Jkbose previous year solved question papers


11

Binomial Theorem – Specific Term

Find the 4th term in the expansion of \((x – 2y)^{12}\).

Show Answer (2 Marks)

✅ Solution

Step 1: General term formula

In the expansion of \((a + b)^n\), the (r+1)th term is:
\(T_{r+1} = ^nC_r \cdot a^{n-r} \cdot b^r\)

Here: \(a = x\), \(b = -2y\), \(n = 12\)
For 4th term: \(r + 1 = 4\) ⇒ \(r = 3\)

Step 2: Calculate the 4th term

\(T_4 = T_{3+1} = ^{12}C_3 \cdot x^{12-3} \cdot (-2y)^3\)
= \(\frac{12 \times 11 \times 10}{3 \times 2 \times 1} \cdot x^9 \cdot (-8y^3)\)
= \(220 \cdot x^9 \cdot (-8y^3)\)
= \(-1760 x^9 y^3\)

Answer: \(-1760 x^9 y^3\)

12

Sequences – First Five Terms

Write the first 5 terms of the sequence whose nth term is \(a_n = 2^n\).

Show Answer (2 Marks)

✅ Solution

Given: \(a_n = 2^n\)

Step 1: Find first term (n = 1)

\(a_1 = 2^1 = 2\)

Step 2: Find subsequent terms

\(a_2 = 2^2 = 4\)
\(a_3 = 2^3 = 8\)
\(a_4 = 2^4 = 16\)
\(a_5 = 2^5 = 32\)

First 5 terms: 2, 4, 8, 16, 32

Note: This is a geometric progression with first term = 2 and common ratio = 2.


13

Complex Numbers – Powers of i

Find the value of \(2i^2 + 3i^4 + i^8\).

Show Answer (2 Marks)

✅ Solution

Step 1: Recall powers of i

\(i^1 = i\)
\(i^2 = -1\)
\(i^3 = -i\)
\(i^4 = 1\)
The pattern repeats every 4 powers.

Step 2: Simplify each term

\(i^2 = -1\) ⇒ \(2i^2 = 2(-1) = -2\)
\(i^4 = 1\) ⇒ \(3i^4 = 3(1) = 3\)
\(i^8 = (i^4)^2 = 1^2 = 1\)

Sum = \(-2 + 3 + 1 = 2\)

Answer: 2

14

Limits – Rational Function

Evaluate: \(\lim_{x\to 1}\left(\frac{x^2 + 1}{x + 100}\right)\)

Show Answer (2 Marks)

✅ Solution

Step 1: Direct substitution

Since the function is continuous at x = 1 (denominator ≠ 0), we can directly substitute x = 1.

Step 2: Substitute and evaluate

\(\lim_{x\to 1}\left(\frac{x^2 + 1}{x + 100}\right) = \frac{1^2 + 1}{1 + 100}\)
= \(\frac{1 + 1}{101}\)
= \(\frac{2}{101}\)

Answer: \(\frac{2}{101}\)

15

Differentiation – Trigonometric Function

Find \(\frac{dy}{dx}\), when \(y = \tan^2 x\).

Show Answer (2 Marks)

✅ Solution

Step 1: Apply chain rule

\(y = \tan^2 x = (\tan x)^2\)
Let \(u = \tan x\), then \(y = u^2\)

Step 2: Differentiate using chain rule

\(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
\(\frac{dy}{du} = 2u = 2\tan x\)
\(\frac{du}{dx} = \sec^2 x\)

\(\frac{dy}{dx} = 2\tan x \cdot \sec^2 x\)
= \(2\sec^2 x \tan x\)

Answer: \(2\sec^2 x \tan x\)

16

Binomial Theorem – Approximation

Using binomial theorem, evaluate \((98)^5\).

Show Answer (2 Marks)

✅ Solution

Step 1: Express 98 as (100 – 2)

\((98)^5 = (100 – 2)^5\)

Step 2: Apply binomial theorem

\((100 – 2)^5 = ^{5}C_0 (100)^5 (-2)^0 + ^{5}C_1 (100)^4 (-2)^1 + ^{5}C_2 (100)^3 (-2)^2 + ^{5}C_3 (100)^2 (-2)^3 + ^{5}C_4 (100)^1 (-2)^4 + ^{5}C_5 (100)^0 (-2)^5\)

= \(1 \times 10^{10} \times 1 + 5 \times 10^8 \times (-2) + 10 \times 10^6 \times 4 + 10 \times 10^4 \times (-8) + 5 \times 100 \times 16 + 1 \times 1 \times (-32)\)
= \(10^{10} – 10 \times 10^8 + 40 \times 10^6 – 80 \times 10^4 + 8000 – 32\)
= \(10^{10} – 10^9 + 4 \times 10^7 – 8 \times 10^5 + 7968\)
= \(9039207968\)

Answer: \(9039207968\)

Topics covered:
· Sets and Relations (difference of sets, domain and range)
· Trigonometry (radian-degree conversion)
· Complex Numbers (cube expansion, powers of i)
· Limits (trigonometric limits, rational functions)
· Differentiation (chain rule, trigonometric functions)
· Binomial Theorem (specific terms, approximations)
· Sequences and Series (first terms of sequences)
· Probability (sample space)