NCERT Solutions Class 12
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NCERT Solutions-Mathematics
- Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 1)
- Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 2)
- Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 3)
- Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 4)
- Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 5)
- Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 1)
- Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 2)
- Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 3)
- Matrices : NCERT Solutions – Class 12 Maths (Ex 1)
- Matrices : NCERT Solutions – Class 12 Maths (Ex 2)
- Matrices : NCERT Solutions – Class 12 Maths (Ex 3)
- Matrices : NCERT Solutions – Class 12 Maths (Ex 4)
- Matrices : NCERT Solutions – Class 12 Maths (Ex 5)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 1)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 2)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 3)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 4)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 5)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 6)
- Determinants : NCERT Solutions – Class 12 Maths (Ex 7)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 1)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 2)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 3)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 4)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 5)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 6)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 7)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 8)
- Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 9)
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NCERT Solutions-Chemistry
- Aldehydes, Ketones and Carboxylic Acids : NCERT Solutions – Class 12 Chemistry
- Alcohols, Phenols and Ethers : NCERT Solutions – Class 12 Chemistry
- Amines : NCERT Solutions – Class 12 Chemistry
- Biomolecules : NCERT Solutions – Class 12 Chemistry
- Chemical Kinetics : NCERT Solutions – Class 12 Chemistry
- Chemistry in Everyday Life : NCERT Solutions – Class 12 Chemistry
- Coordination Compounds : NCERT Solutions – Class 12 Chemistry
- Electrochemistry : NCERT Solutions – Class 12 Chemistry
- General Principles and Processes of Isolation of Elements : NCERT Solutions – Class 12 Chemistry
- Haloalkanes and Haloarenes : NCERT Solutions – Class 12 Chemistry
- Polymers : NCERT Solutions – Class 12 Chemistry
- Surface Chemistry : NCERT Solutions – Class 12 Chemistry
- The d-and f-Block Elements : NCERT Solutions – Class 12 Chemistry
- The p-Block Elements : NCERT Solutions – Class 12 Chemistry
- The Solid State : NCERT Solutions – Class 12 Chemistry
- Solutions : NCERT Solutions – Class 12 Chemistry
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NCERT Solutions-Biology
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NCERT Solutions-Physics
- Electrostatic Potential And Capacitance : NCERT Solutions – Class 12 Physics
- Electric Charges And Fields : NCERT Solutions – Class 12 Physics
- Semiconductor Electronics: Materials, Devices And Simple Circuits : NCERT Solutions – Class 12 Physics
- Ray Optics And Optical Instruments : NCERT Solutions – Class 12 Physics
- Nuclei : NCERT Solutions – Class 12 Physics
- Moving Charges And Magnetism : NCERT Solutions – Class 12 Physics
- Magnetism And Matter : NCERT Solutions – Class 12 Physics
- Electromagnetic Induction : NCERT Solutions – Class 12 Physics
- Dual Nature Of Radiation And Matter : NCERT Solutions – Class 12 Physics
- Current Electricity : NCERT Solutions – Class 12 Physics
- Communication Systems : NCERT Solutions – Class 12 Physics
- Atoms : NCERT Solutions – Class 12 Physics
- Alternating Current : NCERT Solutions – Class 12 Physics
Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 5)
Exercise 5.5
Differentiate the functions with respect to in Exercise 1 to 5.
1.
Ans. Let ……….(i)
Taking logs on both sides, we have
=
[From eq. (i)]
2.
Ans. Let =
……….(i)
Taking logs on both sides, we have
[From eq. (i)]
3.
Ans. Let ……….(i)
Taking logs on both sides, we have
[By Product rule]
=
4.
Ans. Let
Putting and
……….(i)
Now,
=
……….(ii)
Again,
……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
5.
Ans. Let …….(i)
Taking logs on both sides, we have
[From eq. (i)
Differentiate the functions with respect to in Exercise 6 to 11.
6.
Ans. Let
Putting and
……….(i)
Now
= ……….(ii)
Again
……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
7.
Ans. Let =
where
and
……….(i)
Now
……….(ii)
Again
……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
8.
Ans. Let =
where
and
……….(i)
Now
……….(ii)
Again
=
= ……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
9.
Ans. Let
Putting and
, we get
……….(i)
Now
=
…..(ii)
Again
=
……….(iii)
Putting values from eq. (ii) and (iii) in eq. (i),
10.
Ans. Let
Putting and
, we have
……….(i)
Now
=
……….(ii)
Again
……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i),
11.
Ans. Let
Putting and
, we have
……….(i)
Now
=
……….(ii)
Again
=
……….(iii)
Putting the values from eq. (ii) and (iii) in eq. (i)
,
Find in the following Exercise 12 to 15
12.
Ans. Given:
where
and
……….(i)
Now
……….(ii)
Again
……….(iii)
Putting values from eq. (ii) and (iii) in eq. (i),
13.
Ans. Given:
14.
Ans. Given:
15.
Ans. Given:
16. Find the derivative of the function given by and hence
Ans. Given: ……….(i)
Putting the value of from eq. (i),
= 8 x 15 = 120
17. Differentiate in three ways mentioned below:
(i) by using product rule.
(ii) by expanding the product to obtain a single polynomial
(iii) by logarithmic differentiation.
Do they all give the same answer?
Ans. Let ……….(i)
(i)
……….(ii)
(ii)
……….(iii)
(iii)
[From eq. (i)]
……….(iv)
From eq. (ii), (iii) and (iv), we can say that value of is same obtained by three different methods.
18. If and
are functions of
then show that
in two ways – first by repeated application of product rule, second by logarithmic differentiation.
Ans. Given: and
are functions of
To prove:
(i) By repeated application of product rule:
L.H.S. =
=
=
=
=
=
= R.H.S Hence proved.
(ii) By Logarithmic differentiation:
Let
Putting , we get
Hence proved.