NCERT Solutions for Class 12 Maths Exercise Miscellenous Chapter 5 Continuity and Differentiability – FREE PDF Download
Free PDF download of NCERT Solutions for Class 12 Maths Chapter 5 Exercise Miscellenous (Ex Misc) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 5 Continuity and Differentiability Exercise Miscellenous Questions with Solutions to help you to revise complete Syllabus and Score More marks.
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise Miscellenous (Ex Misc.)
Differentiate with respect to 
the functions in Exercises 1 to 5.
1. 
= 
2. 
= 
= 
3.
……….(i) 
4.
= 
= 
= 
= 
5.
dydx=−2x+7√.24−x2√.12−12.cos−1x22x+7√×2(2x+7)dydx=−2x+7.24−x2.12−12.cos−1x22x+7×2(2x+7)
= 
= 
= −[14−x2√2x+7√+cos−1x2(2x+7)32]−
Differentiate with respect to the functions in Exercises 6 to 11.
6. 
……….(i) Now, 
= 
= 
And 
= 
= 
From eq. (i), 
= 
= 
7.
……….(i) Taking log both sides, we get
= 
=>1ydydx=logx1logx.1x+log(logx)1x=>1ydydx=logx1logx.1x+log(logx)1x
=>1ydydx=1x+log(logx)1x=>1ydydx=1x+log(logx)1x
=>1ydydx=[1+log(logx)x]=>1ydydx=
= 
8. for some constants and
for some constants
for some constants 
and 
dydx=(asinx−bcosx)sin(acosx+bsinx)dydx=(asinx−bcosx)sin(acosx+bsinx)
9.
……….(i) Taking log Both Sides, we get
=>logy=(sinx−cosx)log(sinx−cosx)=>logy=(sinx−cosx)log(sinx−cosx)
=> ddxlogy=(sinx−cosx)ddxlog(sinx−cosx)+log(sinx−cosx)ddx(sinx−cosx)ddxlogy=(sinx−cosx)ddxlog(sinx−cosx)+log(sinx−cosx)ddx(sinx−cosx)
10. for some fixed and
for some fixed 
and 
= 
…….(i)
Now taking 
, let 
……….(ii)
Taking log both sides, we get
= 
= 
From eq. (ii),
From eq. (i),
=> dydx=xx(1+logx)+axa−1+axlogadydx=xx(1+logx)+axa−1+axloga
11. for
for 
for 
Putting 
and 
……….(i)
Now
= 
= 
……….(ii)
Again
= 
= 
……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
12. Find if and
and 
and dxdt=10ddt(t−sint)dxdt=10ddt(t−sint)
=>dxdt=10(1−cost)=>dxdt=10(1−cost)
= 
= 
= 
13. Find if
= 
= 
= 0
14. If for prove that
for 
prove that 
Squaring both sides,We get
dydx=(1+x)ddx(−x)−(−x)ddx(1+x)(1+x)2dydx=(1+x)ddx(−x)−(−x)ddx(1+x)(1+x)2
=> dydx=−(1+x)+x(1+x)2dydx=−(1+x)+x(1+x)2
= 
Proved.
15. If for some prove that is a constant independent of a and b.
prove that
……….(i) 
……….(ii)
Again 
[From eq. (ii)
= 
= 
……….(iii)
Putting values of 
and 
in the given expression,
= 
= 
= 
= 
which is a constant and is independent of and 
16. If with prove that
prove that 
= 
= 
17. If and find
find 
and 
Differentiating both sides with respect to 
and 
and 
and 
and 
Now 
Again 
= 
= 
= 
= 
18. If show that exists for all real and find it.
show that 
exists for all real
= 
Now, L.H.D. at 
= 
= 
L.H.D. at = R.H.D. at
Therefore, 
is differentiable at .
Now, L.H.D. at
= 
= 
And R.H.D. at
=
limx→0−−3x=0limx→0−−3x=0
Again L.H.D. at = R.H.D. at
19. Using mathematical induction, prove that for all positive integers
for all positive integers 
be the given statement in the problem. 
…..(i)
= 
,
which is true as 
Now we suppose 
is true.
…….(ii)
To establish the truth of 
we prove,
= 
= 
Therefore, 
is true if is true but 
is true.
By Principal of Induction is true for all 
N.
20. Using the fact that and the differentiation, obtain the sum formula for cosines.
and the differentiation, obtain the sum formula for cosines.
Consider A and B as function of 
and differentiating both sides w.r.t. 
21. Does there exist a function which is continuous everywhere but not differentiable at exactly two points?
is continuous everywhere but it is not differentiable at and 
22. If prove that
prove that 
= 
23. If show that
show that 
= 
= 
Differentiating both sides with respect to
Proved.