TS Inter 1st Year Maths 1A Question Paper May 2018

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TS Inter 1st Year Maths 1A Question Paper May 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 questions.
• Each question carries TWO marks.

Question 1.
If f: R – {0} → R is defined by f(x) = x³ – \(\frac{1}{x^3}\), then show that f(x) + f(\(\frac{1}{x}\)) = 0.
Solution:

Question 2.
Find the domain of the real valued function f(x) = \(\frac{1}{\left(x^2-1\right)(x+3)}\).
Solution:
Given f(x) = \(\frac{1}{\left(x^2-1\right)(x+3)}\)
(x² – 1) (x + 3) ≠ 0
⇒ (x + 1) (x – 1) (x + 3) ≠ 0
⇒ x ≠ -1, x ≠ 1, x ≠ -3
∴ Domain of f = R – [-3, -1, 1]

Question 3.
If \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]=\left[\begin{array}{rr}
5 & 2 \\
-2 & a-4
\end{array}\right]\), then find the values of x, y, z and a.
Solution:
Given \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]=\left[\begin{array}{rr}
5 & 2 \\
-2 & a-4
\end{array}\right]\)
x – 3 = 5 ⇒ x = 8
2y – 8 = 2 ⇒ 2y = 10 ⇒ y = 5
z + 2 = -2 ⇒ z = -4
6 = a – 4 ⇒ a = 10
∴ x = 8, y = 5, z = -4, a = 10

Question 4.
If A = \(\left[\begin{array}{rrr}
-1 & 2 & 3 \\
2 & 5 & 6 \\
3 & x & 7
\end{array}\right]\) is a symmetric matrix, then find x.
Solution:
Given A = \(\left[\begin{array}{rrr}
-1 & 2 & 3 \\
2 & 5 & 6 \\
3 & x & 7
\end{array}\right]\)
Since A is a symmetric matrix.
∴ A^T = A
\(\left[\begin{array}{rrr}
-1 & 2 & 3 \\
2 & 5 & x \\
3 & 6 & 7
\end{array}\right]=\left[\begin{array}{rrr}
-1 & 2 & 3 \\
2 & 5 & 6 \\
3 & x & 7
\end{array}\right]\)
∴ x = 6

Question 5.
If the vectors \(-3 \vec{i}+4 \vec{j}+\lambda \vec{k}\) and \(\mu \vec{i}+8 \vec{j}+6 \vec{k}\) are collinear vectors, then find λ and µ.
Solution:

Question 6.
Find the vector equation of the plane passing through the points \(\vec{i}-2 \vec{j}+5 \vec{k},-5 \vec{j}-\vec{k}\) and \(-3 \vec{i}+5 \vec{j}\).
Solution:

Question 7.
Find the angle between the planes \(\vec{r} \cdot(2 \vec{i}-\vec{j}+2 \vec{k})\) = 3 and\(\vec{r} \cdot(3 \vec{i}+6 \vec{j}+\vec{k})\) = 4.
Solution:

Question 8.
If a cos θ – b sin θ = c, then show that a sin θ + b cos θ = ±\(\sqrt{a^2+b^2-c^2}\).
Solution:
Given a cos θ – b sin θ = c
Let a sin θ + b cos θ = λ
By squaring and adding, we get
(a cos θ – b sin θ)² + (a sin θ + b cos θ)² = c² + λ²
⇒ a² cos²θ – 2ab sin θ cos θ + b² sin²θ + a² sin²θ + 2ab sin θ cos θ + b² cos²θ = c² + λ²
⇒ a² (cos²θ + sin²θ) + b² (sin²θ + cos²θ) = c² + λ²
⇒ a² (1) + b² (1) = c² + λ²
⇒ a² + b² = c² + λ²
⇒ λ² = a² + b² – c²
⇒ λ = ±\(\sqrt{a^2+b^2-c^2}\)
∴ a sin θ + b cos θ = ±\(\sqrt{a^2+b^2-c^2}\)

Question 9.
Find the minimum and maximum values of 3 cos x + 4 sin x.
Solution:
Let f(x) = 3 cos x + 4 sin x
Here a = 3, b = 4, c = 0.
Minimum value of f(x) = c – \(\sqrt{a^2+b^2}\)
= 0 – \(\sqrt{9+16}\)
= 0 – \(\sqrt{25}\)
= -5
Maximum value of f(x) = c + \(\sqrt{a^2+b^2}\)
= 0 + \(\sqrt{9+16}\)
= 0 + \(\sqrt{25}\)
= 5

Question 10.
Show that \(\tan h^{-1}\left(\frac{1}{2}\right)=\frac{1}{2} \log _e^3\).
Solution:

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

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

Question 11.
If 3A = \(\left[\begin{array}{rrr}
1 & 2 & 2 \\
2 & 1 & -2 \\
-2 & 2 & -1
\end{array}\right]\), then show that A^-1 = A’.
Solution:

Question 12.
a, b, c are non-coplanar vectors. Prove that the following four points are coplanar:
-a + 4b – 3c, 3a + 2b – 5c, -3a + 8b – 5c, -3a + 2b + c.
Solution:

Question 13.
The vectors AB = \(3 \vec{i}-2 \vec{j}+2 \vec{k}\) and AD = \(\vec{i}-2 \vec{k}\) represent the adjacent sides of a parallelogram ABCD. Find the angle between the diagonals.
Solution:

Question 14.
Show that \(\cos ^4\left(\frac{\pi}{8}\right)+\cos ^4\left(\frac{3 \pi}{8}\right)+\cos ^4\left(\frac{5 \pi}{8}\right)+\cos ^4\left(\frac{7 \pi}{8}\right)=\frac{3}{2}\).
Solution:


= R.H.S
∴ L.H.S = R.H.S
∴ \(\cos ^4\left(\frac{\pi}{8}\right)+\cos ^4\left(\frac{3 \pi}{8}\right)+\cos ^4\left(\frac{5 \pi}{8}\right)+\cos ^4\left(\frac{7 \pi}{8}\right)=\frac{3}{2}\)

Question 15.
Solve 2 cos²θ – √3 sin θ + 1 = 0.
Solution:
Given 2 cos²θ – √3 sin θ + 1 = 0
⇒ 2(1 – sin²θ) – √3 sin θ + 1 = 0
⇒ 2 – 2 sin²θ – √3 sin θ + 1 = 0
⇒ -2 sin²θ – √3 sin θ + 3 = 0
⇒ 2 sin²θ + √3 sin θ – 3 = 0
⇒ 2 sin²θ + 2√3 sin θ – √3 sin θ – 3 = 0
⇒ 2 sin θ (sin θ + √3) – √3 (sin θ + √3 ) = 0
⇒ (sin θ + √3) (2 sin θ – √3) = 0
since sin θ + √3 ≠ 0
∴ 2 sin θ – √3 = 0
⇒ 2 sin θ = √3
⇒ sin θ = \(\frac{\sqrt{3}}{2}\) = sin \(\frac{\pi}{3}\)
∴ θ = nπ + (-1)ⁿ \(\frac{\pi}{3}\), n ∈ z
∴ Solution set x = [nπ + (-1)ⁿ \(\frac{\pi}{3}\)/n ∈ z]

Question 16.
Solve the following equation for x:
\(3 \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)-4 \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\frac{\pi}{3}\)
Solution:

Question 17.
If sin θ = \(\frac{a}{b+c}\), then show that cos θ = \(\frac{2 \sqrt{b c}}{b+c} \cos \frac{A}{2}\).
Solution:

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

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

Question 18.
If f = {(4, 5), (5, 6), (6, -4)} and g = {(4, -4), (6, 5), (8, 5)} then find:
(i) f + g
(ii) 2f + 4g
(iii) f + 4
(iv) \(\frac{f}{g}\)
(v) \(\sqrt{f}\)
(vi) f²
(vii) |f|
Solution:
Given f = [(4, 5), (5, 6), (6, -4)]
g = [(4, -4), (6, 5), (8, 5)]
∴ Domain of f = [4, 5, 6]
Domain of g = [4, 6, 8]
∴ The domain of f ∩ g = [4, 6]
(i) (f + g) (4) = f(4) + g(4)
= 5 + (-4)
= 1
(f + g) (6) = f(6) + g(6)
= -4 + 5
= 1
∴ f + g = [(4, 1), (6, 1)]
(ii) (2f + 4g) (4) = 2f(4) + 4g(4)
= 2(5) + 4(-4)
= 10 – 16
= -6
(2f + 4g) (6) = 2f(6) + 4g(6)
= 2(-4) + 4(5)
= -8 + 20
= 12
∴ 2f + 4g = [(4, – 6), (6,12)]
(iii) (f + 4) (4) = f(4) + 4
= 5 + 4
= 9
(f + 4) (5) = f(5) + 4
= 6 + 4
= 10
(f + 4) (6) = f(6) + 4
= -4 + 4
= o
∴ f + 4 = [(4, 9), (5, 10), (6, 0)]

(vi) f²(4) = [f(4)]² = 5² = 25
f²(5) = [f(5)]² = 6² = 36
f²(6) = [f(6)]² = (-4)² = 16
∴ f² = [(4, 25), (5, 36), (6, 16)]
(vii) |f|(4) = |f(4)| = |5| = 5
|f| (5) = |f(5)| = |6| = 6
|f| (6) = |f(6)| = |-4| = 4
|∴ f| = [(4, 5), (5, 6), (6, 4)]

Question 19.
Show that ∀ n ∈ N, 12 + (1² + 2²) + (1² + 2² + 3²) + ……. upto n terms = \(\frac{n(n+1)^2(n+2)}{12}\) by Mathematical Induction.
Solution:
The nth term of the given series is (1² + 2² + 3² + …….. + n²)
Let p(n) be the statement:
1² + (1² + 2²) + (1² + 2² + 3²) + …… + (1² + 2² + ……. + n²) = \(\frac{n(n+1)^2(n+2)}{12}\)
and the sum on the L.H.S. is denoted by S(n).
Since s(1) = 1² = \(\frac{1(1+1)^2(1+2)}{12}\) = 1 = 1²
∴ p(1) is true.
Assume that the statement is true for n = k
(i.e.,) S(k) = 1² + (1² + 2²) + …… + (1² + 2² + ….. + k²) = \(\frac{k(k+1)^2(k+2)}{12}\)
We show that S(k + 1) = \(\frac{(k+1)(k+2)^2(k+3)}{12}\)
We observe that
S(k + 1) = 1² + (1² + 2²) + …… + (1² + 2² + 3² + …… + k²) + [1² + 2² + ……. + k² + (k + 1)²]
= S(k) + [1² + 2² + ……. + k² + (k + 1)²]
= \(\frac{k(k+1)^2(k+2)}{12}\) + [1² + 2² + ……. + k² + (k + 1)²]


∴ The statement holds for n = k + 1.
∴ By the principle of mathematical induction, p(n) is true for all n ∈ N.
(i.e.,) 1² + (1² + 2²) + (1² + 2² + 3²) + ……. + (1² + 2² + …… + n²) = \(\frac{n(n+1)^2(n+2)}{12}\)

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 x – y + 3z = 5, 4x + 2y – z = 0, -x + 3y + z = 5 by Cramer’s rule.
Solution:
Given
x – y + 3z = 5
4x + 2y – z = 0
-x + 3y + z = 5
Given system of equations can be written as AX = B

= 1(2 + 3) – (-1) (4 – 1) + 3(12 + 2)
= 5 + 3 + 42
= 50
Δ₁ = \(\left|\begin{array}{rrr}
5 & -1 & 3 \\
0 & 2 & -1 \\
5 & 3 & 1
\end{array}\right|\)
= 5(2 + 3) – (-1) (0 + 5) + 3(0 – 10)
= 25 + 5 – 30
= 0
Δ₂ = \(\left|\begin{array}{rrr}
1 & 5 & 3 \\
4 & 0 & -1 \\
-1 & 5 & 1
\end{array}\right|\)
= 1(0 + 5) – 5(4 – 1) + 3(20 + 0)
= 5 – 15 + 60
= 50
Δ₃ = \(\left|\begin{array}{rrr}
1 & -1 & 5 \\
4 & 2 & 0 \\
-1 & 3 & 5
\end{array}\right|\)
= 1(10 – 0) – (-1) (20 + 0) + 5(12 + 2)
= 10 + 20 + 70
= 100
x = \(\frac{\Delta_1}{\Delta}=\frac{0}{50}\) = 0
y = \(\frac{\Delta_2}{\Delta}=\frac{50}{50}\) = 1
z = \(\frac{\Delta_3}{\Delta}=\frac{100}{50}\) = 2
∴ x = 0, y = 1, z = 2

Question 22.
If \(\vec{a}=\vec{i}-2 \vec{j}+\vec{k}, \vec{b}=2 \vec{i}+\vec{j}+\vec{k}, \vec{c}=\vec{i}+2 \vec{j}-\vec{k}\), find \(\vec{a} \times(\vec{b} \times \vec{c})\) and \(|(\vec{a} \times \vec{b}) \times \vec{c}|\).
Solution:

Question 23.
If A, B, C are angles in a triangle, then prove that cos A + cos B – cos C = -1 + 4 \(\cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}\).
Solution:
Given A + B + C = 180°
⇒ \(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}\) = 90°
L.H.S = cos A + cos B – cos C


= R.H.S
∴ L.H.S = R.H.S
∴ cos A + cos B – cos C = -1 + 4 \(\cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}\).

Question 24.
In ΔABC, if r₁ = 8, r₂ = 12, r₃ = 24, find a, b, c.
Solution:
Given r₁ = 8, r₂ = 12, r₃ = 24

Δ = rs
⇒ 96 = 4s
⇒ s = 24
r₁ = \(\frac{\Delta}{s-a}\)
⇒ s – a = \(\frac{\Delta}{r_1}\)
⇒ 24 – a = \(\frac{96}{8}\)
⇒ 24 – a = 12
⇒ a = 12
r₂ = \(\frac{\Delta}{s-b}\)
⇒ s – b = \(\frac{\Delta}{r_2}\)
⇒ 24 – b = \(\frac{96}{12}\)
⇒ 24 – b = 8
⇒ b = 16
r₃ = \(\frac{\Delta}{s-c}\)
⇒ s – c = \(\frac{\Delta}{r_3}\)
⇒ s – c = \(\frac{96}{24}\)
⇒ 24 – c = 4
⇒ c = 20
∴ a = 12, b = 16, c = 20