Thoroughly analyzing TS Inter 1st Year Maths 1A Model Papers and TS Inter 1st Year Maths 1A Question Paper March 2018 helps students identify their strengths and weaknesses.
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, 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: