TS Inter 1st Year Maths 1A Question Paper May 2019

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TS Inter 1st Year Maths 1A Question Paper May 2019

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 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) = \(\frac{2 x^2-5 x+7}{(x-1)(x-2)(x-3)}\).
Solution:
Given f(x) = \(\frac{2 x^2-5 x+7}{(x-1)(x-2)(x-3)}\)
f(x) ∈ R ⇒ (x – 1) (x – 2) (x – 3) ≠ 0
⇒ x ≠ 1, 2, 3
∴ Domain of f = R – {1, 2, 3}

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & x
\end{array}\right]\) and det A = 45, then find x.
Solution:
Given A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & x
\end{array}\right]\)
det A = 45
⇒ \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
2 & 3 & 4 \\
5 & -6 & x
\end{array}\right]\) = 45
⇒ 1(3x + 24) = 45
⇒ 3x = 45 – 24
⇒ 3x = 21
⇒ x = 7

Question 4.
Find the trace of \(\left[\begin{array}{ccc}
1 & 3 & -5 \\
2 & -1 & 5 \\
2 & 0 & 1
\end{array}\right]\).
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 3 & -5 \\
2 & -1 & 5 \\
2 & 0 & 1
\end{array}\right]\)
Trace of (A) = 1 + (-1) + 1
= 1 – 1 + 1
= 1

Question 5.
Let \(\overline{\mathrm{a}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}+3 \overline{\mathrm{k}}\) and \(\overline{\mathrm{b}}=3 \overline{\mathrm{i}}+\overline{\mathrm{j}}\). Find the unit vector in the direction of \(\overline{\mathrm{a}}+\overline{\mathrm{b}}\).
Solution:
Given \(\overline{\mathrm{a}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}+3 \overline{\mathrm{k}}\)
\(\overline{\mathrm{b}}=3 \overline{\mathrm{i}}+\overline{\mathrm{j}}\)

Question 6.
Find the vector equation of the line passing through the point \(2 \bar{i}+3 \bar{j}+\bar{k}\) and parallel to the vector \(4 \bar{i}-2 \bar{j}+3 \bar{k}\).
Solution:

Question 7.
Find the angle between the vectors \(\overline{\mathrm{i}}+2 \overline{\mathrm{j}}+3 \overline{\mathrm{k}}\) and \(3 \overline{\mathbf{i}}-2 \overline{\mathbf{j}}+2 \overline{\mathrm{k}}\).
Solution:

Question 8.
Express \(\frac{\left(\sqrt{3} \cos 25^{\circ}+\sin 25^{\circ}\right)}{2}\) as a sin of an angle.
Solution:
\(\frac{\left(\sqrt{3} \cos 25^{\circ}+\sin 25^{\circ}\right)}{2}=\frac{\sqrt{3} \cos 25^{\circ}}{2}+\frac{\sin 25^{\circ}}{2}\)
= \(\frac{\sqrt{3}}{2}\) cos 25° + \(\frac{1}{2}\) sin 25°
= sin 60° cos 25° + cos 60° sin 25°
= sin (60° + 25°)
= sin 85°

Question 9.
Express \(\frac{1-\cos \theta+\sin \theta}{1+\cos \theta+\sin \theta}\) in terms of \(\tan \frac{\theta}{2}\).
Solution:

Question 10.
Prove that \(\frac{\tanh x}{{sech} x-1}+\frac{\tanh x}{{sech} x+1}\) = -2 cosech x, for x ≠ 0.
Solution:

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

• Attempt any Five questions.
• Each question carries Four marks.

Question 11.
If A = \(\left[\begin{array}{ccc}
2 & -1 & 2 \\
1 & 3 & -4
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & -2 \\
-3 & 0 \\
5 & 4
\end{array}\right]\), then show that (AB)’ = B’A’.
Solution:

Question 12.
Verify whether the triangle formed by the vectors \(3 \overline{\mathrm{i}}+5 \overline{\mathrm{j}}+2 \overline{\mathrm{k}}\), \(2 \overline{\mathrm{i}}-3 \overline{\mathrm{j}}-5 \bar{k},-5 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}+3 \bar{k}\) is equilateral or not.
Solution:

∴ AB = BC = CA
∴ The given vectors form an equilateral triangle.

Question 13.
If θ is the angle between vectors \(\overline{\mathbf{i}}+\overline{\mathbf{j}}\) and \(\overline{\mathbf{j}}+\overline{\mathbf{k}}\), then find sin θ.
Solution:
Let \(\overline{\mathrm{a}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}\)
\(\bar{b}=\bar{j}+\bar{k}\)

Question 14.
Prove that \(\frac{\cos ^3 \theta-\cos 3 \theta}{\cos \theta}+\frac{\sin ^3 \theta+\sin 3 \theta}{\sin \theta}\) = 3.
Solution:

Question 15.
If α, β are the solutions of the equation a cos θ + b sin θ = c, where a, b, c ∈ R and if a² + b² > 0 and sin α ≠ sin β, then show that sin α + sin β = \(\frac{2 b c}{a^2+b^2}\).
Solution:
Given a cos θ + b sin θ = c
⇒ a cos θ = c – b sin θ
⇒ a² cos²θ = (c – b sin θ)²
⇒ a² (1 – sin²θ) = c² + b² sin²θ – 2bc sin θ
⇒ a² – a² sin²θ = c² + b² sin²θ – 2bc sin θ
⇒ (a² + b²) sin²θ – 2bc sin θ – (c² – a²) = 0 ………(1)
Since α and β are the solutions of the given equation.
∴ sin α, sin β are the roots of (1)
Sum of the roots sin α + sin β = \(\frac{-(-2 b c)}{a^2+b^2}=\frac{2 b c}{a^2+b^2}\).

Question 16.
Prove that \(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\frac{S^2}{\Delta}\).
Solution:

Question 17.
Prove that \(\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right)-\tan ^{-1}\left(\frac{2}{9}\right)\) = 0.
Solution:

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

• Attempt any FIVE questions.
• Each question carries Seven marks.

Question 18.
If f: A → B, g: B → C are any two bijective functions, then prove that gof: A → C is also a bijective function.
Solution:
Given f: A → B, g: B → C are two bijective functions.
f: A → B, g: B → C
⇒ gof: A → C.
(i) To prove gof: A → C is one-one
Let a₁, a₂ ∈ A.
(gof) (a₁) = (gof) (a₂)
⇒ g[f(a₁)] = g[f(a₂)]
⇒ f(a₁) = f(a₂) [∴ g is one-one]
⇒ a₁ = a₂ [∴ f is one-one]
∴ gof: A → C is one-one.
(ii) To prove gof: A → C is onto.
Let c ∈ C.
Since g: B → C is a bijection.
∴ There exists b ∈ B such that g(b) = c
Since f: A → B is a bijection.
∴ There exists a ∈ A such that f(a) = b
C = g(b)
= g(f(a))
= (gof) (a)
∴ There exists a ∈ A such that (gof) (a) = c.
∴ gof: A → C is onto.
∴ gof: A → C is one-one and onto.
∴ gof: A → C is bijection.

Question 19.
Using mathematical induction, prove the statement 1³ + 2³ + 3³ + …….. + n³ = \(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\), ∀ n ∈ N.
Solution:
Let p(n) be the statement that
1³ + 2³ + 3³ + ……… + n³ = \(\frac{n^2(n+1)^2}{4}\)
If n = 1 then
L.H.S = 1³ = 1
R.H.S = \(\frac{1^2(1+1)^2}{4}=\frac{1(4)}{4}\) = 1
∴ L.H.S = R.H.S
∴ P(1) is true.
Assume that P(K) is true.
∴ 1³ + 2³ + 3³ + ……… + k³ = \(\frac{k^2(k+1)^2}{4}\)
1³ + 2³ + 3³ + …….. + k³ + (k + 1)³ = \(\frac{k^2(k+1)^2}{4}\) + (k + 1)³

∴ P(k + 1) is true.
By the principle of finite mathematical induction p(n) is true ∀ n ∈ N.
Hence 1³ + 2³ + 3³ + ……… + n³ = \(\frac{n^2(n+1)^2}{4}\) ∀ n ∈ N.

Question 20.
x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3. Solve the system of equations by using the matrix inversion method.
Solution:
Given system of equations are
x + y + z = 1
2x + 2y + 3z = 6
x + 4y + 9z = 3
Given a system of equations can written as a matrix equation AX = B.

Question 21.
Show that \(\left|\begin{array}{lll}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right|\) = (a – b) (b – c) (c – a) (ab + bc + ca).

Question 22.
Find the shortest distance between the lines \(r=6 \bar{i}+2 \bar{j}+2 \bar{k}+t(\bar{i}-2 \bar{j}+2 \bar{k})\) and \(r=-4 \bar{i}-\bar{k}+s(3 \bar{i}-2 \bar{j}-2 \bar{k})\).
Solution:

Question 23.
If A + B + C = \(\frac{\pi}{2}\), then prove that cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C.
Solution:
Given A + B + C = \(\frac{\pi}{2}\)
L.H.S = cos 2A + cos 2B + cos 2C
= \(2 \cos \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \cos \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\) + 1 – 2 sin²C
= 2 cos (A + B) cos (A – B) + 1 – 2 sin²C
= 1 + 2 cos (90° – C) cos (A – B) – 2 sin²C
= 1 + 2 sin C cos (A – B) – 2 sin²C
= 1 + 2 sin C [cos (A – B) – sin C]
= 1 + 2 sin C [cos (A – B) – sin {90° – (A + B)}]
= 1 + 2 sin C [cos(A – B) – cos (A + B)]
= 1 + 2 sin C [2 sin A sin B]
= 1 + 4 sin A sin B sin C
= R.H.S
∴ L.H.S = R.H.S
∴ cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C.

Question 24.
In a ΔABC, prove that 4(r₁r₂ + r₂r₃+ r₃r₁) = (a + b + c)².
Solution:
L.H.S = 4(r₁r₂ + r₂r₃ + r₃r₁)

= 4s (3s – 2s)
= 4s (s)
= 4s²
= (2s)²
= (a + b + c)²
= R.H.S
∴ L.H.S = R.H.S
∴ 4(r₁r₂ + r₂r₃ + r₃r₁) = (a + b + c)²