Ncert Solutions Maths class 12th

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2 months ago

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P
Payal Gupta

Contributor-Level 10

y = 5x2 + 2x – 25

P(2, -1)

T(p) : T = 0

Þ y – 1 = 10x(2) + 2(x + 2) – 50

y = 2 2 x 4 5 is also tangent to y = x3 – x2 + x at point (a, b)

For y = x3 -x2 + x

d y d x = 3 x 2 2 x + 1 > 0               

y = 22x 1 9 3 9 2 7  which is not tangent to the curve.

3 a 2 2 a + 1 = 2 2 (slope of tangent)

3 a 2 2 a 2 1 = 0 a = 2 ± 4 + 2 5 2 6 = 2 ± 1 6 6 = 3 , 7 3               

b = 27 – 9 + 3 = 21

tangent : y – 21 = 22(x – 3)

 ⇒ y = 22x – 45

a = 3, b = 21

2a + 9b = 6 + 189 = 195

Also, a 3 a 2 + a = b  

For a = 7 3  

b = 3 4 3 2 7 4 9 9 7 3  

= 3 4 3 1 4 7 6 3 2 7 = 5 5 3 2 7  

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

f(x)=4|2x+3|+9[x+12]12[x+20],20<x<20 doubtful points for differentiability : x=32,

f(x)=4(2x+3)+9(1)12(2)240=8x213forx=32+h

8x230 for x = 32h

Not diff. at x = 32

other doubtful points : x+12=integer

20+12<x+12<20+12

x+12=19,18,....,19,20

x=19.5,18.5,17.5,.......,18.5,19.5 total 40 numbers.

No. of number = 19.5 – (19.5)+1=40(1.5)included

20<x<90x=19,18,.....,18,1539points

No. of number = 19 (19) + 1 = 39

Total : 40 + 39 = 79

New question posted

2 months ago

0 Follower 2 Views

New answer posted

2 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

rn?Cr=nn1?Cr1

k=110(k10?Ck)2=k=110(109?CK1)2

10018?C9

22000L = 10018?C9

L =

18!=2163853721111317

9+4+2+1 9!=27345171

6 + 24 + 2 + 1 18!(9!)2=225111317

3 +3 + 1

= 221

New answer posted

2 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

A=[123016001]

A2=[123016001][123016001]

=[106010001]=I+B,B=[006000000]

A4=[106010001][106010001]B2[000000000]=0

A2n=(I+B)n

I + nB + 0 + 0 +……….

A2n=I+[006n000000]=[106n010001]

X'Akx=[33]

=[111]Ak[111]

[111]A2n[111](k=2n)

[1+1+6n+1]=[6n+3]=[33]n=5

k = 10

New question posted

2 months ago

0 Follower 2 Views

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

l i m x 1 0 ( ( x + 2 c o s x ) 3 + 2 ( x + 2 c o s x ) 2 + 3 s i n ( x + 2 c o s x ) ( x + 2 ) 3 ± 2 ( x + 2 ) 2 + 3 s i n ( x + 2 ) ) x

e 1 0 0 1 6 + 3 s i n 2 [ 1 2 3 ( 4 ) + 8 * 1 8 + 3 c o s 2 3 c o s 2 1 ] , using L.H.L Rule, e0 = 1

2 x 2 1 2 x + 7 = 0 o r 3 x 2 + 5 x + 1 2 = x 2 + 3 x + 1 0

x 2 + x + 1 = 0

Sum of roots = 6, no solution.

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

f ( x ) = 0 ( x p ) 2 q = 0  

Roots are  p + q , p q

Now, | f ( a i ) | = 5 0 0  

L e t a 1 , a 2 , a 3 . . . a r a , a + d , a + 2 d , a + 3 d               

=>  9 4 d 2 q = 5 0 0           …….(i)

and | f ( a 1 ) | 2 = | f ( a 2 ) | 2

From equation (i)

9 4 . 4 q 5 q = 5 0 0     

4 q 5 = 5 0 0           

and  2 q = 2 * 5 0 2 = 5 0

New answer posted

2 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

f ' ( a ) = f ' ( b ) = f ' ( c ) = 2

f'' (x) is zero for atleast x 1 ( a , b ) & x 2 ( b , c )

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

R S ( α , , 1 β )

D R o f P Q ( 5 6 1 7 + 2 , 4 3 1 7 + 1 , 1 1 1 1 7 1 )

( 9 0 1 7 , 6 0 1 7 , 9 4 1 7 )

9 0 α + 9 4 β = 6 0

β = 1 5 , α = 1 5 α 2 + β 2 = 4 5 0

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