Maths NCERT Exemplar Solutions Class 11th Chapter Seven

Kindly consider the following Image 

(P? ¹AP - I)²
= (P? ¹AP - I) (P? ¹AP - I)
= P? ¹A (PP? ¹)AP - P? ¹AP - P? ¹AP + I
= P? ¹A²P - 2P? ¹AP + I
= P? ¹ (A² - 2A + I)P = P? ¹ (A - I)²P
| (P? ¹AP - I)²| = |P? ¹ (A - I)²P| = |P? ¹| | (A - I)²| |P| = | (A - I)²| = |A - I|²
A - I = [1, 7, w²], [-1, w², 1], [0, -w, -w]
|A - I| = 1 (-w³ + w) - 7 (w) + w² (w) = -w³ + w - 7w + w³ = -6w.
|A - I|² = (-6w)² = 36w².

Let A = [a? ]? Sum of diagonal elements of A.A? is Tr (A.A? ) = ∑? ∑? a? ² = 9.
where each a? ∈ {0, 1, 2, 3}.
Case I: One of a? = 3 and rest are 0. (3²=9). There are? C? = 9 ways.
Case II: Two of a? are 2, one is 1, and rest are 0. (2² + 2² + 1² = 9). There are? C? *? C? = 36 * 7 = 252 ways.
Case III: One of a? = 2, five are 1, and rest are 0. (2² + 1²+1²+1²+1²+1² = 9). There are? C? *? C? = 9 * 56 = 504 ways.
Case IV: All nine a? = 1. (1² * 9 = 9). There is 1 way.
Total = 9 + 252 + 504 + 1 = 766.

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.

f (x) + f (x + 1) = 2 (1)
replace x with x + 1: f (x + 1) + f (x + 2) = 2 (2)
(2) - (1) ⇒ f (x + 2) = f (x)
∴ f (x) is periodic with period 2.
I? = ∫? f (x)dx = 4 ∫? ² f (x)dx.
I? = ∫? ³ f (x)dx = ∫? f (u-1)du. Let u = x+1.
I? = ∫? f (x-1)dx = 2 ∫? ² f (x-1)dx.
From (1), f (x-1) + f (x) = 2.
I? + 2I? = 4∫? ² f (x)dx + 2 (2∫? ² f (x-1)dx) = 4∫? ² f (x)dx + 4∫? ² (2 - f (x)dx
= 4∫? ² (f (x) + 2 - f (x)dx = 4∫? ² 2 dx = 4 [2x] from 0 to 2 = 16.

y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
a = 1/2 or a = 7.
a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
6b = [4t³ - 45t² + 60t] from 0 to 7 = 4 (7)³ - 45 (7)² + 60 (7) = 1372 - 2205 + 420 = -413.
|a + 6b| = |7 - 413| = |-406| = 406.