Ncert Solutions Maths class 12th

Let f(x) = 8 sin x - sin 2x

$f\text{'}\left(x\right)=8cosx-2cos2x$              

$f\text{'}\text{'}\left(x\right)=-8sinx+4sin2x$               

= -8 (sin x) (1 – cos x)

f’’(x) < 0, f’(x) is a decreasing function

  ^{$f\text{'}\left(\frac{\pi }{3}\right)<f\text{'}\left(x\right)<f\text{'}\left(\frac{\pi }{a}\right)$}               

  $5<f\text{'}\left(x\right)<\frac{8}{\sqrt[]{2}}$

${\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}dx<{\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}\frac{f\left(x\right)}{x}<{\displaystyle \underset{\pi /4}{\overset{\pi /3}{\int }}\frac{8}{\sqrt[]{2}}}}}$

$5\pi 12<l<\frac{8}{\sqrt[]{2}}.\frac{\pi }{\sqrt[]{2}}$        

Find RHL, x = -1 + h

$\underset{h\to 0}{lim}a.sin\left(\frac{\pi \left[-1+h\right]}{2}\right)+\left[2+1-h\right]=-a+2$               

 Find LHL, x = -1 – h

$\underset{h\to 0}{lim}asin\frac{\left(\pi \left[-1-h\right]\right)}{2}=\left[2+1+h\right]=3$               

${\displaystyle \underset{0}{\overset{4}{\int }}f\left(x\right)dx={\displaystyle \underset{0}{\overset{1}{\int }}1dx+{\displaystyle \underset{1}{\overset{2}{\int }}\left(-1\right)dx+{\displaystyle \underset{2}{\overset{3}{\int }}\left(-1\right)dx+{\displaystyle \underset{3}{\overset{4}{\int }}\left(-1\right)dx=-2}}}}}$               

$V={\left|z\right|}^2+{\left|z-3\right|}^2+{\left|z-6i\right|}^2$  

Put z = x + iy

$V=3\left[{\left(x-1\right)}^2+{\left(y-2\right)}^2+10\right]$                

For minimum x = 1, y = 2, V₀ = 30

So, z₀ = 1 + 2i

f (a) = a, a is max of powers of prime P such that P^a divides a.

f (2) = 1; g (2) = 3

f (3) = 1; g (3) = 4

f (4) = 2; g (4) = 5

f (5) = 1; g (5) = 6

f (2) + g (2) = 4

f (3) + g (3) = 5

f (4) + g (4) = 7

f (5) + g (5) = 7

So, f (x) + g (x) is many-one-and an into function. We won't get 'I' in the range.

  $\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-2\left(3a-4b+12c+19\right)}{3^2+{\left(-4\right)}^2+12^2}$

$\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-6a+8b-24c-38}{169}$               

$\left(x,y,z\right)\equiv \left(a-6,\beta ,\gamma \right)$               

  $\frac{\left(a-b\right)-a}{3}=\frac{\beta -b}{-4}=\frac{\gamma -c}{12}=\frac{-6a+8b-24c-38}{169}$              

  $\frac{\beta -b}{-4}=-2$              

$\Rightarrow \beta =8+b$               Views|Posted a year ago