Ncert Solutions Maths class 11th

x⁴ + x² + 1 = 0

x⁴ + 2x² + 1 – x² = 0

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

$x=\pm \omega ,?{\omega }^2$

Now, = ${\alpha }^{1011}+{\alpha }^{2022}-{\alpha }^{3033}$

= ${\omega }^{1011}+{\omega }^{2022}-{\omega }^{3033}$

= 1 + 1 – 1 = 1

 $\frac{dy}{dx}=\frac{1}{1+sin2x}$

$dy=\frac{se{c}^{2}xdx}{{\left(1+tanx\right)}^2}$

$\Rightarrow y=-\frac{1}{1+tanx}+c$

When

$x=\frac{\pi }{4},y=\frac{1}{2}$ gives c = 1

So

$x+\frac{\pi }{4}=\frac{5\pi }{6}or\frac{13\pi }{6}\Rightarrow x=\frac{7\pi }{12}or\frac{23\pi }{12}$

sum of all solutions =

$\pi +\frac{7\pi }{12}+\frac{23\pi }{12}=\frac{42\pi }{12}$

Hence k = 42

Each element of ordered pair (i, j) is either present in A or in B.

So, A + B = Sum of all elements of all ordered pairs {i, j} for $1\le i\le 10$ and $1\le j\le 10$

= 20 (1 + 2 + 3 + … + 10) = 1100

Sum of all elements of $A\cap B=2$  [Sum of natural number upto 100 which are neither divisible by 3 nor by 5]

$=2\left[\frac{100*101}{2}-3\left(\frac{33*34}{2}\right)-5\left(\frac{20*21}{2}\right)+15\left(\frac{6*7}{2}\right)\right]$

= 10100 – 3366 – 2100 + 630

= 5264

$\left(sin10°.sin50°.sin70°\right).\left(sin10°.sin20°.sin40°\right)$

$=\left(\frac{1}{4}sin30°\right).\left[\frac{1}{2}sin10°\left(cos20°-cos60°\right)\right]$

$=\frac{1}{32}\left[sin30°-sin10°-sin10°\right]$

$\frac{1}{64}-\frac{1}{16}sin10°$

Clearly $\alpha =\frac{1}{64}$

Hence 16 + a^-1 = 80

 $\left(4+x^2\right)dy-2x\left(x^2+3y+4\right)dx=0$

$\Rightarrow \frac{dy}{dx}=\left(\frac{6x}{x^2+4}\right)y+2x$

$e^{-3\mathcal{l}n\left(x^2+4\right)}=\frac{1}{{\left(x^2+4\right)}^3}$

So

$\frac{y}{{\left(x^2+4\right)}^3}={\displaystyle \int \frac{2x}{{\left(x^2+4\right)}^3}dx+c}$

$\Rightarrow y=-\frac{1}{2}\left(x^2+4\right)+c{\left(x^2+4\right)}^3$

When x = 0, y = 0 gives

$c=\frac{1}{32}$

So, for x = 2, y = 12

$\therefore {\displaystyle \underset{-6}{\overset{0}{\int }}f\left(x\right)dx=2*\frac{1}{2}\left(2+5\right)*3=21}$

Let y = mx + c is the common tangent

$soc=\frac{1}{m}=\pm \frac{3}{2}\sqrt[]{1+m^2}\Rightarrow m^2=\frac{1}{3}$

so equation of common tangents will be

$y=\pm \frac{1}{\sqrt[]{3}}x\pm \sqrt[]{3}$

which intersects at Q (3, 0)

Major axis and minor axis of ellipse are 12 and 6. So eccentricity

$e^2=1-\frac{1}{4}=\frac{3}{4}$

and length of latus rectum

$=\frac{2b^2}{a}=3$

Hence

$\frac{\mathcal{l}}{e^2}=\frac{3}{3/4}=4$

 $f\left(x\right)=2co{s}^{-1}x+4co{t}^{-1}x-3x^2-2x+10\forall x\in \left[-1,1\right]$

$\Rightarrow f\text{'}\left(x\right)=-\frac{2}{\sqrt[]{1-x^2}}-\frac{4}{1+x^2}-6x-2<0\forall x\in \left[-1,1\right]$

So, f (x) is decreasing function and range of f (x) is

$\left[f\left(1\right),f\left(-1\right)\right],$ which is $\left[\pi +5,5\pi +9\right]$

Now 4a – b = 4 ( + 5) (5 + 9) = 11 - π

Pascal's Triangle is a triangular array of binomial coefficients where each number is the sum of the two numbers directly above it, and its rows provide the coefficients for binomial expansions according to the Binomial Theorem. This le man language for defition of the passcl triangle.