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.

• Answer ALL the 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)

II. Short Answer Type Questions.

• Answer ANY FIVE questions.
• 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)

III. Long Answer Type Questions.

• Answer ANY FIVE questions.
• Each question carries SEVEN marks.

Question 18.
Let f: A → B, I_A and IB be identify functions on A and B respectively, then prove that f₀I_A = f = I_Bof.
Solution:
Given f: A → B and I_A, I_B are identity functions on A, B.
I_A : A → A, I_B : B → B.
I_A : A → A, f : A → A ⇒ f₀I_A : A → B
(f₀I_A) (x) = f[I_A(x)] = f(x) ∀ x ∈ A.
∴ f₀I_A = f
f : A → B, I_B : B → B ⇒ I_B of : A → B.
(I_Bof) (x) = I_B [f(x)] = f(x) ∀ x ∈ A.
∴ I_Bof = f.
∴ f₀I_A = I_Bof = 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
49^k + 16k – 1 = 64t where t ∈ n.
⇒ 49^k = 64t – 16k + 1
∴ 49^k+1 + 16(k + 1) – 1 = 49^k . 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: