Ncert Solutions Maths class 11th

2x + y = 4
2x + 6y = 14
} y=2, x=3
B (1, 2)
Let C (k, 4–2k)
Now AB² = AC²
=> 5k² – 24k + 19 = 0
α = (6+1+10/5)/3 = 18/5
Now 15 (α+β)
15 (17/5) = 51

f (x) = x? – 4x + 1 = 0
f' (x) = 4x³ – 4
= 4 (x–1) (x²+1+x)
=> Two solution

dy/dx = (ax-by+a)/ (bx+cy+a)
=> bxdy + cydy + ady = axdx – bydx + adx
cy²/2 + ay – ax²/2 – ax + bxy = k
ax² + ay² + 2ax – 2ay = k
=> x² + y² + 2x – 2y = λ
Short distance of (11,6)
= √12²+5² – 5
= 13 – 5
= 8

x = Σ a? = 1/ (1-a); y = Σ b? = 1/ (1-b); z = Σ c? = 1/ (1-c)

Now,
a, b, c → AP
1-a, 1-b, 1-c → AP
1/ (1-a), 1/ (1-b), 1/ (1-c) → HP
x, y, z → HP

z? = iz²
Let z = x + iy
x – iy = I (x² – y² + 2xiy)

Case-I
x = 0
–y² = –y
y = 0, 1

Case - II
y = – 1/2
=> x² – 1/4 = 1/2 => x = ±√3/2

Area of polygon
= 1/2 | (0, 1, 1), (√3/2, -1/2, 1), (-√3/2, -1/2, 1) |
= 1/2 | -√3/2 - √3/2 | = 3√3/4

Kindly consider the following Image 

lim (x→0) [a e? - b cos (x) + c e? ] / (x sin (x) = 2
Using Taylor expansions around x=0:
lim (x→0) [a (1+x+x²/2!+.) - b (1-x²/2!+.) + c (1-x+x²/2!+.)] / (x * x) = 2
lim (x→0) [ (a-b+c) + x (a-c) + x² (a/2+b/2+c/2) + O (x³)] / x² = 2
For the limit to exist, the coefficients of lower powers of x in the numerator must be zero.
a - b + c = 0
a - c = 0 ⇒ a = c
Substituting a=c into the first equation: 2a - b = 0 ⇒ b = 2a.
The limit becomes: lim (x→0) [x² (a/2 + b/2 + c/2)] / x² = (a+b+c)/2
(a + b + c) / 2 = 2 ⇒ a + b + c = 4.

|z+i|/|z-3i| = 1 ⇒ |z+i| = |z-3i|. This means z is on the perpendicular bisector of the segment from -i to 3i. The midpoint is i, so z = x+i.
w = z? - 2z + 2. Let z = x + iy.
w = (x² + y²) - 2 (x + iy) + 2 = (x² - 2x + 2 + y²) - 2iy.
Re (w) = x² - 2x + 2 + y² = (x - 1)² + 1 + y².
From the first condition, y=1. Re (w) = (x - 1)² + 1 + 1 = (x - 1)² + 2.
Re (w) is minimum for x = 1.
The common z is z = 1 + i.
w = (1+i) (1-i) - 2 (1+i) + 2 = 2 - 2 - 2i + 2 = 2 - 2i.
w² = (2 - 2i)² = 4 (1 - 2i - 1) = -8i.
w? = (-8i)² = -64 ∈ R.
∴ least n ∈ N for which w? ∈ R is n=4.