\displaystyle \lambda\). Since \(\displaystyle 0\neq 1\), the limit at \(\displaystyle x=0\) never exists.Answer: there is no value of \(\displaystyle \lambda\) for which \(\displaystyle f\) is continuous at \(\displaystyle x=0\).At \(\displaystyle x=1\): every point near \(\displaystyle 1\) satisfies \(\displaystyle x>0\), so \(\displaystyle f(x)=4x+1\) throughout a neighbourhood of \(\displaystyle 1\). Hence \(\displaystyle \lim_{x\to 1}f(x)=5=f(1)\), and \(\displaystyle f\) is continuous at \(\displaystyle x=1\) for every value of \(\displaystyle \lambda\). (Likewise \(\displaystyle f\) is continuous at every point other than \(\displaystyle x=0\).)

Let \(\displaystyle c\) be any integer. Recall the definition of the greatest integer function: \(\displaystyle [x]\) is the greatest integer \(\displaystyle \le x\). Hence \[c-1\le x<c\ \Rightarrow\ [x]=c-1,\qquad c\le x<c+1\ \Rightarrow\ [x]=c.\]Left-hand limit at \(\displaystyle x=c\): for \(\displaystyle x\) just below \(\displaystyle c\) we have \(\displaystyle [x]=c-1\), so \[\lim_{x\to c^{-}}g(x)=\lim_{x\to c^{-}}\left(x-(c-1)\right)=c-c+1=1.\]Right-hand limit at \(\displaystyle x=c\): for \(\displaystyle x\) just above \(\displaystyle c\) we have \(\displaystyle [x]=c\), so \[\lim_{x\to c^{+}}g(x)=\lim_{x\to c^{+}}(x-c)=0.\]Value: \(\displaystyle g(c)=c-[c]=c-c=0\).Since LHL \(\displaystyle =1\neq 0=\) RHL, \(\displaystyle \lim_{x\to c}g(x)\) does not exist, so \(\displaystyle g\) is discontinuous at \(\displaystyle x=c\).As \(\displaystyle c\) was an arbitrary integer, \(\displaystyle g(x)=x-[x]\) is discontinuous at all integral points.

Use the algebra of continuous functions: if \(\displaystyle g\) and \(\displaystyle h\) are continuous at \(\displaystyle c\), then \(\displaystyle g\pm h\) is continuous at \(\displaystyle c\).Here \(\displaystyle f(x)=x^{2}-\sin x+5\), where \(\displaystyle x\mapsto x^{2}\) is a polynomial (continuous on \(\displaystyle \mathbf{R}\)), \(\displaystyle x\mapsto \sin x\) is continuous on \(\displaystyle \mathbf{R}\), and \(\displaystyle x\mapsto 5\) is constant. Hence \(\displaystyle f\) is continuous at every real point, in particular at \(\displaystyle x=\pi\).Checking directly at \(\displaystyle x=\pi\), put \(\displaystyle x=\pi+h\) so that \(\displaystyle h\to 0\): \[\lim_{x\to\pi}f(x)=\lim_{h\to 0}\left[(\pi+h)^{2}-\sin(\pi+h)+5\right]=\lim_{h\to 0}\left[(\pi+h)^{2}+\sin h+5\right]=\pi^{2}+0+5.\] And \(\displaystyle f(\pi)=\pi^{2}-\sin\pi+5=\pi^{2}+5\).Since \(\displaystyle \lim_{x\to\pi}f(x)=\pi^{2}+5=f(\pi)\), the answer is yes: \(\displaystyle f\) is continuous at \(\displaystyle x=\pi\).