Continuity and Differentiability - Test Papers

 CBSE Test Paper 01

Chapter 5 Continuity and Differentiability

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1. Let f (x + y) = f(x) + f(y) ∀$\mathrm{\forall }$ x, y ∈R$\in \mathbf{R}$. Suppose that f (6) = 5 and f ‘ (0) = 1, then f ‘ (6) is equal to 

   1. 1
   2. 30
   3. None of these
   4. 25

2. Derivative of log|x| w.r.t. |x| is 

   1. None of these
   2. 1x$\frac{1}{x}$
   3. ±1x$\pm \frac{1}{x}$
   4. 1|x|$\frac{1}{\left|x\right|}$

3. The function f (x) = 1 + |sin x| is 

   1. differentiable everywhere
   2. continuous everywhere
   3. differentiable nowhere
   4. continuous nowhere

4. Ltx→01−cosxxsinx$\underset{x\to 0}{Lt}\frac{1-\cos x}{x\sin x}$ is equal to 

   1. 1
   2. 2
   3. 0
   4. 12$\frac{1}{2}$

5. Ltx→π/4cosx−sinxx−π4$\underset{x\to \pi /4}{Lt}\frac{\cos x-\sin x}{x-\frac{\pi }{4}}$ is equal to 

   1. −22√$-\frac{2}{\sqrt{2}}$
   2. -1
   3. −12√$-\frac{1}{\sqrt{2}}$
   4. 22√$\frac{2}{\sqrt{2}}$
6. The value of c in Mean value theorem for the function f(x) = x(x - 2), x ∈$\in$[1, 2] is ________.
7. The set of points where the function f given by f(x) = |2x - 1| sin x is differentiable is ________.
8. Differential coefficient of sec (tan-1x) w.r.t. x is ________.

9. Discuss the continuity of the function f(x)=sinx.cosx$f(x)=\sin x.\cos x$. 

10. Determine the value of 'k' for which the following function is continuous at x = 3 : f(x) = {(x+3)2−36x−3,x≠3k,x=3$\{\begin{array}{l}\frac{(x+3{)}^2-36}{x-3},x\ne 3\\ k,x=3\end{array}$. 

11. Determine the value of the constant 'k' so that the function f(x) = {kx|x|,3, if x<0 if x≥0$\{\begin{array}{ll}\frac{kx}{|x|},& \text{ if }x<0\\ 3,& \text{ if }x\ge 0\end{array}$ is continuous at x= 0. 

12. Find dydx,$\frac{dy}{dx},$ y=cos−1(1−x21+x2),0<x<1$y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right),0<x<1$ 

13. Show that the function defined by f(x)=cos(x2)$f\left(x\right)=\cos \left(x^2\right)$ is a continuous function. 

14. Determine if f defined by f(x)=⎧⎩⎨x2sin1x,ifx≠00,ifx=0$f\left(x\right)=\{\begin{array}{c}{x}^2\sin \frac{1}{x},ifx\ne 0\\ 0,ifx=0\end{array}$ is a continuous function. 

15. Find the value of k so that the following function is continuous at x = 2.
    f(x) = {x3+x2−16x+20(x−2)2,k,x≠2x=2}$\left\{\begin{array}{cc}\frac{x^3+x^2-16x+20}{{(x-2)}^2},& x\ne 2\\ k,& x=2\end{array}\right\}$

16. If xy + yx = ab, then find dydx$\frac{dy}{dx}$. 

17. If ey(x + 1) = 1, then show that d2ydx2=(dydx)2$\frac{d^{2}y}{d{x}^2}={\left(\frac{dy}{dx}\right)}^2$. 

18. Find dydx$\frac{dy}{dx}$ if yx+xy+xx=ab$y^x+x^y+x^x=a^b$.

CBSE Test Paper 01
Chapter 5 Continuity and Differentiability

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Solution

1. 1. 1
      Explanation: f′(6)=limh→0f(6+h)−f(6)h$\begin{array}{c}{f}^{\prime }(6)=\underset{h\to 0}{lim}\frac{f(6+h)-f(6)}{h}\end{array}$ = limh→0f(6+h)−f(6+0)h$\underset{h\to 0}{lim}\frac{f(6+h)-f(6+0)}{h}$
      =limh→0f(6)+f(h)−{f(6)+f(0)}h$=\underset{h\to 0}{lim}\frac{f(6)+f(h)-\left\{f(6)+f(0)\right\}}{h}$
      =limh→0f(h)−f(0)h=f′(0)=1$=\underset{h\to 0}{lim}\frac{f(h)-f(0)}{h}=f^{\prime }(0)=1$
2. 4. 1|x|$\frac{1}{\left|x\right|}$
      Explanation: dd|x|(log|x|)=1|x|$\frac{d}{d\left|x\right|}\left(\log \left|x\right|\right)=\frac{1}{\left|x\right|}$
3. 2. continuous everywhere
      Explanation: f(x) = 1 + |sinx| is not derivable at those x for which x for which sinx = 0, however, 1 + |sinx| is continuous everywhere (being the sum of two continuous functions)
4. 4. 12$\frac{1}{2}$
      Explanation: limx→01−cosxxsinx$\underset{x\to 0}{lim}\frac{1-\cos x}{x\sin x}$=limx→01−cos2xxsinx(1+cosx)limx→0sinxx.11+cosx=1.11+1=12$=\underset{x\to 0}{lim}\frac{1-{\cos }^{2}x}{x\sin x(1+\cos x)}\underset{x\to 0}{lim}\frac{\sin x}{x}.\frac{1}{1+\cos x}=1.\frac{1}{1+1}=\frac{1}{2}$
5. 1. −22√$-\frac{2}{\sqrt{2}}$
      Explanation: limx→π4cosx−sinxx−π4$\underset{x\to \frac{\pi }{4}}{lim}\frac{\cos x-\sin x}{x-\frac{\pi }{4}}$ =limx→π4−sinx−cosx1=−sinπ4−cosπ4=−22√=−2–√$=\underset{x\to \frac{\pi }{4}}{lim}\frac{-\sin x-\cos x}{1}=-\sin \frac{\pi }{4}-\cos \frac{\pi }{4}=-\frac{2}{\sqrt{2}}=-\sqrt{2}$
6. 32$\frac{3}{2}$
7. R - {12}$\left\{\frac{1}{2}\right\}$
8. x1+x2√$\frac{x}{\sqrt{1+x^2}}$
9. Since sin x and cos x are continuous functions and product of two continuous function is a continuous function, therefore f(x)=sinx.cosx$f(x)=\sin x.\cos x$ is a continuous function.
10. Given, f(x) = {(x+3)2−36x−3,x,x≠3x=3$\{\begin{array}{ll}\frac{(x+3{)}^2-36}{x-3},& x\ne 3\\ x,& x=3\end{array}$
    We shall use definition of continuity to find the value of k.
    If f(x) is continuous at x = 3,
    Then, we have limx→3f(x)=f(3)$\underset{x\to 3}{lim}f(x)=f(3)$
    ⇒limx→3(x+3)2−36x−3=k$\Rightarrow \underset{x\to 3}{lim}\frac{(x+3{)}^2-36}{x-3}=k$
    ⇒limx→3(x+3)2−62x−3=k$\Rightarrow \underset{x\to 3}{lim}\frac{(x+3{)}^2-6^2}{x-3}=k$
    ⇒limx→3(x+3−6)(x+3+6)x−3=k$\Rightarrow \underset{x\to 3}{lim}\frac{(x+3-6)(x+3+6)}{x-3}=k$ [ ∵$\because$ a2 - b2 = (a - b)(a + b)]
    ⇒limx→3(x−3)(x+9)(x−3)=k$\Rightarrow \underset{x\to 3}{lim}\frac{(x-3)(x+9)}{(x-3)}=k$
    ⇒limx→3(x+9)=k$\Rightarrow \underset{x\to 3}{lim}(x+9)=k$
    ⇒$\Rightarrow$ 3 + 9 = k ⇒$\Rightarrow$ k = 12
11. Let f(x) ={kx|x|,3, if x<0 if x≥0$\{\begin{array}{ll}\frac{kx}{|x|},& \text{ if }x<0\\ 3,& \text{ if }x\ge 0\end{array}$ be continuous at x = 0
    Then,limx→0+f(x)=limx→0−f(x)=f(0)$\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{-}}{lim}f(x)=f(0)$
    ⇒limh→0f(0+h)=limh→0f(0−h)=f(0)$\Rightarrow \underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}f(0-h)=f(0)$
    ⇒3=limh→0k(−h)|−h|=3$\Rightarrow 3=\underset{h\to 0}{lim}\frac{k(-h)}{|-h|}=3$
    ⇒limh→0(−khh)=3$\Rightarrow \underset{h\to 0}{lim}\left(\frac{-kh}{h}\right)=3$
    limh→0(−k)=3$\underset{h\to 0}{lim}(-k)=3$
    ∴$\therefore$ k = - 3
12. Given: y=cos−1(1−x21+x2),0<x<1$y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right),0<x<1$
    Putting x=tanθ$x=\tan \theta$
    y=cos−1(1−tan2θ1+tan2θ)$y={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$
    =cos−1(cos2θ)=2θ=2tan−1x$={\cos }^{-1}\left(\cos 2\theta \right)=2\theta =2{\tan }^{-1}x$
    ∴dydx=2.11+x2=21+x2$\therefore \frac{dy}{dx}=2.\frac{1}{1+x^2}=\frac{2}{1+x^2}$
13. Let f(x)=x2$f\left(x\right)=x^2$ and g(x)=cosx$g\left(x\right)=\cos x$, then
    (gof)(x)=g[f(x)]=g(x2)=cosx2$\left(gof\right)\left(x\right)=g\left[f\left(x\right)\right]=g\left(x^2\right)=\cos x^2$
    Now f and g being continuous it follows that their composite (gof) is continuous.
    Hence cosx2$\cos x^2$ is continuous function.
14. Here, limx→0f(x)=limx→0x2sin1x=0$\underset{x\to 0}{lim}f(x)=\underset{x\to 0}{lim}{x}^2\sin \frac{1}{x}=0$ x a finite quantity = 0
    [∵sin1xlies between - 1 and 1]$\left[\because \sin \frac{1}{x}\text{lies between - 1 and 1}\right]$
    Also f(0) = 0
    Since, limx→0f(x)=f(0)$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$ therefore, the function f is continuous at x = 0.
    Also,when x≠0$x\ne 0$ ,then f(x) is the product of two continuous functions and hence Continuous.Hence,f(x) is continuous everywhere.
15. According to the question, f(x) = {x3+x2−16x+20(x−2)2,k,x≠2x=2}$\left\{\begin{array}{cc}\frac{x^3+x^2-16x+20}{{(x-2)}^2},& x\ne 2\\ k,& x=2\end{array}\right\}$ is continuous at x=2.$x=2.$
    Now, we have f(2) = k
    limx→2f(x)=limx→2x3+x2−16x+20(x−2)2$\underset{x\to 2}{lim}f(x)=\underset{x\to 2}{lim}\frac{x^3+x^2-16x+20}{(x-2{)}^2}$
    =limx→2(x−2)(x2+3x−10)(x−2)2$=\underset{x\to 2}{lim}\frac{(x-2)\left(x^2+3x-10\right)}{(x-2{)}^2}$
    =limx→2(x−2)(x+5)(x−2)(x−2)2$=\underset{x\to 2}{lim}\frac{(x-2)(x+5)(x-2)}{(x-2{)}^2}$
    =limx→2(x+5)$=\underset{x\to 2}{lim}(x+5)$ = 2+ 5 = 7
    f(x) is continuous at x = 2.
    ∴limx→2f(x)$\therefore \underset{x\to 2}{lim}f(x)$= f(2) ⇒$\Rightarrow$ 7 = k⇒$\Rightarrow$k = 7
16. We have, xy + yx = ab.........(i)
    Let xy = v and yx = u......(ii)
    Therefore,on putting these values in Eq. (i), we get,
    v + u = ab
    Therefore,on differentiating both sides w.r.t. x, we get,
    dvdx+dudx=0$\frac{dv}{dx}+\frac{du}{dx}=0$........(iii)
    Now consider, xy = v [ from Eq.(ii)]
    Therefore,on taking log both sides, we get,
    log xy = logv
    ⇒$\Rightarrow$ y log x = log v
    Therefore,on differentiating both sides w.r.t. x, we get,
    y⋅1x+logx⋅dydx=1vdvdx$y\cdot \frac{1}{x}+\log x\cdot \frac{dy}{dx}=\frac{1}{v}\frac{dv}{dx}$
    ⇒v(yx+logx⋅dydx)=dvdx$\Rightarrow v\left(\frac{y}{x}+\log x\cdot \frac{dy}{dx}\right)=\frac{dv}{dx}$
    ⇒dvdx=xy(yx+logxdydx)$\Rightarrow \frac{dv}{dx}=x^y\left(\frac{y}{x}+\log x\frac{dy}{dx}\right)$.........(iv) [ From Eq.(ii)]
    Also, yx = u [From Eq(ii)]
    Therefore,on taking log both sides, we get,
    log yx = log u ⇒$\Rightarrow$ x log y = log u
    Therefore,on differentiating both sides w.r.t. 'x', we get,
    x⋅1ydydx+1⋅logy=1ududx$x\cdot \frac{1}{y}\frac{dy}{dx}+1\cdot \log y=\frac{1}{u}\frac{du}{dx}$
    ⇒xydydx+logy=1ududx$\Rightarrow \frac{x}{y}\frac{dy}{dx}+\log y=\frac{1}{u}\frac{du}{dx}$
    ⇒u[xydydx+logy]=dydx$\Rightarrow u\left[\frac{x}{y}\frac{dy}{dx}+\log y\right]=\frac{dy}{dx}$
    ⇒yx[xydydx+logy]=dudx$\Rightarrow y^x\left[\frac{x}{y}\frac{dy}{dx}+\log y\right]=\frac{du}{dx}$........(v) [ From Eq(ii)]
    Therefore,on substituting the values of dvdx and dudx$\frac{dv}{dx}\text{ and }\frac{du}{dx}$ from Eqs. (iv) and (v) respectively in Eq. (iii), we get
    xy(yx+logx⋅dydx)+yx(xydydx+logy)=0$x^y\left(\frac{y}{x}+\log x\cdot \frac{dy}{dx}\right)+y^x\left(\frac{x}{y}\frac{dy}{dx}+\log y\right)=0$
    ⇒xyyx+xylogx⋅dydx$\Rightarrow x^y\frac{y}{x}+x^y\log x\cdot \frac{dy}{dx}$+yx⋅xydydx+yxlogy=0$+y^x\cdot \frac{x}{y}\frac{dy}{dx}+y^x\log y=0$
    ⇒xylogx⋅dydx+yxxy⋅dydx$\Rightarrow x^y\log x\cdot \frac{dy}{dx}+y^x\frac{x}{y}\cdot \frac{dy}{dx}$=−xyyx−yxlogy$=-x^y\frac{y}{x}-y^x\log y$
    ⇒dydx[xylogx+yx⋅xy]$\Rightarrow \frac{dy}{dx}\left[x^y\log x+y^x\cdot \frac{x}{y}\right]$=−xy⋅yx−yxlogy$=-x^y\cdot \frac{y}{x}-y^x\log y$
    ∴dydx=−xy−1⋅y−yxlogyxylogx+yx−1⋅x$\therefore \frac{dy}{dx}=\frac{-x^{y-1}\cdot y-y^x\log y}{x^y\log x+y^{x-1}\cdot x}$
17. According to the question, ey(x+1)=1$e^y(x+1)=1$
    Taking log both sides,
    ⇒log[ey(x+1)]=log1$\Rightarrow log[e^y(x+1)]=log1$
    ⇒logey+log(x+1)=log1$\Rightarrow log{e}^y+log(x+1)=log1$
    ⇒$\Rightarrow$ y+log(x+1)=log1$y+log(x+1)=log1$ [∵$\because$ log ey = y]
    differentiating both sides w.r.t. x,
    ⇒dydx+1x+1=0$\Rightarrow \frac{dy}{dx}+\frac{1}{x+1}=0$.........(i)
    Differentiating both sides w.r.t. 'x',
    ⇒d2ydx2−1(x+1)2=0$\Rightarrow \frac{d^{2}y}{d{x}^2}-\frac{1}{(x+1{)}^2}=0$
    ⇒d2ydx2−(−dydx)2=0$\Rightarrow \frac{d^{2}y}{d{x}^2}-{\left(-\frac{dy}{dx}\right)}^2=0$ [ From Equation(i)]
    ⇒d2ydx2−(dydx)2=0$\Rightarrow \frac{d^{2}y}{d{x}^2}-{\left(\frac{dy}{dx}\right)}^2=0$
    ⇒d2ydx2=(dydx)2$\Rightarrow \frac{d^{2}y}{d{x}^2}={\left(\frac{dy}{dx}\right)}^2$
18. Let u=yx,v=xy,w=xx$u=y^x,v=x^y,w=x^x$
    u+v+w=ab$u+v+w=a^b$
    Therefore dudx+dwdx+dvdx=0$\frac{du}{dx}+\frac{dw}{dx}+\frac{dv}{dx}=0$ ....(1)
    u=yx$u=y^x$
    Taking log both side
    logu=logyx$\log u=\log y^x$
    logu=x.logy$\log u=x.\log y$
    Differentiate both side w.r.t. to x
    1u.dudx=x.1y.dydx+logy.1$\frac{1}{u}.\frac{du}{dx}=x.\frac{1}{y}.\frac{dy}{dx}+\log y.1$
    dudx=u[xy.dydx+logy]$\frac{du}{dx}=u\left[\frac{x}{y}.\frac{dy}{dx}+\log y\right]$
    dudx=yx[xy.dydx+logy]$\frac{du}{dx}=y^x\left[\frac{x}{y}.\frac{dy}{dx}+\log y\right]$.... (2)
    v=xy$v=x^y$
    Taking log both side
    logv=logxy$\log v=\log x^y$
    logv=y.logx$\log v=y.\log x$
    1v.dvdx=y.1x+logx.dydx$\frac{1}{v}.\frac{dv}{dx}=y.\frac{1}{x}+\log x.\frac{dy}{dx}$
    dvdx=v[yx+logx.dydx]$\frac{dv}{dx}=v\left[\frac{y}{x}+\log x.\frac{dy}{dx}\right]$
    dvdx=xy[yx+logx.dydx]$\frac{dv}{dx}=x^y\left[\frac{y}{x}+\log x.\frac{dy}{dx}\right]$.... (3)
    w=xx$w=x^x$
    Taking log both side
    logw=logxx$\log w=\log x^x$
    logw=xlogx$\log w=x\log x$
    1w.dwdx=x.1x+logx.1$\frac{1}{w}.\frac{dw}{dx}=x.\frac{1}{x}+\log x.1$
    1w.dwdx=1+logx$\frac{1}{w}.\frac{dw}{dx}=1+\log x$
    dwdx=w(1+logx)$\frac{dw}{dx}=w(1+\log x)$
    dwdx=xx(1+logx)$\frac{dw}{dx}=x^x(1+\log x)$.... (4)
    dydx=−xx(1+logx)−y.xy−1−yxlogyx.yx−1+xylogx.$\frac{dy}{dx}=\frac{-x^x(1+\log x)-y.x^{y-1}-y^x\log y}{x.y^{x-1}+x^y\log x.}$ (by putting 2,3 and 4 in 1)