In a Δ A B C , prove that tan B - C 2 = b - c b + c cot A 2 .

To prove, tan B - C 2 = b - c b + c cot A 2 From sine rule, we know that a sin A = b sin B = c sin C = 2 R , where R is circumradius. Now, b - c b + c = 2 R sin B - 2 R sin C 2 R sin B + 2 R sin C = sin B - sin C sin B + sin C = 2 cos B + C 2 · sin B - C 2 2 sin B + C 2 · cos B - C 2 = cot B + C 2 · tan B - C 2 ⇒ b - c b + c = tan A 2 · tan B - C 2 ∵ B + C 2 = 90 ° - A 2 ∴ tan B - C 2 = b - c b + c cot A 2 Hence, proved.

Let f : A → B , g : B → C be bijections. Then prove that ( g o f ) - 1 = f - 1 o g - 1 .

Given, f : A → B , g : B → C are bijections. ⇒ g o f : A → C is a bijection Also g - 1 : C → B and f - 1 : B → A are bijections ⇒ f - 1 o g - 1 : C → A is a bijection. Let, c be any element of C . Then ∃ an element b ∈ B such that g ( b ) = c ⇒ b = g - 1 ( c ) Also ∃ an element a ∈ A such that f ( a ) = b ⇒ a = f - 1 ( b ) Now, ( g o f ) ( a ) = g ( f ( a ) ) = g ( b ) = c ⇒ a = ( g o f ) - 1 ( c ) ⇒ ( g o f ) - 1 ( c ) = a ...(i) Also f - 1 o g - 1 ( c ) = f - 1 g - 1 ( c ) = f - 1 ( b ) ⇒ f - 1 o g - 1 ( c ) = a ...(ii) From (i) and (ii), ( g o f ) - 1 ( c ) = f - 1 o g - 1 ( c ) ∴ ( g o f ) - 1 = f - 1 o g - 1 Hence, proved.

By using mathematical induction show that ∀ n ∈ N , 1 1 · 4 + 1 4 · 7 + 1 7 · 10 + … . upto n terms = n 3 n + 1 .

We can see that, 1 , 4 , 7 , … … forms an AP with first term a = 1 and common difference d = 3 . n t h term of an AP is a n = a 1 + ( n - 1 ) d , where a 1 is the first term and d is the common difference. n th term = 3 n - 2 We can see that, 4 , 7 , 10 , … … forms an AP with first term a = 4 and common difference d = 3 . n th term = 3 n + 1 n th term of 1 1 · 4 + 1 4 · 7 + 1 7 · 10 + … is 1 { ( 3 n - 2 ) ( 3 n + 1 ) } Let S n be the statement that 1 1 · 4 + 1 4 · 7 + 1 7 · 10 + … + 1 { ( 3 n - 2 ) ( 3 n + 1 ) } = n 3 n + 1 If n = 1 , LHS = 1 1 · 4 RHS = 1 3 + 1 = 1 4 LHS = RHS S 1 is true Let us assume that S k is true. 1 1 · 4 + 1 4 · 7 + 1 7 · 10 + … + 1 { ( 3 k - 2 ) ( 3 k + 1 ) } = k 3 k + 1 Adding both sides 1 3 k + 1 3 k + 4 , we get 1 1 · 4 + 1 4 · 7 + 1 7 · 10 + … + 1 { ( 3 k - 2 ) ( 3 k + 1 ) } + 1 { ( 3 k + 1 ) ( 3 k + 4 ) } = k 3 k + 1 + 1 { ( 3 k + 1 ) ( 3 k + 4 ) } = 3 k 2 + 4 k + 1 ( 3 k + 1 ) ( 3 k + 4 ) = ( k + 1 ) ( 3 k + 1 ) ( 3 k + 1 ) ( 3 k + 4 ) = k + 1 3 ( k + 1 ) + 1 S k + 1 is true. S ( 1 ) is true and S ( k + 1 ) is true whenever S ( k ) is true. By principle of finite mathematical induction S n is true for all n ∈ N 1 1 · 4 + 1 4 · 7 + 1 7 · 10 + … + n terms = n 3 n + 1 for all n ∈ N .

If A = 1 - 2 3 0 - 1 4 - 2 2 1 , then find A ' - 1 .

Given, A = 1 - 2 3 0 - 1 4 - 2 2 1 ⇒ A ' = 1 0 - 2 - 2 - 1 2 3 4 1 A ' = 1 - 1 - 8 - 0 + - 2 - 8 + 3 = - 9 + 10 = 1 Minors of matrix A ' M 11 = - 1 2 4 1 = - 9 , M 12 = - 2 2 3 1 = - 8 , M 13 = - 2 - 1 3 4 = - 5 M 21 = 0 - 2 4 1 = 8 , M 22 = 1 - 2 3 1 = 7 , M 23 = 1 0 3 4 = 4 M 31 = 0 - 2 - 1 2 = - 2 , M 32 = 1 - 2 - 2 2 = - 2 , M 33 = 1 0 - 2 - 1 = - 1 Cofactors of matrix is C = M 11 - M 12 M 13 - M 21 M 22 - M 23 M 31 - M 32 M 33 ⇒ C = - 9 8 - 5 - 8 7 - 4 - 2 2 - 1 Now, A ' - 1 = C T A ' = - 9 - 8 - 2 8 7 2 - 5 - 4 - 1 1 ∴ A ' - 1 = - 9 - 8 - 2 8 7 2 - 5 - 4 - 1

If A , B , C are angles of a triangle, then prove that sin 2 A 2 + sin 2 B 2 - sin 2 C 2 = 1 - 2 cos A 2 cos B 2 sin C 2 ..

To prove, sin 2 A 2 + sin 2 B 2 - sin 2 C 2 = 1 - 2 cos A 2 cos B 2 sin C 2 LHS = sin 2 A 2 + sin 2 B 2 - sin 2 C 2 = 1 - cos A 2 + 1 - cos B 2 - sin 2 C 2 = 1 2 2 - cos A + cos B - sin 2 C 2 = 1 2 2 - 2 cos A + B 2 · cos A - B 2 - sin 2 C 2 = 1 - cos 90 ° - C 2 cos A - B 2 - sin 2 C 2 ∵ A + B + C = 180 ° = 1 - sin C 2 · cos A - B 2 - sin 2 C 2 = 1 - sin C 2 cos A - B 2 + sin C 2 = 1 - sin C 2 cos A - B 2 + sin 90 ° - A + B 2 = 1 - sin C 2 cos A - B 2 + cos A + B 2 = 1 - sin C 2 2 cos A 2 cos B 2 = 1 - 2 cos A 2 cos B 2 sin C 2 = RHS Hence, proved.

E M B I B E

Intermediate First Year Mathematics paper - 1A>Hyderabad Model Question Paper (2012-13)>Chapter 11 - Hyderabad Model Question Paper (2012-13)>HYDERABAD MODEL QUESTION PAPER (2012-13)

Telangana Board Solutions for Exercise 1: HYDERABAD MODEL QUESTION PAPER (2012-13)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 1: HYDERABAD MODEL QUESTION PAPER (2012-13)

Attempt the practice questions from Exercise 1: HYDERABAD MODEL QUESTION PAPER (2012-13) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics paper - 1A solutions are prepared by Experienced Embibe Experts.

Questions from Telangana Board Solutions for Exercise 1: HYDERABAD MODEL QUESTION PAPER (2012-13) with Hints & Solutions

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 11 - Hyderabad Model Question Paper (2012-13)>HYDERABAD MODEL QUESTION PAPER (2012-13)>Q 17

In a $Δ A B C$ , prove that $\tan (\frac{B - C}{2}) = \frac{b - c}{b + c} \cot \frac{A}{2}$ .

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 11 - Hyderabad Model Question Paper (2012-13)>HYDERABAD MODEL QUESTION PAPER (2012-13)>Q 18

Let $f : A \to B, g : B \to C$ be bijections. Then prove that $(g o f)^{- 1} = f^{- 1} o g^{- 1}$ .

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 11 - Hyderabad Model Question Paper (2012-13)>HYDERABAD MODEL QUESTION PAPER (2012-13)>Q 19

By using mathematical induction show that $\forall n \in N, \frac{1}{1 \cdot 4} + \frac{1}{4 \cdot 7} + \frac{1}{7 \cdot 10} + \dots .$ upto $n$ terms $= \frac{n}{3 n + 1}$ .

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 11 - Hyderabad Model Question Paper (2012-13)>HYDERABAD MODEL QUESTION PAPER (2012-13)>Q 20

If $A = [\begin{matrix} 1 & - 2 & 3 \\ 0 & - 1 & 4 \\ - 2 & 2 & 1 \end{matrix}]$ , then find ${(A^{'})}^{- 1}$ .

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 11 - Hyderabad Model Question Paper (2012-13)>HYDERABAD MODEL QUESTION PAPER (2012-13)>Q 21

Solve the following equations by Gauss-Jordan method $3 x + 4 y + 5 z = 18, 2 x - y + 8 z = 13$ and $5 x - 2 y + 7 z = 20$ .

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 11 - Hyderabad Model Question Paper (2012-13)>HYDERABAD MODEL QUESTION PAPER (2012-13)>Q 22

If $A = (1, - 2, - 1), B = (4, 0, - 3), C = (1, 2, - 1)$ and $D = (2, - 4, - 5),$ find the distance between lines $\bar{A B}$ and $\bar{C D}$ . [Enter the value excluding units]

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 11 - Hyderabad Model Question Paper (2012-13)>HYDERABAD MODEL QUESTION PAPER (2012-13)>Q 23

If $A, B, C$ are angles of a triangle, then prove that $\sin^{2} \frac{A}{2} + \sin^{2} \frac{B}{2} - \sin^{2} \frac{C}{2} = 1 - 2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$ ..