Class 12th

We are given bounds for a function f (t) on two intervals and need to find the range of g (3) = ∫? ³ f (t) dt.
We split the integral: g (3) = ∫? ¹ f (t)dt + ∫? ³ f (t)dt.
For the first interval t ∈ [0, 1], we have 1/3 ≤ f (t) ≤ 1. Integrating from 0 to 1 gives:
∫? ¹ (1/3) dt ≤ ∫? ¹ f (t)dt ≤ ∫? ¹ 1 dt ⇒ 1/3 ≤ ∫? ¹ f (t)dt ≤ 1.
For the second interval t ∈ (1, 3], we have 0 ≤ f (t) ≤ 1/2. Integrating from 1 to 3 gives:
∫? ³ 0 dt ≤ ∫? ³ f (t)dt ≤ ∫? ³ (1/2) dt ⇒ 0 ≤ ∫? ³ f (t)dt ≤ (1/2) (3-1) = 1.
Adding the inequalities for the two parts of the integral:
1/3 + 0 ≤ g (3) ≤ 1 + 1.
1/3 ≤ g (3) ≤ 2.
So, g (3) ∈ [1/3, 2].

The general term Tr+1 in the expansion of (a/x + x)^n (assuming from context, as it is not explicitly stated) is:
Tr+1 = nCr * (a/x)^r * x^ (n-r) = nCr * a^r * x^ (n-2r)
(The text shows (a/x^2) leading to x^ (n-3r)

Assuming the term is (x + a/x^2)^n:
Tr+1 = nCr * x^ (n-r) * (a/x^2)^r = nCr * a^r * x^ (n-3r)
T3 = T (2+1) = nC2 * a^2 * x^ (n-6)
T4 = T (3+1) = nC3 * a^3 * x^ (n-9)
T5 = T (4+1) = nC4 * a^4 * x^ (n-12)

The ratio of the coefficients is given:
(nC2 * a^2) / (nC3 * a^3) = 12/8 = 3/2
(n (n-1)/2) * a^2) / (n (n-1) (n-2)/6) * a^3) = 3/2
(3 / (n-2) * (1/a) = 3/2 => a (n-2) = 2 (i)

(nC3 * a^3) / (nC4 * a^4) = 8/3
(n (n-1) (n-2)/6) * a^3) / (n (n-1) (n-2) (n-3)/24) * a^4) = 8/3
(4 / (n-3) * (1/a) = 8/3 => 8a (n-3) = 12 => 2a (n-3) = 3 => a (n-3) = 3/2 (ii)

From (i), a = 2 / (n-2). Substitute into (ii):
(2 / (n-2) * (n-3) = 3/2
4 (n-3) = 3 (n-2)
4n - 12 = 3n - 6
n = 6

Substitute n=6 back into a (n-2) = 2:
a (6-2) = 2 => 4a = 2 => a = 1/2.

The term independent of x is when the power of x is 0.
n - 3r = 0 => 6 - 3r = 0 => r = 2.
The term is T (2+1) = T3.

The value of this term is nC2 * a^2 = 6C2 * (1/2)^2 = 15 * (1/4) = 15/4 = 3.75.
The text approximates this to 4.