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Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.7, Exercise 5.8

Exericse 5.7

Question 1.
x² + 3x + 2 = y(say)
Solution:
\frac { dy }{ dx } =2x+3\quad and\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2

 Question 2.
x20 = y(say)
Solution:
\frac { dy }{ dx } ={ 20 }x^{ 19 }\quad =>\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =20\times { 19x }^{ 18 }={ 380 }x^{ 18 }\qquad

Question 3.
x.cos x = y(say)
Solution:
\frac { dy }{ dx } =x(-sinx)+cosx.1,=-xsinx+cosx
\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-xcosx-sinx-sinx=-xcosx-2sinx

 Question 4.
log x = y (say)
Solution:
\frac { dy }{ dx } =\frac { 1 }{ x } =>\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-\frac { 1 }{ { x }^{ 2 } }

 Question 5.
x3 log x = y (say)
Solution:
x3 log x = y
=>\frac { dy }{ dx } ={ x }^{ 3 }.\frac { 1 }{ x } +logx\times { 3x }^{ 2 }={ x }^{ 2 }+{ 3x }^{ 2 }logx
\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2x+{ 3x }^{ 2 }.\frac { 1 }{ x } +logx.6x=x(5+6logx)

 Question 6.
ex sin5x = y
Solution:
ex sin5x = y
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 6

 Question 7.
e6x cos3x = y
Solution:
e6x cos3x = y
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 7

Question 8.
tan-1 x = y
Solution:
\frac { dy }{ dx } =\frac { 1 }{ 1+{ x }^{ 2 } } =>\frac { { d }^{ 2y } }{ { dx }^{ 2 } } =\frac { -2x }{ { ({ 1+x }^{ 2 }) }^{ 2 } }

 Question 9.
log(logx) = y
Solution:
log(logx) = y
\frac { dy }{ dx } =\frac { 1 }{ logx } .\frac { 1 }{ x }
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 9

 Question 10.
sin(log x) = y
Solution:
sin(log x) = y
\frac { dy }{ dx } =\frac { cos(logx) }{ x }
and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { x.\left[ -sin(logx) \right] .\frac { 1 }{ x } -cos(logx).1 }{ { x }^{ 2 } }
=\frac { \left[ sin(logx)+cos(logx) \right] }{ { x }^{ 2 } }

 Question 11.
If y = 5 cosx – 3 sin x, prove that \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0
Solution:
\frac { dy }{ dx } =-5sinx-3cosx
\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-5cosx+3sinx=-y
\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0
Hence proved

 Question 12.
If y = cos-1 x, Find \frac { { d }^{ 2 }y }{ { dx }^{ 2 } }  in terms of y alone.
Solution:
\frac { dy }{ dx } =-{ \left( { 1-x }^{ 2 } \right) }^{ -\frac { 1 }{ 2 } }
\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { -cosy }{ { \left( { sin }^{ 2 }y \right) }^{ \frac { 3 }{ 2 } } } =-coty\quad { cosec }^{ 2 }y

Question 13.
If y = 3 cos (log x) + 4 sin (log x), show that
{ x }^{ 2 }{ y }_{ 2 }+{ xy }_{ 1 }+y=0
Solution:
Given that
y = 3 cos (log x) + 4 sin (log x)
tiwari academy class 12 maths Chapter 5 Continuity and Differentiability 13

 Question 14.
If\quad y=A{ e }^{ mx }+B{ e }^{ nx },\quad show\quad that\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -(m+n)\frac { dy }{ dx } +mny=0
Solution:
Given that
\quad y=A{ e }^{ mx }+B{ e }^{ nx },\quad
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 14

Question 15.
If y = 500e7x + 600e-7x, show that \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =49y.
Solution:
we have
y = 500e7x + 600e-7x
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 15

 Question 16.
If ey(x+1) = 1,show that \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }
Solution:
{ e }^{ y }(x+1)=1=>{ e }^{ y }=\frac { 1 }{ x+1 }
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 16

Question 17.
If y=(tan-1 x)² show that (x²+1)²y2+2x(x²+1)y1=2
Solution:
we have
y=(tan-1 x)²
tiwari academy class 12 maths Chapter 5 Continuity and Differentiability 17

 

Exercise 5.8

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Question 1.
Verify Rolle’s theorem for the function
f(x) = x² + 2x – 8,x∈ [-4,2]
Solution:
Now f(x) = x² + 2x – 8 is a polynomial
∴ it is continuous and derivable in its domain x∈R.
Hence it is continuous in the interval [-4,2] and derivable in the interval (- 4,2)
f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 = 0,
f(2) = 2² + 4 – 8 = 8 – 8 = 0
Conditions of Rolle’s theorem are satisfied.
f'(x) = 2x + 2
∴ f’ (c) = 2c + 2 = 0

or c = – 1, c = – 1 ∈ [-4,2]

Question 2.
Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?
(i) f(x) = [x] for x ∈ [5,9]
(ii) f (x) = [x] for x ∈ [-2,2]
(iii) f (x) = x² – 1 for x ∈ [1,2]
Solution:
(i) In the interval [5, 9], f (x) = [x] is neither continuous nor derivable at x = 6,7,8 Hence Rolle’s theorem is not applicable
(ii) f (x) = [x] is not continuous and derivable at -1, 0, 1. Hence Rolle’s theorem is not applicable.
(iii) f(x) = (x² – 1),f(1) = 1 – 1 = 0,
f(2) = 22 – 1 = 3
f(a)≠f(b)
Though it is continous and derivable in the interval [1,2].
Rolle’s theorem is not applicable.
In case of converse if f (c)=0, c ∈ [a, b] then conditions of rolle’s theorem are not true.
(i) f (x) = [x] is the greatest integer less than or equal to x.
∴f(x) = 0, But fis neither continuous nor differentiable in the interval [5,9].
(ii) Here also, theough f (x) = 0, but f is neither continuous nor differentiable in the interval [-2,2].
(iii) f (x)=x² – 1, f'(x)=2x. Here f'(x) is not zero in the [1,2], So f (2) ≠ f’ (2).

 Question 3.
If f: [-5,5] –>R is a differentiable function and if f (x) does not vanish anywhere then prove that f (- 5) ≠ f (5).
Solution:
For Rolle’s theorem
If (i) f is continuous in [a, b]
(ii) f is derivable in [a, b]
(iii) f (a) = f (b)
then f’ (c)=0, c e (a, b)
∴ f is continuous and derivable
but f (c) ≠ 0 =>f(a) ≠ f(b) i.e., f(-5)≠f(5)

 Question 4.
Verify Mean Value Theorem, if
f (x) = x² – 4x – 3 in the interval [a, b], where a = 1 and b = 4.
Solution:
f (x) = x² – 4x – 3. It being a polynomial it is continuous in the interval [1,4] and derivable in (1,4), So all the condition of mean value theorem hold.
then f’ (x) = 2x – 4,
f’ (c) = 2c – 4
f(4)= 16 – 16 – 3 = – 3,
f(1)= 1 – 4 – 3 = – 6
Then there exist a value c such that
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 4

Question 5.
Verify Mean Value Theorem, if f (x)=x3 – 5x2 – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1,3) for which f’ (c) = 0.
Solution:
f (x)=x3 – 5x2 – 3x,
It is a polynomial. Therefore it is continuous in the interval [1,3] and derivable in the interval (1,3)
Also, f'(x)=3x²-10x-3
tiwari academy class 12 maths Chapter 5 Continuity and Differentiability 5

 Question 6.
Examine the applicability of Mean Value theroem for all three functions given in the above Question 2.
Solution:
(i) F (x)= [x] for x ∈ [5,9], f (x) = [x] in the interval [5, 9] is neither continuous, nor differentiable.
(ii) f (x) = [x], for x ∈ [-2,2],
Again f (x) = [x] in the interval [-2,2] is neither continous, nor differentiable.
(iii) f(x) = x²-1 for x ∈ [1,2], It is a polynomial. Therefore it is continuous in the interval [1,2] and differentiable in the interval (1,2)
f (x) = 2x, f(1) = 1 – 1 = 0 ,
f(2) = 4 – 1 = 3, f'(c) = 2c
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability 6

Thus f’ (c) = 0 at c = – 1.

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