Ncert Solutions Maths class 11th

x² - |x| - 12 = 0
Case 1: x ≥ 0, |x| = x
x² - x - 12 = 0
(x-4) (x+3) = 0, x=4 (x=-3 is rejected)
Case 2: x < 0, |x| = -x
x² + x - 12 = 0
(x+4) (x-3) = 0, x=-4 (x=3 is rejected)
Two real solutions: 4 and -4.

x² - 4xy – 5y² = 0
Equation of pair of straight line bisectors is (x²-y²)/ (a-b) = xy/h
(x²-y²)/ (1- (-5) = xy/ (-2)
(x²-y²)/6 = xy/ (-2)
x²-y² = -3xy
x² + 3xy - y² = 0

$f\left(x\right)=x^4-4x+1=0$

$f\text{'}\left(x\right)=4x^3-4$

$=4\left(x-1\right)\left(x^2+1+x\right)$

$\Rightarrow$ Two solution

Given $ta{n}^{-1}a+t{a}^{-1}b=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ ${}^{}{}^{}\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.$

$ta{n}^{-1}\left(\frac{a+b}{1-ab}\right)=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

$\frac{a+b}{1-ab}=1...........\left(i\right)$

OR a + b = 1 – ab .(ii)

Now, $\left(a+b\right)-\left(\frac{a^2+b^2}{2}\right)+\left(\frac{a^3+b^3}{3}\right)-\left(\frac{a^4+b^4}{4}\right)+........$ $\frac{{}^{}{}^{}}{}\frac{{}^{}{}^{}}{}\frac{{}^{}{}^{}}{}$

$\left(a-\frac{a^2}{2}+\frac{a^3}{3}+\frac{a^3}{3}-\frac{a^4}{4}+......\right)+\left(b-\frac{b^2}{2}+\frac{b^3}{3}-\frac{b^4}{4}+.......\right)$

log (1 + a) + log(1 + b) = log (1 + a) (1 + b) = log {1 + a + b + ab} = log_e2

$P{A}^2+P{B}^2=co{s}^2\theta +{\left(sin\theta -3\right)}^2+co{s}^2\theta +{\left(sin\theta +6\right)}^2$

= 2 cos² θ + 2 sin² θ + 6 sin q + 45

= 6 sin θ + 47

for maximum of PA² + PB², sin θ = 1

then P (1, 2)

Hence P, A & B will lie on a straight line.

Given    n = 2^x. 3^y. 5^z . (i)

$y+z=5\&\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

${\displaystyle \sum _{n=1}^{\infty }\frac{n^2+6n+10}{\left(2n+1\right)!}}$

put 2n + 1 = r, r = 3, 5, 7,.

so n = $\frac{r-1}{2}$

$Now{\displaystyle \sum _{r=\left(3,5,7,.....\right)}^{\infty }\frac{n^2+6n+10}{\left(2n+1\right)!}}={\displaystyle \sum _{r=\left(3,5,7.....\right)}^{\infty }\left(\frac{\frac{{\left(r-1\right)}^2}{4}+3r-3+10}{r!}\right)}={\displaystyle \sum _{r=\left(3,5,7,......\right)}^{\infty }\left(\frac{r^2+10r+29}{4.r!}\right)}$

$=\frac{1}{8}\{41e-\frac{19}{e}-80)=\frac{41}{8}e-\frac{19}{8}{e}^{-1}-10$

$h=\frac{cos\theta +3}{2}$

=> $k=\frac{sin\theta +2}{2}$

=> $cos\theta =2h-3\&sin\theta =2k-2$

${\left(h-3/2\right)}^2+{\left(k-1\right)}^2=\frac{1}{4}$

$\therefore$ circle of radius r = $\frac{1}{2}$

Given $P_n={\alpha }^n+{\beta }^n,P_{n-1}=11\&P_{n+1}=29$

    $P_n={\alpha }^{n-2}.{\alpha }^2+{\beta }^{n-2}.{\beta }^2.........\left(i\right)$                          

Now quadratic equation having roots α & β will be x² – (α + β)x + αβ = 0

i.e.         x² – x – 1 = 0,     put x = α and put x = β

So          α² = α + 1           & β² = β + 1

(i)     $P_n={\alpha }^{n-2}\left(\alpha +1\right)+{\beta }^{n-2}\left(\beta +1\right)$

$P_n=P_{n-1}+P_{n-2}$          

= > ${P_n}^2=234$