NCERT solution Class 12 application of derivatives

NCERT Solutions

Class 12 Maths

Application of Derivatives

NCERT Questions

Ex.6.1 Q.1
Ex.6.1 Q.2
Ex.6.1 Q.3
Ex.6.1 Q.4
Ex.6.1 Q.5
Ex.6.1 Q.6
Ex.6.1 Q.7
Ex.6.1 Q.8
Ex.6.1 Q.9
Ex.6.1 Q.10
Ex.6.1 Q.11
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Ex.6.1 Q.14
Ex.6.1 Q.15
Ex.6.1 Q.16
Ex.6.1 Q.17
Ex.6.1 Q.18
Ex.6.2 Q.1
Ex.6.1 Q.2
Ex.6.2 Q.3
Ex.6.2 Q.4
Ex.6.2 Q.5
Ex.6.2 Q.6
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Ex.6.2 Q.8
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Ex.6.2 Q.10
Ex.6.2 Q.11
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Ex.6.2 Q.13
Ex.6.2 Q.14
Ex.6.2 Q.15
Ex.6.2 Q.16
Ex.6.2 Q.17
Ex.6.2 Q.18
Ex.6.2 Q.19
Ex.6.3 Q.1
Ex.6.3 Q.2
Ex.6.3 Q.3
Ex.6.3 Q.4
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Ex.6.3 Q.6
Ex.6.3 Q.7
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Ex.6.3 Q.24
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Ex.6.3 Q.26
Ex.6.3 Q.27
Ex.6.4 Q.1
Ex.6.4 Q.2
Ex.6.4 Q.3
Ex.6.4 Q.4
Ex.6.4 Q.5
Ex.6.4 Q.6
Ex.6.4 Q.7
Ex.6.4 Q.8
Ex.6.4 Q.9
Ex.6.5 Q.1
Ex.6.5 Q.2
Ex.6.5 Q.3
Ex.6.5 Q.4
Ex.6.5 Q.5
Ex.6.5 Q.6
Ex.6.5 Q.7
Ex.6.5 Q.8
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Ex.6.5 Q.11
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Ex.6.5 Q.13
Ex.6.5 Q.14
Ex.6.5 Q.15
Ex.6.5 Q.16
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Ex.6.5 Q.18
Ex.3.5 Q.19
Ex.6.5 Q.20
Ex.6.5 Q.21
Ex.6.5 Q.22
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Ex.6.5 Q.24
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Ex.Misc.Q.1
Ex.Misc.Q.2
Ex.Misc.Q.3
Ex.Misc.Q.4
Ex.Misc.Q.5
Ex.Misc.Q.6
Ex.Misc.Q.7
Ex.Misc.Q.8
Ex.Misc.Q.9
Ex.Misc.Q.10
Ex.Misc.Q.11
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Ex.Misc.Q.13
Ex.Misc.Q.14
Ex.Misc.Q.15
Ex.Misc.Q.16
Ex.Misc.Q.17
Ex.Misc.Q.18
Ex.Misc.Q.19
Ex.Misc.Q.20
Ex.Misc.Q.21
Ex.Misc.Q.22
Ex.Misc.Q.23
Ex.Misc.Q.24
Ex.2.2 Q.21

Ex.6.3 Q.18

For the curve y = 4x³ − 2x⁵, find all the points at which the tangents passes through the origin.

The equation of the given curve is y = 4x³ − 2x⁵

So, = 12x² − 10x⁴

Therefore, the slope of the tangent at a point (x, y) is 12x² − 10x⁴

The equation of the tangent at (x, y) is given by,

Y – y = (12x² − 10x⁴)(X – x) …………1

When the tangent passes through the origin (0, 0), then X = Y = 0

Therefore, equation 1 reduces to:

-y = (12x² − 10x⁴)(0 – x)

⟹ y = 12x³ − 10x⁵

Also, we have y = 4x³ − 2x⁵

So, 12x³ − 10x⁵ = 4x³ − 2x⁵

⟹ 8x⁵ − 8x³ = 0

⟹ x⁵ − x³ = 0

⟹ x³(x³ - 1) = 0

⟹ x = 0, ±1

When x = 1, y = 4 (0)³ − 2 (0)⁵ = 0

When x = 1, y = 4 (1)³ − 2 (1)⁵ = 4 – 2 = 2

When x = −1, y = 4 (−1)³ − 2 (−1)⁵ = -4 + 2 = -2

Hence, the required points are (0, 0), (1, 2), and (−1, −2).

Video solution:

Application of Derivatives Class 12 Maths NCERT Chapter 6 NCERT Solutions

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