Ncert Solutions Maths class 12th

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New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

(a) r = 3cm

When r = 3 cm,

ddr = 2 * π * 3 cm = 6π.

Thus, the area of the circle is changing at the rate of 6π cm cm25.

(b) r = 4cm

whenr = 4cm,

ddr = 2 * π * 4cm = 8π.

Thus, the area of the circle is changing at the rate of 8 cm25.

New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the solution

New answer posted

10 months ago

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V
Vishal Baghel

Contributor-Level 10

|x+2x+3x+2ax+3x+4x+2bx+4x+5x+2c|=0

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Vishal Baghel

Contributor-Level 10

Given equations are :-

2x+3y+10z=4

4x6y+5z=1

6x+9y20z=2

This system of equation can be written, in matrix form, as AX= B, Where

⇒ x = 2, y = 3, z = 5.

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

L . H . S . = | s i n α c o s α c o s ( α + δ ) s i n β c o s β c o s ( β + δ ) s i n γ c o s γ c o s ( γ + δ ) | = 1 s i n δ c o s δ | s i n α s i n δ c o s α c o s δ c o s ( α + δ ) s i n β s i n δ c o s β c o s δ c o s ( β + δ ) s i n γ s i n δ c o s γ c o s δ c o s ( γ + δ ) |

Applying cos(A+B)=cosAcosBsinAsinB in c3

c1c1+c3?=1sinδcosδ|cosαcosδcosαcosδcosαcosδsinαsinδcosβcosδcosβcosδcosβcosδsinβsinδcosγcosδcosγcosδcosγcosδsinγsinδ|c1=c2?=0=R.H.S.

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

L . H . S . = | 1 1 + p 1 + p + q 2 3 + 2 p 4 + 3 p + 2 q 3 6 + 3 p 1 0 + 6 p + 3 q | R 2 R 2 2 R 1 R 3 R 3 3 R 1 = | 1 1 + p 1 + p + q 0 1 2 + p 0 3 7 + 3 p | R 3 R 3 3 R 2 = | 1 1 + p 1 + p + q 0 1 2 + p 0 0 1 |

Expanding along c1

?=|12+p01|=1=R.H.S.

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

L . H . S . = | 3 a a + b a + c b + a 3 b b + c c + a c + b 3 c | c 1 c 1 + c 2 + c 3 = | a + b + c a + b a + c a + b + c 3 b b + c a + b + c c + b 3 c | ( a + b + c ) | 1 a + b a + c 1 3 b b + c 1 c + b 3 c |

R2R2R1and R3R3R1

=(a+b+c)|1a+ba+c03b+abb+a0a+b+ab3c+ac|

Expanding along c1

=(a+b+c)|2b+aabac2c+a|=(a+b+c)[4bc+2ab+2ac+a2+ac+abbc]=(a+b+c)(3ab+3bc+3ac)=3(a+b+c)(ab+bc+ac)=R.H.S.

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

L . H . S . = | x x 2 1 + p x 3 y y 2 1 + p y 3 z z 2 1 + p z 3 | = | x x 2 1 y y 2 1 z z 2 1 | + | x x 2 p x 3 y y 2 p y 3 z z 2 p z 3 | = ? 1 + ? 2 ( 1 ) N o w ? 2 = | x x 2 p x 3 y y 2 p y 3 z z 2 p z 3 | = p x y z | 1 x x 2 1 y y 2 1 z z 2 | c 1 c 3 = p x y z | x 2 x 1 y 2 y 1 z 2 z 1 | c 1 c 2 = p x y z | x x 2 1 y y 2 1 z z 2 1 | = p x y z ? 1

Putting value in (1)

?1+pxyz?1=(1+pxyz)?1(2)Now?1=|xx21yy21zz21|

R2R2R1 and R3R3R1

=|xx21yxy2x20zxz2x20|

Expanding along c3

?1=|(yx)(yx)(y+x)(zx)(zx)(z+x)|=(yx)(zx)|1y+x1z+x|=(yx)(zx)(z+xyx)=(xy)(yz)(zx)

Putting ?1 in (2)

L.H.S.=(1+pxyz)(xy)(yz)(zx)=R.H.S

New answer posted

10 months ago

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Vishal Baghel

Contributor-Level 10

  L . H . S . = | α α 2 β + γ β β 2 γ + α γ γ 2 α + β | c 3 c 3 + c 1 = | α α 2 α + β + γ β β 2 α + β + γ γ γ 2 α + β + γ | = ( α + β + γ ) | α α 2 1 β β 2 1 γ γ 2 1 |

R 2 R 2 R 1 and R 3 R 3 R 1

= ( α + β + γ ) | α α 2 1 β α β 2 α 2 0 γ α γ 2 α 2 0 |

Expanding along c 3

= ( α + β + γ ) | β α β 2 α 2 γ α γ 2 α 2 | = ( α + β + γ ) | β α ( β α ) ( β + α ) γ α ( γ α ) ( γ + α ) | ( α + β + γ ) ( β α ) ( γ α ) | 1 β + γ 1 γ + α | ( α + β + γ ) ( β α ) ( γ α ) { γ + 2 ( β γ ) } ( α + β + γ ) ( β α ) ( γ α ) ( γ β ) ( α + β + γ ) ( α β ) ( β γ ) ( γ α ) = R . H . S .

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