NCERT Solutions · Class 12 Mathematics Continuity and Differentiability

Take as known that \(\displaystyle \sin x\) and \(\displaystyle \cos x\) are continuous on \(\displaystyle \mathbf{R}\), and use the quotient rule: if \(\displaystyle g,h\) are continuous at \(\displaystyle c\) and \(\displaystyle h(c)\neq 0\), then \(\displaystyle g/h\) is continuous at \(\displaystyle c\).Cosine. At any real \(\displaystyle c\), with \(\displaystyle x=c+h\), \[\lim_{h\to 0}\cos(c+h)=\lim_{h\to 0}\left(\cos c\cos h-\sin c\sin h\right)=\cos c.\] So \(\displaystyle \cos\) is continuous on \(\displaystyle \mathbf{R}\) (domain \(\displaystyle \mathbf{R}\), no point of discontinuity).Cosecant. \(\displaystyle \mathrm{cosec}\,x=\dfrac{1}{\sin x}\), defined where \(\displaystyle \sin x\neq 0\), i.e. on \(\displaystyle \mathbf{R}-\{n\pi: n\in\mathbf{Z}\}\). At any \(\displaystyle c\) in this domain, \(\displaystyle \sin c\neq 0\), so the quotient rule gives continuity. Hence \(\displaystyle \mathrm{cosec}\) is continuous on its whole domain \(\displaystyle \mathbf{R}-\{n\pi\}\).Secant. \(\displaystyle \sec x=\dfrac{1}{\cos x}\), defined where \(\displaystyle \cos x\neq 0\), i.e. on \(\displaystyle \mathbf{R}-\left\{(2n+1)\dfrac{\pi}{2}: n\in\mathbf{Z}\right\}\). By the quotient rule it is continuous at every point of that domain.Cotangent. \(\displaystyle \cot x=\dfrac{\cos x}{\sin x}\), defined on \(\displaystyle \mathbf{R}-\{n\pi: n\in\mathbf{Z}\}\), and continuous there by the quotient rule.Answer: each of \(\displaystyle \cos\), \(\displaystyle \mathrm{cosec}\), \(\displaystyle \sec\) and \(\displaystyle \cot\) is continuous at every point of its own domain; the excluded points \(\displaystyle n\pi\) (for \(\displaystyle \mathrm{cosec}\), \(\displaystyle \cot\)) and \(\displaystyle (2n+1)\pi/2\) (for \(\displaystyle \sec\)) are not points of discontinuity, because the functions are not defined there.

For \(\displaystyle x<0\): \(\displaystyle f(x)=\dfrac{\sin x}{x}\) is a quotient of two functions continuous on \(\displaystyle \mathbf{R}\) with denominator \(\displaystyle x\neq 0\), so \(\displaystyle f\) is continuous at every \(\displaystyle c<0\).For \(\displaystyle x>0\): \(\displaystyle f(x)=x+1\) is a polynomial, so \(\displaystyle f\) is continuous at every \(\displaystyle c>0\).Only \(\displaystyle x=0\) remains. Using the standard limit \(\displaystyle \displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1\), \[\lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{-}}\frac{\sin x}{x}=1,\qquad \lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}(x+1)=1,\] and \(\displaystyle f(0)=0+1=1\).Since LHL \(\displaystyle =\) RHL \(\displaystyle =f(0)=1\), \(\displaystyle f\) is continuous at \(\displaystyle x=0\) too.Answer: \(\displaystyle f\) has no point of discontinuity; it is continuous on \(\displaystyle \mathbf{R}\).

For \(\displaystyle x\neq 0\): \(\displaystyle x\mapsto x^{2}\) is continuous, \(\displaystyle x\mapsto \dfrac{1}{x}\) is continuous where \(\displaystyle x\neq 0\), and \(\displaystyle \sin\) is continuous on \(\displaystyle \mathbf{R}\); so \(\displaystyle \sin\dfrac{1}{x}\) is continuous at every \(\displaystyle c\neq 0\) (composition of continuous functions), and the product \(\displaystyle x^{2}\sin\dfrac{1}{x}\) is continuous at every \(\displaystyle c\neq 0\).At \(\displaystyle x=0\) the factor \(\displaystyle \sin\dfrac1x\) has no limit, so use the sandwich (squeeze) theorem instead. For all \(\displaystyle x\neq 0\), \[-1\le \sin\frac{1}{x}\le 1\ \Longrightarrow\ -x^{2}\le x^{2}\sin\frac{1}{x}\le x^{2}.\] Since \(\displaystyle \lim_{x\to 0}\left(-x^{2}\right)=0=\lim_{x\to 0}x^{2}\), the sandwich theorem gives \[\lim_{x\to 0}x^{2}\sin\frac{1}{x}=0.\] And \(\displaystyle f(0)=0\), so \(\displaystyle \lim_{x\to 0}f(x)=f(0)\).Answer: yes, \(\displaystyle f\) is continuous at every real point, i.e. \(\displaystyle f\) is a continuous function.

For \(\displaystyle x\neq 0\): \(\displaystyle f(x)=\sin x-\cos x\) is the difference of two functions continuous on \(\displaystyle \mathbf{R}\), hence continuous at every \(\displaystyle c\neq 0\).At \(\displaystyle x=0\): using continuity of \(\displaystyle \sin\) and \(\displaystyle \cos\), \[\lim_{x\to 0}f(x)=\lim_{x\to 0}(\sin x-\cos x)=\sin 0-\cos 0=0-1=-1,\] while the definition gives \(\displaystyle f(0)=-1\).Since \(\displaystyle \lim_{x\to 0}f(x)=-1=f(0)\), \(\displaystyle f\) is continuous at \(\displaystyle x=0\) as well. (The second branch simply supplies the value the first branch was heading towards, so the definition creates no break.)Answer: \(\displaystyle f\) is continuous at every point of \(\displaystyle \mathbf{R}\).

Continuity at \(\displaystyle x=\dfrac{\pi}{2}\) requires \(\displaystyle \displaystyle\lim_{x\to \pi/2}f(x)=f\!\left(\frac{\pi}{2}\right)=3\).The expression \(\displaystyle \dfrac{k\cos x}{\pi-2x}\) is of the form \(\displaystyle \dfrac{0}{0}\) at \(\displaystyle x=\pi/2\), so substitute \(\displaystyle x=\dfrac{\pi}{2}+h\), where \(\displaystyle h\to 0\) as \(\displaystyle x\to \pi/2\). Then \[\cos x=\cos\!\left(\frac{\pi}{2}+h\right)=-\sin h,\qquad \pi-2x=\pi-\pi-2h=-2h.\] Hence \[\lim_{x\to \pi/2}\frac{k\cos x}{\pi-2x}=\lim_{h\to 0}\frac{k(-\sin h)}{-2h}=\frac{k}{2}\lim_{h\to 0}\frac{\sin h}{h}=\frac{k}{2}\cdot 1=\frac{k}{2}.\]Setting this equal to the value \(\displaystyle 3\): \[\frac{k}{2}=3\ \Longrightarrow\ k=6.\]Answer: \(\displaystyle k=6\).

Continuity at \(\displaystyle x=2\) requires LHL \(\displaystyle =\) RHL \(\displaystyle =f(2)\).Value: \(\displaystyle f(2)=k(2)^{2}=4k\) (the point \(\displaystyle x=2\) lies in the branch \(\displaystyle x\le 2\)).Left-hand limit: \(\displaystyle \displaystyle\lim_{x\to 2^{-}}kx^{2}=4k\).Right-hand limit: \(\displaystyle \displaystyle\lim_{x\to 2^{+}}3=3\).Since LHL already equals \(\displaystyle f(2)\), the condition reduces to \[4k=3\ \Longrightarrow\ k=\frac{3}{4}.\]Answer: \(\displaystyle k=\dfrac{3}{4}\).

Continuity at \(\displaystyle x=\pi\) requires LHL \(\displaystyle =\) RHL \(\displaystyle =f(\pi)\).Value: \(\displaystyle f(\pi)=k\pi+1\) (the point \(\displaystyle x=\pi\) lies in the branch \(\displaystyle x\le\pi\)).Left-hand limit: \(\displaystyle \displaystyle\lim_{x\to \pi^{-}}(kx+1)=k\pi+1\).Right-hand limit: \(\displaystyle \displaystyle\lim_{x\to \pi^{+}}\cos x=\cos\pi=-1\) (using continuity of \(\displaystyle \cos\)).Equating, \[k\pi+1=-1\ \Longrightarrow\ k\pi=-2\ \Longrightarrow\ k=-\frac{2}{\pi}.\]Answer: \(\displaystyle k=-\dfrac{2}{\pi}\).

Continuity at \(\displaystyle x=5\) requires LHL \(\displaystyle =\) RHL \(\displaystyle =f(5)\).Value: \(\displaystyle f(5)=5k+1\) (the point \(\displaystyle x=5\) lies in the branch \(\displaystyle x\le 5\)).Left-hand limit: \(\displaystyle \displaystyle\lim_{x\to 5^{-}}(kx+1)=5k+1\).Right-hand limit: \(\displaystyle \displaystyle\lim_{x\to 5^{+}}(3x-5)=15-5=10\).Equating, \[5k+1=10\ \Longrightarrow\ 5k=9\ \Longrightarrow\ k=\frac{9}{5}.\]Answer: \(\displaystyle k=\dfrac{9}{5}\).

Each branch (a constant, a linear function, a constant) is continuous on the open interval where it applies, so \(\displaystyle f\) can fail to be continuous only at the joining points \(\displaystyle x=2\) and \(\displaystyle x=10\). Impose continuity at both.At \(\displaystyle x=2\): \(\displaystyle f(2)=5\), LHL \(\displaystyle =\displaystyle\lim_{x\to 2^{-}}5=5\), RHL \(\displaystyle =\displaystyle\lim_{x\to 2^{+}}(ax+b)=2a+b\). Continuity gives \[2a+b=5.\qquad (1)\]At \(\displaystyle x=10\): \(\displaystyle f(10)=21\), LHL \(\displaystyle =\displaystyle\lim_{x\to 10^{-}}(ax+b)=10a+b\), RHL \(\displaystyle =\displaystyle\lim_{x\to 10^{+}}21=21\). Continuity gives \[10a+b=21.\qquad (2)\]Subtract ($\displaystyle 1$) from ($\displaystyle 2$): \[8a=16\ \Longrightarrow\ a=2,\] and then from ($\displaystyle 1$), \(\displaystyle b=5-2a=5-4=1\).Answer: \(\displaystyle a=2\) and \(\displaystyle b=1\) (so the middle branch is \(\displaystyle 2x+1\)), and with these values \(\displaystyle f\) is continuous on \(\displaystyle \mathbf{R}\).