Ncert Solutions Maths class 11th

Kindly consider the following figure

  ${\displaystyle \underset{0}{\overset{1.5}{\int }}\left[x^2\right]dx={\displaystyle \underset{0}{\overset{1}{\int }}0dx+{\displaystyle \underset{1}{\overset{\sqrt[]{2}}{\int }}1dx+{\displaystyle \underset{\sqrt[]{2}}{\overset{1.5}{\int }}2dx=2-\sqrt[]{2}}}}}$

  $L=\underset{x\to \infty }{lim}\frac{{\left(2+\frac{1}{x}\right)}^{40}{\left(4-\frac{1}{x}\right)}^5}{{\left(2+\frac{3}{x}\right)}^{45}}=32$

Any point on x + y = 1 can be taken as (t, 1 – t) The equation of chord with this as mid-point is y (1 – t) -2a (x + t) = (1 – t)² – 4at

It passes through (a, 2a)

So, t² – 2t + 2a² – 2a + 1 = 0

This should have two distinct real roots.

So D > 0

$\begin{array}{l}\Rightarrow a^2-a<0\\ \Rightarrow 0<a<1\end{array}$          

So, length of latus rectum < 4 and 0 < a < 1

$\Rightarrow LR=1,2,3$            

Let P is a root   $\left(\in I\right)$

Case-I : p is odd

p¹⁰ + ap⁹ + b   $\ne 0$

Because LHS odd

Case-II : p is even

$p^{10}+a{p}^9+b\ne 0$  

because, LHs is odd

PQ is focal chord.
Quadrilateral PTQR is square.

Area = (PQ * TR) / 2 = (4 * 4) / 2 = 8

lim (x→0) (ae²? - bcosx + c) / (x sinx)
= lim (x→0) (ae²? - bcosx + c) / x² * lim (x→0) x/sinx
For limit to exist
a - b + c = 0
2a + b = 0
(4a-b)/2 = 1
c = 2, b = 2, a = -1
a+b+c = 3

Family of circle through (2, 2) and (9, 9), (x-2) (x-9) + (y-2) (y-9) + λ (y-x) = 0
Touches y=0 (x-axis)
(x-2) (x-9) + 18 - λx = 0
x² - 11x - λx + 36 = 0
x² - (11+λ)x + 36 = 0 has repeated roots
D = 0 ⇒ λ = 1, λ = -23
x = 6 or x = -6
Difference = 12

All even + 2 odd 1 even
¹? C? + ¹? C? * ¹? C?
(10*9*8)/6 + (10*9)/2 * 10
120 + 450 = 570

tan? ¹ (x) = t
t² - 4t + 3 > 0
t ∈ (-∞, 1) U (3, ∞)
tan? ¹ (x) ∈ (-π/2, 1)
x ∈ (-∞, tan1)
the largest integral x = 1