# TS Inter 1st Year Maths 1A Question Paper March 2018

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## TS Inter 1st Year Maths 1A Question Paper March 2018

Time: 3 Hours
Maximum Marks: 75

Note: This question paper consists of three sections A, B, and C.

Section – A
(10 × 2 = 20 Marks)

I. Very Short Answer Type Questions.

• Each Question carries TWO marks.

Question 1.
If a, b ∈ R, f: R → R defined by f(x) = ax + b (a ≠ 0), find f-1.
Solution:
Given f: R → R defined by f(x) = ax + b (a ≠ 0)
Let f(x) = y then ax + b = y
⇒ ax = y – b
⇒ x = $$\frac{y-b}{a}$$ [∵ a ≠ 0]
⇒ f-1(y) = $$\frac{y-b}{a}$$
∴ f-1(x) = $$\frac{x-b}{a}$$

Question 2.
Find the domain of real valaued function f(x) = $$\sqrt{4 x-x^2}$$.
Solution:
Given f(x) = $$\sqrt{4 x-x^2}$$
f(x) ∈ R ⇒ 4x – x2 ≥ 0
⇒ x2 – 4x ≤ 0
⇒ x(x – 4) ≤ 0
⇒ x ∈ [0, 4]
∴ Domain of f = [0, 4]

Question 3.
If $$\left[\begin{array}{rr} 2 & 4 \\ -1 & k \end{array}\right]$$ and A2 = 0, then find the value of ‘k’.
Solution:

Question 4.
Find the rank of the following matrix $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$.
Solution:
Let A = $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$
|A| = $$\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right|$$ = 0
∴ Rank of A < 3.
All the 2-rowed minors of A is zero.
∴ Rank of A < 2.
Since ‘A’ is a non-zero matrix.
∴ Rank of A = 1.

Question 5.
Find the vector equation of the line joining the points $$2 \bar{i}+\bar{j}+3 \bar{k}$$ and $$-4 \bar{i}+3 \bar{j}-\bar{k}$$.
Solution:

Question 6.
If $$\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-3 \overline{\mathbf{j}}+\overline{\mathrm{k}}$$ and $$\overline{\mathrm{b}}=4 \overline{\mathrm{i}}+\mathrm{m} \overline{\mathrm{j}}+n \overline{\mathrm{k}}$$ are collinear vectors, then find ‘m’ and ‘n’.
Solution:

Question 7.
If $$\overline{\mathbf{a}}=2 \overline{\mathbf{i}}-3 \overline{\mathbf{j}}+\overline{\mathrm{k}}$$ and $$\overline{\mathrm{b}}=\overline{\mathrm{i}}+4 \overline{\mathrm{j}}-2 \overline{\mathrm{k}}$$, then find $$(\bar{a}+\bar{b}) \times(\bar{a}-\bar{b})$$.
Solution:

Question 8.
Prove that $$\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}$$ = cot 36°.
Solution:
We know $$\frac{\cos A+\sin A}{\cos A-\sin A}$$ = tan (45° + A)
∴ $$\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}$$ = tan (45° + 9°)
= tan 54°
= tan (90° – 36°)
= cot 36°
∴ $$\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}$$ = cot 36°

Question 9.
Prove that sin 50° – sin 70° + sin 10° = 0.
Solution:
L.H.S. = sin 50° – sin 70° + sin 10°
= 2 cos$$\left(\frac{50^{\circ}+70^{\circ}}{2}\right)$$ sin$$\left(\frac{50^{\circ}-70^{\circ}}{2}\right)$$ + sin 10°
= 2 cos 60° sin (-10°) + sin 10°
= 2 ($$\frac{1}{2}$$) sin (-10°) + sin 10°
= -sin 10° + sin 10°
= 0
= R.H.S.
∴ L.H.S. = R.H.S.
∴ sin 50° – sin 70° + sin 10° = 0.

Question 10.
Show that (cosh x + sinh x)n = cosh (nx) + sinh (nx), for any n ∈ R.
Solution:

Section – B
(5 × 4 = 20 Marks)

• Each question carries FOUR marks.

Question 11.
If A = $$\left[\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right]$$ is a non-singular matrix, then prove that A-1 = $$\frac{{Adj} A}{{det} A}$$.
Solution:

Question 12.
If $$\bar{a}, \bar{b}, \bar{c}$$ are non-coplanar vectors, then prove that $$-\bar{a}+4 \bar{b}-3 \bar{c}, 3 \bar{a}+2 \bar{b}-5 \bar{c},-3 \bar{a}+8 \bar{b}-5 \bar{c},-3 \bar{a}+2 \bar{b}+$$$$\bar{c}$$ are coplanar.
Solution:

Question 13.
For any two vectors $$\bar{a}$$ and $$\bar{b}$$, show that $$\left(1+|\bar{a}|^2\right)\left(1+|\bar{b}|^2\right)=|1-\bar{a} \cdot \bar{b}|^2+|\bar{a}+\bar{b}+\bar{a} \times \bar{b}|^2$$.
Solution:

Question 14.
Prove that $$\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}$$ = 4.
Solution:

Question 15.
Solve the equation sin x + √3 cos x = √2.
Solution:
Given sin x + √3 cos x = √2
⇒ $$\frac{1}{2} \sin x+\frac{\sqrt{3}}{2} \cos x=\frac{\sqrt{2}}{2}$$
⇒ sin 30° sin x + cos 30° cos x = $$\frac{1}{\sqrt{2}}$$
⇒ cos (x – 30°) = $$\frac{1}{\sqrt{2}}$$
⇒ x – 30° = 2nπ ± 45°, n ∈ z
⇒ x – 30 = 2nπ + 45°, x – 30° = 2nπ – 45°, n ∈ z
⇒ x = 2nπ + 75°, x = 2nπ – 15°, n ∈ z
⇒ x = 2nπ + $$\frac{5\pi}{12}$$, x = 2nπ – $$\frac{\pi}{12}$$, n ∈ z
∴ Solution set x = {2nπ + $$\frac{5\pi}{12}$$, n ∈ z} ∪ {2nπ – $$\frac{\pi}{12}$$, n ∈ z}

Question 16.
Prove that $$\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{1}{5}\right)+\tan ^{-1}\left(\frac{1}{8}\right)=\frac{\pi}{4}$$.
Solution:

Question 17.
Prove that cot A + cot B + cot C = $$\frac{a^2+b^2+c^2}{4 \Delta}$$.
Solution:

Section – C
(5 × 7 = 35 Marks)

• Each question carries SEVEN marks.

Question 18.
Let f: A → B, IA and IB be identify functions on A and B respectively, then prove that f0IA = f = IBof.
Solution:
Given f: A → B and IA, IB are identity functions on A, B.
IA : A → A, IB : B → B.
IA : A → A, f : A → A ⇒ f0IA : A → B
(f0IA) (x) = f[IA(x)] = f(x) ∀ x ∈ A.
∴ f0IA = f
f : A → B, IB : B → B ⇒ IB of : A → B.
(IBof) (x) = IB [f(x)] = f(x) ∀ x ∈ A.
∴ IBof = f.
∴ f0IA = IBof = f

Question 19.
Show that 49n + 16n – 1 is divisible by 64 for all positive integers ‘n’.
Solution:
Let p(n) be the statement that 49n + 16n – 1 is divisible by 64.
If n = 1 then 49n + 16n – 1 = 491 + 16(1) – 1
= 49 + 16 – 1
= 64
= 64 (1)
∴ P(1) is true.
Assume that p(k) is true
49k + 16k – 1 = 64t where t ∈ n.
⇒ 49k = 64t – 16k + 1
∴ 49k+1 + 16(k + 1) – 1 = 49k . 49 + 16k + 16 – 1
= 49(64t – 16k + 1) + 16k + 15
= 49 . 64t – 49 . 16k + 49 + 16k + 15
= 49 . 64t – 48 . 16k + 64
= 64(49t – 12k + 1)
∴ P(k + 1) is true
∴ By the principle of finite mathematical induction p(n) is true for all n ∈ N.
∴ 49n – 16n + 1 is divisible by 64 ∀ n ∈ N.

Question 20.
Show that $$\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|^2=\left|\begin{array}{ccc} 2 b c-a^2 & c^2 & b^2 \\ c^2 & 2 a c-b^2 & a^2 \\ b^2 & a^2 & 2 a b-c^2 \end{array}\right|$$ = (a3 + b3 + c3 – 3abc)2.
Solution:

Question 21.
Solve the following system of equations by Cramer’s rule:
2x – y + 3z = 8
-x + 2y + z = 4
3x + y – 4z = 0
Solution:
Given system of equations are
2x – y + 3z = 8
-x + 2y + z = 4
3x + y – 4z = 0
The given system of equations can be expressed as AX = B.
where A = $$\left[\begin{array}{rrr} 2 & -1 & 3 \\ -1 & 2 & 1 \\ 3 & 1 & -4 \end{array}\right]$$, X = $$\left[\begin{array}{l} x \\ y \\ z \end{array}\right]$$ and B = $$\left[\begin{array}{l} 8 \\ 4 \\ 0 \end{array}\right]$$
Δ = $$\left|\begin{array}{rrr} 2 & -1 & 3 \\ -1 & 2 & 1 \\ 3 & 1 & -4 \end{array}\right|$$
= 2(-8 – 1) – (-1) (4 – 3) + 3(-1 – 6)
= 2(-9) + 1(1) + 3(-7)
= -18 + 1 – 21
= -38 ≠ 0
Δ1 = $$\left|\begin{array}{rrr} 8 & -1 & 3 \\ 4 & 2 & 1 \\ 0 & 1 & -4 \end{array}\right|$$
= 8(-8 – 1) – (-1) (-16 – 0) + 3(4 – 0)
= 8(-9) + 1(-16) + 12
= -72 – 16 + 12
= -76
Δ2 = $$\left|\begin{array}{rrr} 2 & 8 & 3 \\ -1 & 4 & 1 \\ 3 & 0 & -4 \end{array}\right|$$
= 2(-16 – 0) – 8(4 – 3) + 3(0 – 12)
= 2(-16) – 8(1) + 3(-12)
= -32 – 8 – 36
= -76
Δ3 = $$\left|\begin{array}{rrr} 2 & -1 & 8 \\ -1 & 2 & 4 \\ 3 & 1 & 0 \end{array}\right|$$
= 2(0 – 4) – (-1) (0 – 12) + 8(-1 – 6)
= 2(-4) + 1(-12) + 8(-7)
= -8 – 12 – 56
= -76
By Cramer’s Rule
x = $$\frac{\Delta_1}{\Delta}$$
= $$\frac{-76}{-38}$$
= 2
y = $$\frac{\Delta_2}{\Delta}$$
= $$\frac{-76}{-38}$$
= 2
z = $$\frac{\Delta_3}{\Delta}$$
= $$\frac{-76}{-38}$$
= 2
∴ x = 2, y = 2, z = 2.

Question 22.
If $$\bar{a}=\bar{i}-2 \bar{j}+3 \bar{k}, \bar{b}=2 \bar{i}+\bar{j}+\bar{k}, \bar{c}=\bar{i}+\bar{j}+2 \bar{k}$$, then find $$|(\bar{a} \times \bar{b}) \times \bar{c}|$$ and $$|\overrightarrow{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})|$$.
Solution:

Question 23.
If A + B + C = 2S, then prove that cos (s – A) + cos (s – B) + cos (s – C) + cos S = $$4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$$.
Solution:
Given A + B + C = 2S
L.H.S. = cos (S – A) + cos (S – B) + cos (S – C) + cos S

Question 24.
If P1, P2, P3 are altitudes drawn from vertices A, B, C to the opposite sides of a triangle respectively, then show that:
(i) $$\frac{1}{P_1}+\frac{1}{P_2}+\frac{1}{P_3}=\frac{1}{r}$$
(ii) P1P2P3 = $$\frac{(a b c)^2}{8 R^3}=\frac{8 \Delta^3}{a b c}$$
Solution: