Maths

Total 4 digit number

4 digit divisible by 7

1001, 1008, -9996

9996 = 1001 + (n₁ – 1) 7

n₁ = 1286

4 digit no divisible by 3

1002, 1005, -9999

9999 = 1002 + (n₂ – 1)3

n₂ = 3000

4 digit number visible by 21

1008, 1031, -9996

n₃ = 429

4 digit number divisible by 7 or 3

= 9000 – 1286 – 3000 + 429

= 5143

$AP=\sqrt[]{9+1-6+4+1}$

AP = 3 = AQ

$r=\sqrt[]{1+4-1}=2$        

$tan\theta =\frac{3}{2}$  

$\frac{Areaof\Delta APQ}{Areaof\Delta BPQ}=\frac{AR}{RB}=\frac{3sin\theta }{2cos\theta }=\frac{9}{4}$

$f\left(2\right)=f\left(4\right)=0andf\left(x\right)=x^3-6x^2+ax+b$  

$\Rightarrow f\left(x\right)=\left(x-2\right)\left(x-4\right)x=x^3-6x^2+8x$               

->a = 8, b = 0

$f\text{'}\left(x\right)=0\Rightarrow x=2\pm \frac{2}{\sqrt[]{3}},x_4=2+\frac{2}{\sqrt[]{3}},f\left(x_4\right)=\frac{-16}{3\sqrt[]{3}}$           

$f\text{'}\left(x_3\right)=\frac{\sqrt[]{3}}{2}f\left(x_4\right)=\frac{-8}{3}$               

The tangent to the parabola

$y^2=8xat\left(2,-4\right)is-4y=4\left(x+2\right)$  

x + y + 2 = 0

$OA=\sqrt[]{a}$         

$\Rightarrow \left|\frac{0+0+2}{\sqrt[]{2}}\right|=\sqrt[]{a}$  

$\sqrt[]{2}=\sqrt[]{a}$      

a = 2

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

$=\frac{1}{2}{\left[x+2x^2-\frac{x^3}{3}\right]}_0^{3/2}=\frac{39}{8}$

${\displaystyle \underset{0}{\overset{3/2}{\int }}mx=\frac{9m}{8}}$              

(as per question)

$\Rightarrow \frac{39}{8}=\frac{9m}{4}$

$m=\frac{39}{18}$

12m = 26

Required number =   $\frac{4*2+24*3+36*4}{2}=112$

${\displaystyle \int \frac{sinx}{si{n}^{3}x+co{s}^{3}x}dx}$

$I={\displaystyle \int \frac{tanxse{c}^{2}x}{ta{n}^{3}x+1}dx}$

Let, tan x = t

sec² xdx = dt

${\displaystyle \int \frac{tdt}{t^3+1}={\displaystyle \int \frac{t}{\left(t+1\right)\left(t^2-t+1\right)}dt}}$    

$\frac{t}{\left(t+1\right)\left(t^2-t+1\right)}=\frac{A}{\left(t+1\right)}+\frac{Bt+c}{t^2-t+1}$

$l=\frac{-1}{3}{\displaystyle \int \frac{1}{t+1}dt+\frac{1}{3}{\displaystyle \int \frac{t+1}{t^2-t+1}dt}}$

The value of 18 $\left(\alpha +\beta +{\gamma }^2\right)$ is equal to 3.

$f\left(x\right)=a{x}^3+b{x}^2+cx+d$   

  $f\text{'}\left(x\right)=3a{x}^2+2bx+c$              

$f\text{'}\text{'}\left(x\right)=6ax+2b$

$f\text{'}\text{'}\left(-1\right)=0$       

-6a + 2b = 0

=> b = 3a

$f\text{'}\left(1\right)=0$      

 -5 + d = -10 . (i)

 f (-1) = 6

11a + d = 6 . (ii)

$a=1,d=-5,b=3,c=-9$

$f\left(3\right)=27+27-27-5=22$

f (x) = x + a sin x

$wherea={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cosyf\left(y\right)dy}$                

$a={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cosy\left(y+asiny\right)dy}$                

-> a = p - 2

(2, -2) lie in a plane

=>2 + 6 + 4 + β = 0

=> b = -12

Line is perpendicular to normal

α (1) – 5 (3) + 2 (-2) = 0

α = 19

α + β = 7