Ncert Solutions Maths class 11th

$a^2\left(e^2-1\right)=b^2$

$e=\sqrt[]{\frac{5}{2}}\Rightarrow b^2=\frac{3a^2}{2}$

Length of latus rectum $\frac{2b^2}{a}=6\sqrt[]{2}$  

$3a=6\sqrt[]{2}\Rightarrow a=2\sqrt[]{2}$  

$b=2\sqrt[]{3}$  

y = 2x + c is tangent to hyperbola

$\therefore c^2=a^2{m}^2-b^2=20$  

$f\left(x\right)=x^7+5x^3+3x+1$

$f\text{'}\left(x\right)=7x^6+15x^4+3>0\forall x\in R$

$\therefore f\left(x\right)$ is increasing

 for x $\infty$ - , f (x) $\infty$

x = 0, f (x) = 1

$\therefore f\left(x\right)$ = 0 has only one real root.

Let $\frac{A_2}{A_1}=\frac{A_3}{A_2}=...=r$  

$\therefore A_1{A}_3{A}_5{A}_7=\frac{1}{1296}$

$\Rightarrow A_1{r}^3=\frac{1}{6}.......\left(i\right)$  

Again, A₂ + A₄ = $\frac{7}{36}$  

$A_{1}r=\frac{7}{36}-\frac{1}{6}=\frac{1}{36}........\left(ii\right)$

$\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$  

  $\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$  

Sum of digits

1 + 2 + 3 + 5 + 6 + 7 = 24

So, either 3 or 6 rejected at a time

Case 1 Last digit is 2

……….2

no. of cases = 

  Case 2 Last digit is 6

……….6

= 4! = 24

Total cases = 72

sin x = 1 – sin² x

sin x =  $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$  

draw y = sin x

y =  $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.

First common term to both AP's is 9

t₇₈ of $\left(3,6,9,......\right)=78*3=234$  

t₅₉ of $\left(5,9,13,........\right)=5+\left(5-1\right)4=237$  

n^th common term $\le 234$  

9 + (n – 1) 12   234

n <  $\frac{237}{12}\Rightarrow n=19$  

Now sum of 19 terms with a = 9, d = 12

$=\frac{19}{2}\left(2.9+\left(19-1\right)\cdot 12\right)=2223$  

Slope of AH = $\frac{a+2}{1}$ slope of BC = $-\frac{1}{p}\therefore p=a+2$ $\therefore C\left(\frac{18p-30}{p+1},\frac{15p-33}{p+1}\right)$  

slope of HC =  $\frac{16p-p^2-31}{16p-32}$  

slope of BC * slope of HC = -1 Þ p = 3 or 5

hence p = 3 is only possible value.

$x^8-x^7-x^6+x^5+3x^4-4x^3-2x^2+4x-1=0$  

$\Rightarrow x^7\left(x-1\right)-x^5\left(x-1\right)+3x^3\left(x-1\right)-x\left(x^2-1\right)+2x\left(1-x\right)+\left(x-1\right)=0$  

$\Rightarrow \left(x-1\right)\left(x^2-1\right)\left(x^5+3x-1\right)=0\therefore x=\pm 1$  are roots of above equation and x⁵ + 3x – 1 is a monotonic term hence vanishes at exactly one value of x other then 1 or -1.

 3 real roots.