📚 JKBOSE Class 11th

✅ Proof

Step 1: Express tan 4x as tan(2×2x)

Using tan 2A formula: \(\tan 2A = \frac{2\tan A}{1 – \tan^2 A}\)
Let A = 2x

Step 2: Apply tan 2A formula

\(\tan 4x = \tan(2×2x) = \frac{2\tan 2x}{1 – \tan^2 2x}\)

Step 3: Substitute tan 2x formula

\(\tan 2x = \frac{2\tan x}{1 – \tan^2 x}\)

So, \(\tan 4x = \frac{2\left(\frac{2\tan x}{1 – \tan^2 x}\right)}{1 – \left(\frac{2\tan x}{1 – \tan^2 x}\right)^2}\)

Step 4: Simplify the expression

Numerator: \(2 × \frac{2\tan x}{1 – \tan^2 x} = \frac{4\tan x}{1 – \tan^2 x}\)

Denominator: \(1 – \frac{4\tan^2 x}{(1 – \tan^2 x)^2} = \frac{(1 – \tan^2 x)^2 – 4\tan^2 x}{(1 – \tan^2 x)^2}\)

So, \(\tan 4x = \frac{\frac{4\tan x}{1 – \tan^2 x}}{\frac{(1 – \tan^2 x)^2 – 4\tan^2 x}{(1 – \tan^2 x)^2}}\)

= \(\frac{4\tan x}{1 – \tan^2 x} × \frac{(1 – \tan^2 x)^2}{(1 – \tan^2 x)^2 – 4\tan^2 x}\)

= \(\frac{4\tan x(1 – \tan^2 x)}{(1 – \tan^2 x)^2 – 4\tan^2 x}\)

Step 5: Expand denominator

\((1 – \tan^2 x)^2 – 4\tan^2 x = 1 – 2\tan^2 x + \tan^4 x – 4\tan^2 x\)
= \(1 – 6\tan^2 x + \tan^4 x\)

∴ \(\tan 4x = \frac{4\tan x(1 – \tan^2 x)}{1 – 6\tan^2 x + \tan^4 x}\)

Hence proved.

✅ Solution

Given: \(Z = 1 + \sqrt{3}i\)
Here, \(a = 1\), \(b = \sqrt{3}\)

Step 1: Find modulus (r)

\(r = \sqrt{a^2 + b^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2\)

Step 2: Find amplitude (θ)

\(\tan θ = \frac{b}{a} = \frac{\sqrt{3}}{1} = \sqrt{3}\)
Since a > 0 and b > 0, θ lies in 1st quadrant
\(\tan θ = \sqrt{3} = \tan \frac{\pi}{3}\)
∴ \(θ = \frac{\pi}{3}\) radians or 60°

Step 3: Write polar form

Polar form: \(Z = r(\cos θ + i\sin θ)\)
= \(2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)\)

Answers:
Modulus (r) = 2
Amplitude (θ) = \(\frac{\pi}{3}\) radians or 60°
Polar form: \(2\left(\cos \frac{\pi}{3} + i\sin \frac{\pi}{3}\right)\)

✅ Solution

Part 1: Derivative of \(\frac{\sin(x + a)}{\cos x}\)

Step 1: Apply quotient rule

Let \(y = \frac{\sin(x + a)}{\cos x}\)
Using quotient rule: \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} – u\frac{dv}{dx}}{v^2}\)
Here, \(u = \sin(x + a)\), \(v = \cos x\)

Step 2: Find derivatives

\(\frac{du}{dx} = \cos(x + a)\)
\(\frac{dv}{dx} = -\sin x\)

\(\frac{dy}{dx} = \frac{\cos x \cdot \cos(x + a) – \sin(x + a) \cdot (-\sin x)}{\cos^2 x}\)
= \(\frac{\cos x \cos(x + a) + \sin(x + a) \sin x}{\cos^2 x}\)

Using identity: \(\cos A \cos B + \sin A \sin B = \cos(A – B)\)
= \(\frac{\cos[(x + a) – x]}{\cos^2 x} = \frac{\cos a}{\cos^2 x}\)

Answer: \(\frac{\cos a}{\cos^2 x}\)

Part 2: Evaluate \(\lim_{x \to 0} \frac{\sqrt{1 + x} – 1}{x}\)

Step 1: Rationalize numerator

\(\lim_{x \to 0} \frac{\sqrt{1 + x} – 1}{x} × \frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}\)

Step 2: Simplify

= \(\lim_{x \to 0} \frac{(1 + x) – 1}{x(\sqrt{1 + x} + 1)}\)
= \(\lim_{x \to 0} \frac{x}{x(\sqrt{1 + x} + 1)}\)
= \(\lim_{x \to 0} \frac{1}{\sqrt{1 + x} + 1}\)

Step 3: Substitute x = 0

= \(\frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}\)

Answer: \(\frac{1}{2}\)

✅ Solution

Step 1: Understand the problem

The circle passes through (0,0) and makes intercepts a and b on x-axis and y-axis respectively.
This means the circle passes through (a,0) and (0,b).

Step 2: General equation of circle

Let the equation be: \(x^2 + y^2 + 2gx + 2fy + c = 0\)

Step 3: Use given conditions

1. Passes through (0,0):
\(0 + 0 + 0 + 0 + c = 0\) ⇒ \(c = 0\)

2. Passes through (a,0):
\(a^2 + 0 + 2g(a) + 0 + 0 = 0\)
\(a^2 + 2ga = 0\)
\(2ga = -a^2\)
\(g = -\frac{a}{2}\)

3. Passes through (0,b):
\(0 + b^2 + 0 + 2f(b) + 0 = 0\)
\(b^2 + 2fb = 0\)
\(2fb = -b^2\)
\(f = -\frac{b}{2}\)

Step 4: Write final equation

Substituting \(g = -\frac{a}{2}\), \(f = -\frac{b}{2}\), \(c = 0\):
\(x^2 + y^2 + 2\left(-\frac{a}{2}\right)x + 2\left(-\frac{b}{2}\right)y + 0 = 0\)
\(x^2 + y^2 – ax – by = 0\)

Answer: \(x^2 + y^2 – ax – by = 0\)

✅ Solution

Part 1: Prove coefficients of \(a^m\) and \(a^n\) are equal

Step 1: General term in expansion

In \((1 + a)^{m+n}\), the (r+1)th term is:
\(T_{r+1} = ^{m+n}C_r \cdot 1^{m+n-r} \cdot a^r = ^{m+n}C_r \cdot a^r\)

Step 2: Coefficient of \(a^m\)

For \(a^m\), r = m
Coefficient = \({}^{m+n}C_m\)

Step 3: Coefficient of \(a^n\)

For \(a^n\), r = n
Coefficient = \({}^{m+n}C_n\)

Step 4: Prove equality

We know: \({}^{m+n}C_m = \frac{(m+n)!}{m!n!}\)
and \({}^{m+n}C_n = \frac{(m+n)!}{n!m!}\)

Since both are equal to \(\frac{(m+n)!}{m!n!}\),
\({}^{m+n}C_m = {}^{m+n}C_n\)

Hence, coefficients of \(a^m\) and \(a^n\) are equal.

Part 2: Prove \((6^n – 5n)\) leaves remainder 1 when divided by 25

Step 1: Express 6 as (1 + 5)

\(6^n = (1 + 5)^n\)

Step 2: Apply binomial theorem

\((1 + 5)^n = ^nC_0 + ^nC_1(5) + ^nC_2(5^2) + ^nC_3(5^3) + … + ^nC_n(5^n)\)
= \(1 + 5n + 25( ^nC_2 + 5^nC_3 + … + 5^{n-2} ^nC_n)\)

Step 3: Consider \(6^n – 5n\)

\(6^n – 5n = [1 + 5n + 25( ^nC_2 + 5^nC_3 + …)] – 5n\)
= \(1 + 25( ^nC_2 + 5^nC_3 + …)\)

This is of the form \(1 + 25k\), where \(k = ^nC_2 + 5^nC_3 + …\) is an integer.

When divided by 25, \(1 + 25k\) gives remainder 1.

Hence, \((6^n – 5n)\) always leaves remainder 1 when divided by 25.

✅ Solution

Part 1: Three coins tossed

Step 1: Sample space for 3 coins

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Total outcomes = 8

Step 2: Probability of 3 heads

Favorable outcome: HHH (1 outcome)
P(3 heads) = \(\frac{1}{8}\)

Step 3: Probability of 2 heads

Favorable outcomes: HHT, HTH, THH (3 outcomes)
P(2 heads) = \(\frac{3}{8}\)

Step 4: Probability of at least 2 heads

At least 2 heads means 2 or 3 heads
Favorable outcomes: HHH, HHT, HTH, THH (4 outcomes)
P(at least 2 heads) = \(\frac{4}{8} = \frac{1}{2}\)

Part 2: Mutually exclusive events

Given: \(P(A) = \frac{3}{5}\), \(P(B) = \frac{1}{5}\), A and B are mutually exclusive

For mutually exclusive events: \(P(A \cap B) = 0\)

Formula: \(P(A \cup B) = P(A) + P(B) – P(A \cap B)\)

\(P(A \text{ or } B) = P(A \cup B) = \frac{3}{5} + \frac{1}{5} – 0\)
= \(\frac{4}{5}\)

Answer: \(\frac{4}{5}\)

✅ Solution

Given: A(-2, 3, 5) and B(1, -4, 6)
Ratio m:n = 2:3

(i) Internal division (2:3)

Internal division formula:
\(P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)\)

Here m=2, n=3
x-coordinate = \(\frac{2×1 + 3×(-2)}{2+3} = \frac{2 – 6}{5} = \frac{-4}{5}\)
y-coordinate = \(\frac{2×(-4) + 3×3}{5} = \frac{-8 + 9}{5} = \frac{1}{5}\)
z-coordinate = \(\frac{2×6 + 3×5}{5} = \frac{12 + 15}{5} = \frac{27}{5}\)

(ii) External division (2:3)

External division formula:
\(P = \left(\frac{mx_2 – nx_1}{m-n}, \frac{my_2 – ny_1}{m-n}, \frac{mz_2 – nz_1}{m-n}\right)\)

Here m=2, n=3, m-n = -1
x-coordinate = \(\frac{2×1 – 3×(-2)}{2-3} = \frac{2 + 6}{-1} = \frac{8}{-1} = -8\)
y-coordinate = \(\frac{2×(-4) – 3×3}{-1} = \frac{-8 – 9}{-1} = \frac{-17}{-1} = 17\)
z-coordinate = \(\frac{2×6 – 3×5}{-1} = \frac{12 – 15}{-1} = \frac{-3}{-1} = 3\)

Answers:
(i) Internal division: \(\left(-\frac{4}{5}, \frac{1}{5}, \frac{27}{5}\right)\)
(ii) External division: (-8, 17, 3)

✅ Solution

Given:
n(X ∪ Y) = 18
n(X) = 8
n(Y) = 15

Step 1: Use the formula for union

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

Step 2: Substitute values

✅ Solution

Given: \(f(x) = x^2\), \(g(x) = 2x + 1\)

1. \((f + g)(x)\):

\((f + g)(x) = f(x) + g(x) = x^2 + (2x + 1) = x^2 + 2x + 1\)

2. \((f – g)(x)\):

\((f – g)(x) = f(x) – g(x) = x^2 – (2x + 1) = x^2 – 2x – 1\)

3. \((f \cdot g)(x)\):

\((f \cdot g)(x) = f(x) \cdot g(x) = x^2 \cdot (2x + 1) = 2x^3 + x^2\)

4. \(\left(\frac{f}{g}\right)(x)\):

\(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2}{2x + 1}\), where \(2x + 1 \neq 0\) i.e., \(x \neq -\frac{1}{2}\)

Answers:
\((f + g)(x) = x^2 + 2x + 1\)
\((f – g)(x) = x^2 – 2x – 1\)
\((f \cdot g)(x) = 2x^3 + x^2\)
\(\left(\frac{f}{g}\right)(x) = \frac{x^2}{2x + 1}, x \neq -\frac{1}{2}\)

✅ Proof

Step 1: Apply sum-to-product formulas

Using formulas:
\(\sin A + \sin B = 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)\)
\(\cos A – \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\)

Step 2: Apply to numerator and denominator

\(\sin 5x + \sin 3x = 2\sin\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right)\)
= \(2\sin 4x \cos x\)

\(\cos 5x – \cos 3x = -2\sin\left(\frac{5x+3x}{2}\right)\sin\left(\frac{5x-3x}{2}\right)\)
= \(-2\sin 4x \sin x\)

Step 3: Substitute in the given expression

LHS = \(\frac{2\sin 4x \cos x}{-2\sin 4x \sin x}\)
= \(\frac{\cos x}{-\sin x}\)
= \(-\cot x\)

Wait, let’s check carefully:
Actually, \(\frac{2\sin 4x \cos x}{-2\sin 4x \sin x} = \frac{\cos x}{-\sin x} = -\cot x\)

But we need \(\tan 4x\). Let me re-examine:

Correct calculation:
LHS = \(\frac{2\sin 4x \cos x}{-2\sin 4x \sin x} = \frac{\cos x}{-\sin x} = -\cot x\)

This doesn’t match \(\tan 4x\). Let me check the original formula again:

Actually, there might be a typo in the question. The correct identity should be:
\(\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x\)

For \(\cos 5x + \cos 3x = 2\cos\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right) = 2\cos 4x \cos x\)

Then \(\frac{2\sin 4x \cos x}{2\cos 4x \cos x} = \frac{\sin 4x}{\cos 4x} = \tan 4x\) ✓

Conclusion: The given identity has a sign error. The correct identity is:
\(\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x\)

✅ Solution

Given: \(\lim_{x \to 0} \left(\frac{\tan 3x – 2x}{3x – \sin^2 x}\right)\)

Step 1: Direct substitution gives 0/0 form

At x = 0: numerator = tan 0 – 0 = 0
denominator = 0 – 0 = 0
So we have 0/0 indeterminate form.

Step 2: Use series expansions or L’Hôpital’s rule

Using expansions as x → 0:
tan 3x = 3x + \(\frac{(3x)^3}{3}\) + … = 3x + 9x³ + …
sin x = x – \(\frac{x^3}{6}\) + … ⇒ sin² x = x² – \(\frac{x^4}{3}\) + …

Step 3: Substitute expansions

Numerator: tan 3x – 2x = (3x + 9x³ + …) – 2x = x + 9x³ + …
Denominator: 3x – sin² x = 3x – (x² – \(\frac{x^4}{3}\) + …) = 3x – x² + …

So the limit becomes:
\(\lim_{x \to 0} \frac{x + 9x^3 + …}{3x – x^2 + …}\)
= \(\lim_{x \to 0} \frac{x(1 + 9x^2 + …)}{x(3 – x + …)}\)
= \(\lim_{x \to 0} \frac{1 + 9x^2 + …}{3 – x + …}\)
= \(\frac{1}{3}\)

Answer: \(\frac{1}{3}\)

✅ Solution

Given:
\(P(E) = \frac{1}{4}\), \(P(F) = \frac{1}{2}\), \(P(E \cap F) = \frac{1}{8}\)

(i) \(P(E \text{ or } F) = P(E \cup F)\):

Using formula: \(P(E \cup F) = P(E) + P(F) – P(E \cap F)\)
= \(\frac{1}{4} + \frac{1}{2} – \frac{1}{8}\)
= \(\frac{2}{8} + \frac{4}{8} – \frac{1}{8}\)
= \(\frac{5}{8}\)

(ii) \(P(\text{not E and not F}) = P(E’ \cap F’)\):

Using De Morgan’s law: \(E’ \cap F’ = (E \cup F)’\)
So \(P(E’ \cap F’) = P((E \cup F)’) = 1 – P(E \cup F)\)
= \(1 – \frac{5}{8} = \frac{3}{8}\)

Answers:
(i) \(P(E \text{ or } F) = \frac{5}{8}\)
(ii) \(P(\text{not E and not F}) = \frac{3}{8}\)

✅ Proof

Step 1: Express terms of AP

Let first term = A, common difference = d

pth term: \(a = A + (p-1)d\) …(1)
qth term: \(b = A + (q-1)d\) …(2)
rth term: \(c = A + (r-1)d\) …(3)

Step 2: Consider the expression

We need to prove: \((q – r)a + (r – p)b + (p – q)c = 0\)

Substitute values from (1), (2), (3):

Step 3: Substitute and simplify

LHS = \((q-r)[A + (p-1)d] + (r-p)[A + (q-1)d] + (p-q)[A + (r-1)d]\)

= \(A[(q-r) + (r-p) + (p-q)] + d[(q-r)(p-1) + (r-p)(q-1) + (p-q)(r-1)]\)

First part: \((q-r) + (r-p) + (p-q) = q-r+r-p+p-q = 0\)

Second part: Expand each term:
\((q-r)(p-1) = pq – q – pr + r\)
\((r-p)(q-1) = rq – r – pq + p\)
\((p-q)(r-1) = pr – p – qr + q\)

Sum = \((pq – q – pr + r) + (rq – r – pq + p) + (pr – p – qr + q)\)
= \((pq – pq) + (-q + q) + (-pr + pr) + (r – r) + (rq – qr) + (-r + r) + (p – p)\)
= 0

∴ LHS = \(A × 0 + d × 0 = 0\)

Hence proved.