Maths

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

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

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

(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

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Let point on ellipse (2sinθ, 3cosθ) and the mid point of line segment joining (-3, -5) and

(2 sin θ, 3cosθ) will be (h, k)

$\frac{2sin\theta -3}{2}=h\frac{3cos\theta -5}{2}=k$

sin² θ + cos² θ = 1

${\left(\frac{2h+3}{2}\right)}^2+{\left(\frac{2k+5}{3}\right)}^2$ = 1

$36x^2+16y^2+108x+80y+145=0$

$S_n={\displaystyle \sum _{r=1}^{n-1}r\left(n-r\right)}$

$=n\sum r-\sum r^2$

$=n\frac{\left(n-1\right)n}{2}-\frac{\left(n-1\right)n\left(2n-1\right)}{6}=\frac{n^3-n}{6}$

$S={\displaystyle \sum _{n=4}^{\infty }\left(\frac{2}{6}.\frac{n^3-n}{n!}-\frac{1}{\left(n-2\right)!}\right)=\frac{1}{3}}{\displaystyle \sum _{n=4}^{\infty }\frac{1}{\left(n-3\right)!}=\frac{e-1}{3}}$