Ncert Solutions Maths class 12th

$f\left(x\right)=\left[x-1\right]cos\left(\frac{2x-}{2}\right)\pi$

$Ifx=k,k\in I$

then f(x) = 0 as $cos\left(\frac{2k-1}{2}\right)\pi =0,\forall k\in I$  

LHL = $\underset{h\to 0}{Lt}\left[k-h-1\right]cos\left(\frac{2k-2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-2\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$  

$RHL=\underset{h\to 0}{Lt}\left[k+h-1\right]cos\left(\frac{2k+2h-1}{2}\right)\pi =\underset{h\to 0}{Lt}\left(k-1\right)cos\left(\frac{2k-1}{2}\right)\pi \to 0$           

  $\therefore f\left(x\right)$  is continuous $\forall x\in R$  

$f\left(x+y\right)=2f\left(x\right)f\left(y\right),x,y\in N$

$f\left(1\right)=2$

${\displaystyle \sum _{k=1}^{10}f\left(\alpha +k\right)={\displaystyle \sum _{k=1}^{10}2f\left(\alpha \right)f\left(k\right)}}$

$=2f\left(\alpha \right){\displaystyle \sum _{k=1}^{10}f\left(k\right)=2.2^{2\alpha -1}.\frac{2}{3}\left(4^{10}-1\right)}$

$=\frac{2^{2\alpha +1}}{3}.\left(4^{10}-1\right)$

$\Rightarrow 2\alpha +1=9$

= 4

  $f\left(x\right)=2x-1g:R-\left\{1\right\}\to R$       

  $g\left(x\right)=\frac{x-1/2}{x-1}$         

  $f\left\{g\left(x\right)\right\}=2\left(\frac{x-1/2}{x-1}\right)-1=\frac{2x-1-x+1}{x-1}=\frac{x}{x-1},x\ne 1$          

$y=\frac{x}{x-1}\Rightarrow x=\frac{y}{y-1}\therefore y\ne 1$           

One – one but not onto

$\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=\frac{2\left(a+b+c\right)}{24}=\frac{c}{5}=\frac{a}{4}=\frac{b}{3}$

$r=\frac{\Delta }{S}=\frac{6k^2}{6k}=k$

$R=\frac{5k}{2}\Rightarrow \frac{R}{r}=\frac{5}{2}$

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 =

 

c = -5, a = 2, b = 1

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

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

| 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

General form Σ? ? C? (1/2)? < 1/2
Σ? ? C? < 2?
⇒? C? +? C? +.? C? < 128
⇒256− (? C? +? C? +.? C? ) < 128
⇒? C? +? C? +.? C? > 128
n≥5