Maths NCERT Exemplar Solutions Class 11th Chapter Five

I = ∫ (π/24 to 5π/24) dx/ (1+³√tan2x). Using King's rule.
2I = ∫ (π/24 to 5π/24) dx = 4π/24=π/6. I=π/12.

dy/dx - 1 = xe^ (y-x). Let y-x=t. dt/dx = xe? e? dt=xdx.
-e? = x²/2+C. y (0)=0⇒t=0⇒-1=C.
-e^ (x-y) = x²/2-1. y=x-ln (1-x²/2).
y'=1+x/ (1-x²/2)=0 ⇒ x=-1. Min value at x=-1.
y (-1)=-1-ln (1/2) = -1+ln2. This differs from solution.

Σ (1/a) (1-rb/a)? ¹ = (1/a)Σ (1+rb/a+r²b²/a²+.)
≈ (1/a)Σ (1+rb/a) = n/a + (b/a²)n (n+1)/2
Compare coeffs: α=1/a, β=b/2a². γ=b²/3a³. This differs from solution.

sinx+sin4x + sin2x+sin3x = 0
2sin (5x/2)cos (3x/2) + 2sin (5x/2)cos (x/2) = 0
2sin (5x/2) [cos (3x/2)+cos (x/2)] = 0
4sin (5x/2)cosxcos (x/2)=0.
sin (5x/2)=0 ⇒ 5x/2=kπ ⇒ x=2kπ/5. x=0, 2π/5, 4π/5, 6π/5, 8π/5, 2π.
cosx=0 ⇒ x=π/2, 3π/2.
cos (x/2)=0 ⇒ x=π.
Sum = 9π.

f' (x) = 12sin³xcosx+30sin²xcosx+12sinxcosx = 3sin2x (2sin²x+5sinx+2) = 3sin2x (2sinx+1) (sinx+2).
In [-π/6, π/2], sinx+2>0. 2sinx+1>0 except at x=-π/6. sin2x>0 for x∈ (0, π/2), <0 for x (-/6,0).
So f' (x)<0 on (-/6,0) (decreasing) and f' (x)>0 on (0, π/2) (increasing).

gogog (3n+1)=gog (3n+2)=g (3n+3)=3n+1. So gogog=I.
If fog=f, then f must map range of g to values consistent with f.
There exists a one-one function f: N→N such that fog=f. e.g. f (x)=x.

Vectors are coplanar. Determinant is zero. Row operations.
This leads to 2a=b+c.

f (x) = ∫? [y]dy. For x∈ [n, n+1), [y]= [x]=n.
f (x) = Σ (k=0 to n-1) ∫? ¹ k dy + ∫? n dy = Σk + n (x-n).
f (x) is continuous at integers. f' (x)=n= [x]. Not differentiable at integers.

S? = 3n/2 [2a+ (3n-1)d]. S? = 2n/2 [2a+ (2n-1)d].
S? =3S? ⇒ 3n/2 [2a+ (3n-1)d] = 3 (2n/2) [2a+ (2n-1)d].
2a+ (3n-1)d = 2 [2a+ (2n-1)d] ⇒ 2a+ (n-1)d=0.
S? /S? = (4n/2 [2a+ (4n-1)d]) / (2n/2 [2a+ (2n-1)d]) = 2 [- (n-1)d+ (4n-1)d]/ [- (n-1)d+ (2n-1)d] = 2 (3n)/ (n)=6.