Ncert Solutions Maths class 12th

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New answer posted•a year ago
S
a year ago

f ( x + y ) = 2 f ( x ) f ( y ) , x , y ∈ N

f ( 1 ) = 2

∑ k = 1 1 0 f ( α + k ) = ∑ k = 1 1 0 2 f ( α ) f ( k )

= 2 f ( α ) ∑ k = 1 1 0 f ( k ) = 2 . 2 2 α − 1 . 2 3 ( 4 1 0 − 1 )

= 2 2 α + 1 3 . ( 4 1 0 − 1 )

⇒ 2 α + 1 = 9

= 4

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New answer posted•a year ago
S
a year ago

  f ( x ) = 2 x − 1         g : R − { 1 } → R       

  g ( x ) = x − 1 / 2 x − 1         

  f { g ( x ) } = 2 ( x − 1 / 2 x − 1 ) − 1 = 2 x − 1 − x + 1 x − 1 = x x − 1 , x ≠ 1          

y = x x − 1 ⇒ x = y y − 1 ∴ y ≠ 1           

One – one but not onto

...Read more

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Let total number of throws = n

Probability of getting 2 times = Probability of getting an even number 3 times.

   

[as probability of getting odd number = probability of getting even number = ]

Probability of getting an odd number for odd number of times =

 

     &

...Read more

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Let five terms in G.P. be a/r², a/r, a, ar, ar²
Then, a (r? ² + r? ¹ + 1 + r + r²) / (1/a) (r? ² + r? ¹ + 1 + r? ¹ + r? ²) = 49
⇒ a² = 49 ⇒ a = ±7
Also, a/r² + a = 35
Therefore, a = -7 is not possible
Now, fifth term = ar² = a (7/28) = p ⇒ 4p = 7

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New answer posted•a year ago
S
a year ago

lim (x→1) [f (1)g (x)-f (1)-g (1)f (x)+g (1)] / [f (1)g (x)-f (x)g (1)], form: 0/0
lim (x→1) [f (1)g' (x)-g (1)f' (x)] / [f (1)g' (x)-f' (x)g (1)] = 1

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New answer posted•a year ago
S
a year ago

| 1 -1 -1 |
| 1 -k | = 0
| k 2 1 |

⇒ 1 (1 + 2k) + 1 (1 + k²) – 1 (2 – k) = 0
2k + 1 + 1 + k² − 2 + k = 0
k² + 3k = 0
k = 0, -3

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New answer posted•a year ago
New answer posted•a year ago