CBSE Test Paper 01
Chapter 5 Continuity and Differentiability
Let f (x + y) = f(x) + f(y) ∀ x, y ∈R. Suppose that f (6) = 5 and f ‘ (0) = 1, then f ‘ (6) is equal to
Derivative of log|x| w.r.t. |x| is
- None of these
- 1x
- ±1x
- 1|x|
The function f (x) = 1 + |sin x| is
- differentiable everywhere
- continuous everywhere
- differentiable nowhere
- continuous nowhere
Ltx→01−cosxxsinx is equal to
Ltx→π/4cosx−sinxx−π4 is equal to
- The value of c in Mean value theorem for the function f(x) = x(x - 2), x ∈[1, 2] is ________.
- The set of points where the function f given by f(x) = |2x - 1| sin x is differentiable is ________.
- Differential coefficient of sec (tan-1x) w.r.t. x is ________.
Discuss the continuity of the function f(x)=sinx.cosx.
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.
Determine the value of the constant 'k' so that the function f(x) = {kx|x|,3, if x<0 if x≥0 is continuous at x= 0.
Find dydx, y=cos−1(1−x21+x2),0<x<1
Show that the function defined by f(x)=cos(x2) is a continuous function.
Determine if f defined by f(x)=⎧⎩⎨x2sin1x,ifx≠00,ifx=0 is a continuous function.
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}
If xy + yx = ab, then find dydx.
If ey(x + 1) = 1, then show that d2ydx2=(dydx)2.
Find dydx if yx+xy+xx=ab.
CBSE Test Paper 01
Chapter 5 Continuity and Differentiability
Solution
- 1
Explanation: f′(6)=limh→0f(6+h)−f(6)h = limh→0f(6+h)−f(6+0)h
=limh→0f(6)+f(h)−{f(6)+f(0)}h
=limh→0f(h)−f(0)h=f′(0)=1
- 1|x|
Explanation: dd|x|(log|x|)=1|x|
- 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)
- 12
Explanation: limx→01−cosxxsinx=limx→01−cos2xxsinx(1+cosx)limx→0sinxx.11+cosx=1.11+1=12
- −22√
Explanation: limx→π4cosx−sinxx−π4 =limx→π4−sinx−cosx1=−sinπ4−cosπ4=−22√=−2–√
- 32
- R - {12}
- x1+x2√
- Since sin x and cos x are continuous functions and product of two continuous function is a continuous function, therefore f(x)=sinx.cosx is a continuous function.
- Given, f(x) = {(x+3)2−36x−3,x,x≠3x=3
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)
⇒limx→3(x+3)2−36x−3=k
⇒limx→3(x+3)2−62x−3=k
⇒limx→3(x+3−6)(x+3+6)x−3=k [ ∵ a2 - b2 = (a - b)(a + b)]
⇒limx→3(x−3)(x+9)(x−3)=k
⇒limx→3(x+9)=k
⇒ 3 + 9 = k ⇒ k = 12 - Let f(x) ={kx|x|,3, if x<0 if x≥0 be continuous at x = 0
Then,limx→0+f(x)=limx→0−f(x)=f(0)
⇒limh→0f(0+h)=limh→0f(0−h)=f(0)
⇒3=limh→0k(−h)|−h|=3
⇒limh→0(−khh)=3
limh→0(−k)=3
∴ k = - 3 - Given: y=cos−1(1−x21+x2),0<x<1
Putting x=tanθ
y=cos−1(1−tan2θ1+tan2θ)
=cos−1(cos2θ)=2θ=2tan−1x
∴dydx=2.11+x2=21+x2 - Let f(x)=x2 and g(x)=cosx, then
(gof)(x)=g[f(x)]=g(x2)=cosx2
Now f and g being continuous it follows that their composite (gof) is continuous.
Hence cosx2 is continuous function. - Here, limx→0f(x)=limx→0x2sin1x=0 x a finite quantity = 0
[∵sin1xlies between - 1 and 1]
Also f(0) = 0
Since, limx→0f(x)=f(0) therefore, the function f is continuous at x = 0.
Also,when x≠0 ,then f(x) is the product of two continuous functions and hence Continuous.Hence,f(x) is continuous everywhere. - According to the question, f(x) = {x3+x2−16x+20(x−2)2,k,x≠2x=2} is continuous at x=2.
Now, we have f(2) = k
limx→2f(x)=limx→2x3+x2−16x+20(x−2)2
=limx→2(x−2)(x2+3x−10)(x−2)2
=limx→2(x−2)(x+5)(x−2)(x−2)2
=limx→2(x+5) = 2+ 5 = 7
f(x) is continuous at x = 2.
∴limx→2f(x)= f(2) ⇒ 7 = k⇒k = 7 - 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........(iii)
Now consider, xy = v [ from Eq.(ii)]
Therefore,on taking log both sides, we get,
log xy = logv
⇒ y log x = log v
Therefore,on differentiating both sides w.r.t. x, we get,
y⋅1x+logx⋅dydx=1vdvdx
⇒v(yx+logx⋅dydx)=dvdx
⇒dvdx=xy(yx+logxdydx).........(iv) [ From Eq.(ii)]
Also, yx = u [From Eq(ii)]
Therefore,on taking log both sides, we get,
log yx = log u ⇒ x log y = log u
Therefore,on differentiating both sides w.r.t. 'x', we get,
x⋅1ydydx+1⋅logy=1ududx
⇒xydydx+logy=1ududx
⇒u[xydydx+logy]=dydx
⇒yx[xydydx+logy]=dudx........(v) [ From Eq(ii)]
Therefore,on substituting the values of dvdx and dudx from Eqs. (iv) and (v) respectively in Eq. (iii), we get
xy(yx+logx⋅dydx)+yx(xydydx+logy)=0
⇒xyyx+xylogx⋅dydx+yx⋅xydydx+yxlogy=0
⇒xylogx⋅dydx+yxxy⋅dydx=−xyyx−yxlogy
⇒dydx[xylogx+yx⋅xy]=−xy⋅yx−yxlogy
∴dydx=−xy−1⋅y−yxlogyxylogx+yx−1⋅x - According to the question, ey(x+1)=1
Taking log both sides,
⇒log[ey(x+1)]=log1
⇒logey+log(x+1)=log1
⇒ y+log(x+1)=log1 [∵ log ey = y]
differentiating both sides w.r.t. x,
⇒dydx+1x+1=0.........(i)
Differentiating both sides w.r.t. 'x',
⇒d2ydx2−1(x+1)2=0
⇒d2ydx2−(−dydx)2=0 [ From Equation(i)]
⇒d2ydx2−(dydx)2=0
⇒d2ydx2=(dydx)2 - Let u=yx,v=xy,w=xx
u+v+w=ab
Therefore dudx+dwdx+dvdx=0 ....(1)
u=yx
Taking log both side
logu=logyx
logu=x.logy
Differentiate both side w.r.t. to x
1u.dudx=x.1y.dydx+logy.1
dudx=u[xy.dydx+logy]
dudx=yx[xy.dydx+logy].... (2)
v=xy
Taking log both side
logv=logxy
logv=y.logx
1v.dvdx=y.1x+logx.dydx
dvdx=v[yx+logx.dydx]
dvdx=xy[yx+logx.dydx].... (3)
w=xx
Taking log both side
logw=logxx
logw=xlogx
1w.dwdx=x.1x+logx.1
1w.dwdx=1+logx
dwdx=w(1+logx)
dwdx=xx(1+logx).... (4)
dydx=−xx(1+logx)−y.xy−1−yxlogyx.yx−1+xylogx. (by putting 2,3 and 4 in 1)