Ncert Solutions Maths class 11th

Truth table analysis shows that (P ∨ Q) ∧ (¬P) is equivalent to Q ∧ ¬P.
Then (Q ∧ ¬P) ⇒ Q. This is a tautology.
The provided solution seems to have an error.
Let's check the options. (P ∨ Q) is a tautology. (P ∧ ¬Q) is a contradiction.
~ (P ⇒ Q) ⇔ P ∧ ¬Q is true.

sinθ + cosθ = 1/2
16 (sin (2θ) + cos (4θ) + sin (6θ)
= 16 [2sin (4θ)cos (2θ) + cos (4θ)]
= 16 [4sin (2θ)cos² (2θ) + 2cos² (2θ) - 1] . (i)
Now, sinθ + cosθ = 1/2, squaring on both sides, we get
1 + sin (2θ) = 1/4
sin (2θ) = -3/4
cos² (2θ) = 1 - sin² (2θ) = 1 - 9/16 = 7/16
From equation (i)
16 [4 (-3/4) (7/16) + 2 (7/16) - 1]
16 [-21/16 + 14/16 - 16/16] = 16 [-23/16] = -23

S? : |z - 3 - 2i|² = 8
|z - (3 + 2i)| = 2√2
(x - 3)² + (y - 2)² = (2√2)²
S? : Re (z) ≥ 5
x ≥ 5
S? : |z - z? | ≥ 8
|2iy| ≥ 8
2|y| ≥ 8
|y| ≥ 4
y ≥ 4 or y ≤ -4
From the graph of the circle (S? ) and the regions (S? and S? ), we can see that there is one point of intersection at (5, 4).
∴ n (S? ∩ S? ∩ S? ) = 1

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