NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

Use the theorem on composition: if \(\displaystyle g\) is continuous at \(\displaystyle c\) and \(\displaystyle h\) is continuous at \(\displaystyle g(c)\), then \(\displaystyle h\circ g\) is continuous at \(\displaystyle c\).Write \(\displaystyle f=h\circ g\) with \[g(x)=x^{2},\qquad h(y)=\cos y,\qquad (h\circ g)(x)=\cos\left(x^{2}\right)=f(x).\]\(\displaystyle g(x)=x^{2}\) is a polynomial, hence continuous at every real \(\displaystyle c\).\(\displaystyle h(y)=\cos y\) is continuous at every real \(\displaystyle y\): with \(\displaystyle y=d+t\), \[\lim_{t\to 0}\cos(d+t)=\lim_{t\to 0}\left(\cos d\cos t-\sin d\sin t\right)=\cos d.\] In particular \(\displaystyle h\) is continuous at \(\displaystyle g(c)=c^{2}\).Therefore \(\displaystyle h\circ g\) is continuous at every real \(\displaystyle c\); that is, \(\displaystyle f(x)=\cos\left(x^{2}\right)\) is a continuous function on \(\displaystyle \mathbf{R}\).

Use the composition theorem again. Write \(\displaystyle f=h\circ g\) with \[g(x)=\cos x,\qquad h(y)=|y|,\qquad (h\circ g)(x)=|\cos x|=f(x).\]\(\displaystyle g(x)=\cos x\) is continuous at every real \(\displaystyle c\).\(\displaystyle h(y)=|y|\) is continuous at every real \(\displaystyle y\): for \(\displaystyle d>0\), \(\displaystyle h(y)=y\) near \(\displaystyle d\); for \(\displaystyle d<0\), \(\displaystyle h(y)=-y\) near \(\displaystyle d\); and at \(\displaystyle d=0\), \[\lim_{y\to 0^{-}}(-y)=0=\lim_{y\to 0^{+}}y=h(0).\] So \(\displaystyle h\) is continuous everywhere, in particular at \(\displaystyle y=g(c)=\cos c\).Since \(\displaystyle g\) is continuous at \(\displaystyle c\) and \(\displaystyle h\) is continuous at \(\displaystyle g(c)\), the composite \(\displaystyle h\circ g\) is continuous at \(\displaystyle c\); \(\displaystyle c\) being arbitrary, \(\displaystyle f(x)=|\cos x|\) is a continuous function on \(\displaystyle \mathbf{R}\).(Note that the corners of the graph, at \(\displaystyle x=(2n+1)\pi/2\) where \(\displaystyle \cos x=0\), are points where \(\displaystyle f\) is continuous but not differentiable.)

Take \(\displaystyle f=h\circ g\) with \[g(x)=|x|,\qquad h(y)=\sin y,\qquad (h\circ g)(x)=\sin|x|.\]\(\displaystyle g(x)=|x|\) is continuous at every real \(\displaystyle c\): for \(\displaystyle c>0\) it agrees with \(\displaystyle x\) near \(\displaystyle c\) and for \(\displaystyle c<0\) with \(\displaystyle -x\) near \(\displaystyle c\); at \(\displaystyle c=0\), \[\lim_{x\to 0^{-}}(-x)=0,\qquad \lim_{x\to 0^{+}}x=0,\qquad g(0)=0,\] so all three agree.\(\displaystyle h(y)=\sin y\) is continuous at every real \(\displaystyle y\), since \(\displaystyle \lim_{t\to 0}\sin(d+t)=\lim_{t\to 0}\left(\sin d\cos t+\cos d\sin t\right)=\sin d\).By the theorem on the composition of continuous functions, \(\displaystyle h\circ g\) is continuous at every real \(\displaystyle c\).Hence \(\displaystyle \sin|x|\) is a continuous function on \(\displaystyle \mathbf{R}\).

Remove both moduli by splitting at the points where the expressions inside change sign, namely \(\displaystyle x=-1\) and \(\displaystyle x=0\).For \(\displaystyle x<-1\): \(\displaystyle |x|=-x\) and \(\displaystyle |x+1|=-(x+1)\), so \(\displaystyle f(x)=-x+(x+1)=1\).For \(\displaystyle -1\le x<0\): \(\displaystyle |x|=-x\) and \(\displaystyle |x+1|=x+1\), so \(\displaystyle f(x)=-x-(x+1)=-2x-1\).For \(\displaystyle x\ge 0\): \(\displaystyle |x|=x\) and \(\displaystyle |x+1|=x+1\), so \(\displaystyle f(x)=x-(x+1)=-1\).Thus \[f(x)=\begin{cases}1,&x<-1\\ -2x-1,&-1\le x<0\\ -1,&x\ge 0.\end{cases}\] Each piece is a polynomial, so only \(\displaystyle x=-1\) and \(\displaystyle x=0\) need testing.At \(\displaystyle x=-1\): \(\displaystyle \displaystyle\lim_{x\to -1^{-}}f(x)=1\), \(\displaystyle \displaystyle\lim_{x\to -1^{+}}(-2x-1)=2-1=1\), and \(\displaystyle f(-1)=-2(-1)-1=1\). All equal, so \(\displaystyle f\) is continuous at \(\displaystyle x=-1\).At \(\displaystyle x=0\): \(\displaystyle \displaystyle\lim_{x\to 0^{-}}(-2x-1)=-1\), \(\displaystyle \displaystyle\lim_{x\to 0^{+}}f(x)=-1\), and \(\displaystyle f(0)=-1\). All equal, so \(\displaystyle f\) is continuous at \(\displaystyle x=0\).(The same conclusion follows at once from the algebra of continuous functions: \(\displaystyle |x|\) and \(\displaystyle |x+1|\) are continuous on \(\displaystyle \mathbf{R}\), so their difference is too.)Answer: \(\displaystyle f\) has no point of discontinuity; it is continuous on the whole of \(\displaystyle \mathbf{R}\).