CBSE Class 9 Mathematics Revision Worksheet Set 01

Question. Prove by the principle of mathematical induction, that for all \( n \in \mathbf{N} \):
a) \( 1 + 4 + 7 + \dots + (3n - 2) = \frac{n(3n - 1)}{2} \)
b) \( 1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n + 1)}{2}\right)^2 \)
c) \( 1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \dots + \frac{1}{1 + 2 + 3 + \dots + n} = \frac{2n}{n + 1} \), for all \( n \in \mathbf{N} \)
d) \( \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \dots + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1} \)
e) \( 7^n - 3^n \) is divisible by 4.
f) \( (1 + x)^n \ge 1 + nx \) whenever \( x \) is +ve and \( n \) is a +ve integer.
g) \( 4^n + 15n - 1 \) is divisible by 9, for \( n \in \mathbf{N} \).
Answer:
a) Let \( P(n): 1 + 4 + 7 + \dots + (3n - 2) = \frac{n(3n - 1)}{2} \)
• For \( n = 1 \): LHS \( = 3(1) - 2 = 1 \); RHS \( = \frac{1(3(1) - 1)}{2} = 1 \). Thus, \( P(1) \) is true.
• Assume \( P(k) \) is true for some \( k \in \mathbf{N} \):
\( 1 + 4 + 7 + \dots + (3k - 2) = \frac{k(3k - 1)}{2} \)
• For \( n = k + 1 \):
LHS \( = [1 + 4 + \dots + (3k - 2)] + [3(k + 1) - 2] = \frac{k(3k - 1)}{2} + (3k + 1) = \frac{3k^2 - k + 6k + 2}{2} = \frac{3k^2 + 5k + 2}{2} = \frac{(k + 1)(3k + 2)}{2} = \frac{(k + 1)(3(k + 1) - 1)}{2} = \) RHS.
Thus, \( P(k + 1) \) is true. By the principle of mathematical induction, \( P(n) \) is true for all \( n \in \mathbf{N} \).

b) Let \( P(n): 1^3 + 2^3 + 3^3 + \dots + n^3 = \left(\frac{n(n + 1)}{2}\right)^2 \)
• For \( n = 1 \): LHS \( = 1^3 = 1 \); RHS \( = \left(\frac{1(2)}{2}\right)^2 = 1 \). Thus, \( P(1) \) is true.
• Assume \( P(k) \) is true for some \( k \in \mathbf{N} \):
\( 1^3 + 2^3 + \dots + k^3 = \left(\frac{k(k + 1)}{2}\right)^2 \)
• For \( n = k + 1 \):
LHS \( = \left(\frac{k(k + 1)}{2}\right)^2 + (k + 1)^3 = (k + 1)^2 \left[\frac{k^2}{4} + (k + 1)\right] = (k + 1)^2 \left[\frac{k^2 + 4k + 4}{4}\right] = \frac{(k + 1)^2(k + 2)^2}{4} = \left(\frac{(k + 1)(k + 2)}{2}\right)^2 = \) RHS.
Thus, \( P(k + 1) \) is true. By PMI, \( P(n) \) is true for all \( n \in \mathbf{N} \).

c) Note that the \( r \)-th term is \( \frac{1}{1 + 2 + \dots + r} = \frac{1}{\frac{r(r + 1)}{2}} = \frac{2}{r(r + 1)} \).
Let \( P(n): \sum_{r=1}^n \frac{2}{r(r + 1)} = \frac{2n}{n + 1} \)
• For \( n = 1 \): LHS \( = \frac{2}{1(2)} = 1 \); RHS \( = \frac{2(1)}{1 + 1} = 1 \). Thus, \( P(1) \) is true.
• Assume \( P(k) \) is true: \( \sum_{r=1}^k \frac{2}{r(r + 1)} = \frac{2k}{k + 1} \)
• For \( n = k + 1 \):
LHS \( = \frac{2k}{k + 1} + \frac{2}{(k + 1)(k + 2)} = \frac{2k(k + 2) + 2}{(k + 1)(k + 2)} = \frac{2(k^2 + 2k + 1)}{(k + 1)(k + 2)} = \frac{2(k + 1)^2}{(k + 1)(k + 2)} = \frac{2(k + 1)}{k + 2} = \) RHS.
Thus, \( P(k + 1) \) is true. By PMI, \( P(n) \) is true for all \( n \in \mathbf{N} \).

d) Let \( P(n): \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \dots + \frac{1}{(2n - 1)(2n + 1)} = \frac{n}{2n + 1} \)
• For \( n = 1 \): LHS \( = \frac{1}{1 \cdot 3} = \frac{1}{3} \); RHS \( = \frac{1}{2(1) + 1} = \frac{1}{3} \). Thus, \( P(1) \) is true.
• Assume \( P(k) \) is true: sum \( = \frac{k}{2k + 1} \)
• For \( n = k + 1 \):
LHS \( = \frac{k}{2k + 1} + \frac{1}{(2k + 1)(2k + 3)} = \frac{k(2k + 3) + 1}{(2k + 1)(2k + 3)} = \frac{2k^2 + 3k + 1}{(2k + 1)(2k + 3)} = \frac{(2k + 1)(k + 1)}{(2k + 1)(2k + 3)} = \frac{k + 1}{2(k + 1) + 1} = \) RHS.
Thus, \( P(k + 1) \) is true. By PMI, \( P(n) \) is true for all \( n \in \mathbf{N} \).

e) Let \( P(n): 7^n - 3^n \) is divisible by 4.
• For \( n = 1 \): \( 7^1 - 3^1 = 4 \), which is divisible by 4. So, \( P(1) \) is true.
• Assume \( P(k) \) is true: \( 7^k - 3^k = 4m \) for some integer \( m \implies 7^k = 4m + 3^k \).
• For \( n = k + 1 \):
\( 7^{k+1} - 3^{k+1} = 7 \cdot 7^k - 3 \cdot 3^k = 7(4m + 3^k) - 3 \cdot 3^k = 28m + (7 - 3)3^k = 28m + 4 \cdot 3^k = 4(7m + 3^k) \), which is divisible by 4.
Thus, \( P(k + 1) \) is true. By PMI, \( 7^n - 3^n \) is divisible by 4 for all \( n \in \mathbf{N} \).

f) Let \( P(n): (1 + x)^n \ge 1 + nx \) for \( x > 0 \).
• For \( n = 1 \): \( (1 + x)^1 = 1 + x \ge 1 + (1)x \), which is true. Thus, \( P(1) \) is true.
• Assume \( P(k) \) is true: \( (1 + x)^k \ge 1 + kx \).
• For \( n = k + 1 \): Since \( x > 0 \implies (1 + x) > 0 \), multiplying both sides by \( (1 + x) \):
\( (1 + x)^{k+1} \ge (1 + kx)(1 + x) = 1 + x + kx + kx^2 = 1 + (k + 1)x + kx^2 \).
Since \( k > 0 \) and \( x^2 > 0 \), \( kx^2 > 0 \), so \( 1 + (k + 1)x + kx^2 > 1 + (k + 1)x \).
Hence, \( (1 + x)^{k+1} \ge 1 + (k + 1)x \).
Thus, \( P(k + 1) \) is true. By PMI, the inequality holds for all positive integers \( n \).

g) Let \( P(n): 4^n + 15n - 1 \) is divisible by 9.
• For \( n = 1 \): \( 4^1 + 15(1) - 1 = 18 = 9 \times 2 \), which is divisible by 9. Thus, \( P(1) \) is true.
• Assume \( P(k) \) is true: \( 4^k + 15k - 1 = 9m \) for some integer \( m \implies 4^k = 9m - 15k + 1 \).
• For \( n = k + 1 \):
\( 4^{k+1} + 15(k + 1) - 1 = 4 \cdot 4^k + 15k + 14 = 4(9m - 15k + 1) + 15k + 14 = 36m - 60k + 4 + 15k + 14 = 36m - 45k + 18 = 9(4m - 5k + 2) \), which is divisible by 9.
Thus, \( P(k + 1) \) is true. By PMI, \( P(n) \) is true for all \( n \in \mathbf{N} \).

Question. Prove that \( \frac{\cos 8A \cos 5A - \cos 12A \cos 9A}{\sin 8A \cos 5A + \cos 12A \sin 9A} = \tan 4A \).
Answer: Multiplying both the numerator and denominator by 2:
\[ \text{LHS} = \frac{2\cos 8A \cos 5A - 2\cos 12A \cos 9A}{2\sin 8A \cos 5A + 2\cos 12A \sin 9A} \]
Using the product-to-sum formulas:
• \( 2\cos 8A \cos 5A = \cos(8A + 5A) + \cos(8A - 5A) = \cos 13A + \cos 3A \)
• \( 2\cos 12A \cos 9A = \cos(12A + 9A) + \cos(12A - 9A) = \cos 21A + \cos 3A \)
• \( 2\sin 8A \cos 5A = \sin(8A + 5A) + \sin(8A - 5A) = \sin 13A + \sin 3A \)
• \( 2\cos 12A \sin 9A = \sin(12A + 9A) - \sin(12A - 9A) = \sin 21A - \sin 3A \)

Substituting these into the expression:
\[ \text{Numerator} = (\cos 13A + \cos 3A) - (\cos 21A + \cos 3A) = \cos 13A - \cos 21A \]
\[ \text{Denominator} = (\sin 13A + \sin 3A) + (\sin 21A - \sin 3A) = \sin 21A + \sin 13A \]

Now applying sum-to-product formulas:
• \( \cos 13A - \cos 21A = 2\sin\left(\frac{13A + 21A}{2}\right)\sin\left(\frac{21A - 13A}{2}\right) = 2\sin(17A)\sin(4A) \)
• \( \sin 21A + \sin 13A = 2\sin\left(\frac{21A + 13A}{2}\right)\cos\left(\frac{21A - 13A}{2}\right) = 2\sin(17A)\cos(4A) \)

Therefore:
\[ \text{LHS} = \frac{2\sin(17A)\sin(4A)}{2\sin(17A)\cos(4A)} = \frac{\sin 4A}{\cos 4A} = \tan 4A = \text{RHS} \]

Question. Prove that \( \tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x \)
Answer: We can write \( 3x = 2x + x \).
Taking tangent on both sides:
\[ \tan 3x = \tan(2x + x) \]
Using the identity \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \):
\[ \tan 3x = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x} \]
Cross-multiplying:
\[ \tan 3x(1 - \tan 2x \tan x) = \tan 2x + \tan x \]
\[ \tan 3x - \tan 3x \tan 2x \tan x = \tan 2x + \tan x \]
Rearranging the terms:
\[ \tan 3x \tan 2x \tan x = \tan 3x - \tan 2x - \tan x \]

Question. If \( A = \{1,2,5\} \), \( B = \{1,2,3,4\} \) and \( C = \{5,6,2\} \), Verify that
a) \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)
b) \( (A - B) \times C = (A \times C) - (B \times C) \)
Answer: Given \( A = \{1,2,5\} \), \( B = \{1,2,3,4\} \), \( C = \{2,5,6\} \).

a) \( B \cap C = \{2\} \)
LHS \( = A \times (B \cap C) = \{1,2,5\} \times \{2\} = \{(1,2), (2,2), (5,2)\} \)
Now, \( A \times B = \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (5,1), (5,2), (5,3), (5,4)\} \)
\( A \times C = \{(1,2), (1,5), (1,6), (2,2), (2,5), (2,6), (5,2), (5,5), (5,6)\} \)
RHS \( = (A \times B) \cap (A \times C) = \{(1,2), (2,2), (5,2)\} \)
Since LHS = RHS, \( A \times (B \cap C) = (A \times B) \cap (A \times C) \) is verified.

b) \( A - B = \{1,2,5\} - \{1,2,3,4\} = \{5\} \)
LHS \( = (A - B) \times C = \{5\} \times \{2,5,6\} = \{(5,2), (5,5), (5,6)\} \)
Now, \( B \times C = \{(1,2), (1,5), (1,6), (2,2), (2,5), (2,6), (3,2), (3,5), (3,6), (4,2), (4,5), (4,6)\} \)
RHS \( = (A \times C) - (B \times C) \): removing elements of \( B \times C \) from \( A \times C \):
\( = \{(5,2), (5,5), (5,6)\} \)
Since LHS = RHS, \( (A - B) \times C = (A \times C) - (B \times C) \) is verified.

Question. What is the value of \( \cos\left(\frac{\pi}{4} - x\right)\cos\left(\frac{\pi}{4} - y\right) - \sin\left(\frac{\pi}{4} - x\right)\sin\left(\frac{\pi}{4} - y\right) \)
Answer: Using the identity \( \cos A \cos B - \sin A \sin B = \cos(A + B) \), where \( A = \frac{\pi}{4} - x \) and \( B = \frac{\pi}{4} - y \):
\[ = \cos\left[\left(\frac{\pi}{4} - x\right) + \left(\frac{\pi}{4} - y\right)\right] \]
\[ = \cos\left(\frac{\pi}{2} - (x + y)\right) \]
Since \( \cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta \):
\[ = \sin(x + y) \]

Question. What is the eccentricity of the curve \( 4x^2 + y^2 = 100 \).
Answer: Dividing the equation by 100:
\[ \frac{4x^2}{100} + \frac{y^2}{100} = 1 \implies \frac{x^2}{25} + \frac{y^2}{100} = 1 \]
This is an ellipse of the form \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) where \( a^2 = 100 \) and \( b^2 = 25 \) (with major axis along the y-axis).
The eccentricity \( e \) is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{25}{100}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

Question. What is the value of \( \cot\left(-\frac{15\pi}{4}\right) \)
Answer: Since \( \cot(-\theta) = -\cot\theta \):
\[ \cot\left(-\frac{15\pi}{4}\right) = -\cot\left(\frac{15\pi}{4}\right) \]
We can rewrite \( \frac{15\pi}{4} \) as \( 4\pi - \frac{\pi}{4} \):
\[ \cot\left(4\pi - \frac{\pi}{4}\right) = -\cot\left(\frac{\pi}{4}\right) = -1 \]
Therefore:
\[ \cot\left(-\frac{15\pi}{4}\right) = -(-1) = 1 \]

Question. Find \( \sin \frac{x}{2} \), \( \cos \frac{x}{2} \) and \( \tan \frac{x}{2} \) if \( \tan x = -\frac{4}{3} \), \( x \) in quadrant II
Answer: Given \( \tan x = -\frac{4}{3} \) and \( x \) lies in Quadrant II, i.e., \( \frac{\pi}{2} < x < \pi \).
Dividing by 2 gives \( \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \), so \( \frac{x}{2} \) lies in Quadrant I (where all trigonometric ratios are positive).

We know that:
\[ \sec^2 x = 1 + \tan^2 x = 1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9} \]
Since \( x \) is in Quadrant II, \( \cos x < 0 \):
\[ \cos x = -\frac{3}{5} \]

Using half-angle formulas:
• \( \sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}} = \sqrt{\frac{1 - (-3/5)}{2}} = \sqrt{\frac{8/5}{2}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \)
• \( \cos \frac{x}{2} = \sqrt{\frac{1 + \cos x}{2}} = \sqrt{\frac{1 + (-3/5)}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \)
• \( \tan \frac{x}{2} = \frac{\sin (x/2)}{\cos (x/2)} = \frac{2/\sqrt{5}}{1/\sqrt{5}} = 2 \)

Question. Find the square root of \( -7 - 24i \)
Answer: Let \( \sqrt{-7 - 24i} = x + iy \), where \( x, y \in \mathbf{R} \).
Squaring both sides:
\[ x^2 - y^2 + 2ixy = -7 - 24i \]
Equating real and imaginary parts:
1) \( x^2 - y^2 = -7 \)
2) \( 2xy = -24 \implies xy = -12 \) (since \( xy < 0 \), \( x \) and \( y \) have opposite signs).

Using the identity \( (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 \):
\[ (x^2 + y^2)^2 = (-7)^2 + (-24)^2 = 49 + 576 = 625 \]
\[ x^2 + y^2 = 25 \quad (\text{since } x^2 + y^2 > 0) \]

Adding \( x^2 - y^2 = -7 \) and \( x^2 + y^2 = 25 \):
\[ 2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3 \]
Subtracting the equations:
\[ 2y^2 = 32 \implies y^2 = 16 \implies y = \pm 4 \]
Since \( xy = -12 \) is negative, \( x \) and \( y \) must have opposite signs:
When \( x = 3 \), \( y = -4 \); and when \( x = -3 \), \( y = 4 \).
Hence, the square roots of \( -7 - 24i \) are \( \pm(3 - 4i) \).

Question. Prove that \( \cos^2 x + \cos^2\left(x + \frac{\pi}{3}\right) + \cos^2\left(x - \frac{\pi}{3}\right) = \frac{3}{2} \)
Answer: Using the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \):
\[ \text{LHS} = \frac{1 + \cos 2x}{2} + \frac{1 + \cos\left(2x + \frac{2\pi}{3}\right)}{2} + \frac{1 + \cos\left(2x - \frac{2\pi}{3}\right)}{2} \]
\[ = \frac{3}{2} + \frac{1}{2}\left[\cos 2x + \cos\left(2x + \frac{2\pi}{3}\right) + \cos\left(2x - \frac{2\pi}{3}\right)\right] \]
Using \( \cos(A + B) + \cos(A - B) = 2\cos A \cos B \):
\[ \cos\left(2x + \frac{2\pi}{3}\right) + \cos\left(2x - \frac{2\pi}{3}\right) = 2\cos 2x \cos\left(\frac{2\pi}{3}\right) \]
Since \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \):
\[ = 2\cos 2x \left(-\frac{1}{2}\right) = -\cos 2x \]
Substituting back:
\[ \text{LHS} = \frac{3}{2} + \frac{1}{2}[\cos 2x - \cos 2x] = \frac{3}{2} + 0 = \frac{3}{2} = \text{RHS} \]

Question. Show that \( (A \cup B)' = A' \cap B' \) and \( (A \cap B)' = A' \cup B' \)
Where \( U = \{1,2,3,4,5,6,7,8\} \), \( A = \{2,3,4\} \) and \( B = \{3,4,5\} \)
Answer: Given:
\( U = \{1,2,3,4,5,6,7,8\} \)
\( A = \{2,3,4\} \implies A' = U - A = \{1,5,6,7,8\} \)
\( B = \{3,4,5\} \implies B' = U - B = \{1,2,6,7,8\} \)

1) To show \( (A \cup B)' = A' \cap B' \):
\( A \cup B = \{2,3,4,5\} \)
LHS \( = (A \cup B)' = U - (A \cup B) = \{1,6,7,8\} \)
RHS \( = A' \cap B' = \{1,5,6,7,8\} \cap \{1,2,6,7,8\} = \{1,6,7,8\} \)
Since LHS = RHS, \( (A \cup B)' = A' \cap B' \) is proved.

2) To show \( (A \cap B)' = A' \cup B' \):
\( A \cap B = \{3,4\} \)
LHS \( = (A \cap B)' = U - (A \cap B) = \{1,2,5,6,7,8\} \)
RHS \( = A' \cup B' = \{1,5,6,7,8\} \cup \{1,2,6,7,8\} = \{1,2,5,6,7,8\} \)
Since LHS = RHS, \( (A \cap B)' = A' \cup B' \) is proved.

Solve the following problems:-

1.Evaluate the exponent:
(i) 5^-3
(ii) (1 /3)^-4
(iii) (5 /2)^-3
(iv) (-2)^-5
(v) (-3/4)^-4

We have:
(i) 5^-3 =1/5³ =1/125

(ii) (1/3)^-4 =(3/1)⁴ =3⁴ =81

(iii) (5/2)^-3 =(2/5)³ =2³/5³ =8/125

(iv) (-2)^-5 =1/(-2)⁵ =1/-25 =1/-32=-1/32

(v) (-3/4)-4 =(4/-3)⁴ =(-4/3)⁴ =(-4)⁴ /3⁴ =4⁴/3⁴ =256/81

2.Evaluate:(-2/7)^-4 ×(-5/7)²
Solution:
(-2/7)^-4×(-5/7)²
=(7/-2)⁴ ×(-5/7)²
=(-7/2)⁴ ×(-5/7)² [Since,(7/-2)=(-7/2)]
=(-7)⁴ /2⁴ ×(-5)² /7²
={7⁴ ×(-5)² }/{2⁴ ×7² } [Since,(-7)⁴ =7⁴]
={72 ×(-5)² }/24
=[49×(-5)×(-5)]/16
=1225/16

3.Evaluate:(-1/4)^-3 ×(-1/4)^-2
Solution:
(-1/4)^-3 ×(-1/4)^-2
=(4/-1)³ ×(4/-1)²
=(-4)³ ×(-4)²
=(-4)⁽³⁺²⁾
=(-4)⁵
=-4⁵
=-1024.

4.Evaluate:{[(-3)/2]²}^-3
Solution:
{[(-3)/2]²}^-3
=(-3/2)²×(-3)
=(-3/2)^-6
=(2/-3)⁶
=(-2/3)⁶
=(-2)⁶ /3⁶
=2⁶/3⁶
=64/729

5.Simplify:
(i)(2^-1 ×5^-1)^-1 ÷4^-1
(ii)(4^-1 +8^-1)÷(2/3)^-1
Solution:
(i)(2^-1 ×5^-1)^-1 ÷4^-1
=(1/2×1/5)^-1 ÷(4/1)^-1
=(1/10)^-1 ÷(1/4)
= 10/1 ÷ 1/4
=(10÷ 1/4)
=(10×4)
=40.

(ii)(4^-1 +8^-1)÷(2/3)^-1
=(1/4+1/8)÷(3/2)
=(2+1)/8÷3/2
=(3/8÷3/2)
=(3/8÷2/3)
=1/4

6.Simplify:(1/2)^-2 +(1/3)^-2 +(1/4)^-2
Solution:
(1/2)^-2 +(1/3)^-2 +(1/4)^-2
=(2/1)² +(3/1)² +(4/1)²
=(2² +3² +4²)
=(4+9+16)
=29

Please refer to attached file for CBSE Class 9 Mathematics Worksheet - Revision Worksheet (1)