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

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AP 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.

• Attempt All questions.
• Each question carries Two marks.

Question 1.
If A = {-2, -1, 0, 1, 2} and f: A → B is a surjection defined by f(x) = x² + x + 1, then find B.
Solution:
Given A = {-2, -1, 0, 1, 2} and f(x) = x² + x + 1
f(-2) = (-2)² + (-2) + 1 = 4 – 2 + 1 = 3
f(-1) = (-1)² + (-1) + 1 = 1 – 1 + 1 = 1
f(0) = 0² + 0 + 1 = 0 + 0 + 1 = 1
f(1) = 1² + 1 + 1 = 1 + 1 + 1 = 3
f(2) = 2² + 2 + 1 = 4 + 2 + 1 = 7
Since f: A → B is a surjection.
∴ B = f(A) = {1, 3, 7}

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

Question 3.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
-5 & 3
\end{array}\right]\), then find A + A’ and AA’.
Solution:
Given A = \(\left[\begin{array}{cc}
2 & -4 \\
-5 & 3
\end{array}\right]\)
A’ = \(\left[\begin{array}{cc}
2 & -5 \\
-4 & 3
\end{array}\right]\)

Question 4.
Find the rank of the matrix \(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\).
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\)
det A = 1(6 – 0) – 2(8 + 1)
= 6 – 18
= -12 ≠ 0
∴ Rank of A = 3

Question 5.
If the position vectors of the points A, B and C are \(-2 \bar{i}+\bar{j}-\bar{k}\), \(-4 \bar{i}+2 \bar{j}+2 \bar{k}\) and \(6 \bar{i}-3 \bar{j}-13 \bar{k}\) respectively and \(\overline{\mathrm{AB}}=\lambda \overline{\mathrm{AC}}\), then find the value of λ.
Solution:

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

Question 7.
\(\bar{a}=2 \bar{i}-\bar{j}+\bar{k}, \bar{b}=\bar{i}-3 \bar{j}-5 \bar{k}\), Find the vector \(\bar{c}\) such that \(\bar{a}\), \(\bar{b}\) and \(\bar{c}\) form the sides of a triangle.
Solution:

Question 8.
Sketch the graph of the function sin 2x in the interval (0, π).
Solution:

Question 9.
Evaluate \(\cos ^2 52 \frac{1}{2}^{\circ}-\sin ^2 22 \frac{1}{2}^{\circ}\).
Solution:

Question 10.
If sinh x = 3, then show that x = log_e(3 + √10).
Solution:
Given sinh x = 3
⇒ x = sinh^-1(3)
= log_e(3 + \(\sqrt{3^2+1}\))
= log_e(3 + \(\sqrt{10}\))

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

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

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

Question 12.
If \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar vectors, prove that the following four points are co-planar \(6 \bar{a}+2 \bar{b}-\bar{c}, 2 \bar{a}-\bar{b}+3 \bar{c},-\bar{a}+2 \bar{b}-4 \bar{c}\), \(-12 \bar{a}-\bar{b}-3 \bar{c}\).
Solution:

Question 13.
If \(\bar{a}+\bar{b}+\bar{c}\) = 0, |\(\bar{a}\)| = 3, |\(\bar{b}\)| = 5 and |\(\bar{c}\)| = 7, then find the angle between \(\bar{a}\) and \(\bar{b}\).
Solution:

Question 14.
If a, b, c are non-zero real numbers and α, β are solutions of the equation a cos θ + b sin θ = c, then show that:
(i) sin α + sin β = \(\frac{2 b c}{a^2+b^2}\)
(ii) sin α . sin β = \(\frac{c^2-a^2}{a^2+b^2}\)
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 C + (c² – a²) = 0 ………(1)
since α and β are solutions of the given equation.
∴ sin α, sin β are the roots of (1)
(i) sum of the roots = sin α + sin β
= \(\frac{-(-2 b c)}{a^2+b^2}\)
= \(\frac{2 b c}{a^2+b^2}\)
(ii) product of the roots = sin α . sin β = \(\frac{c^2-a^2}{a^2+b^2}\)

Question 15.
Solve the equation cot²x – (√3 + 1) cot x + √3 = 0; 0 < x < \(\frac{\pi}{2}\).
Solution:
Given cot²x -(√3 + 1) cot x + √3 = 0
⇒ cot²x – √3 cot x – cot x + √3 = 0
⇒ cot x (cot x – √3) – 1(cot x – √3) = 0
⇒ (cot x – 1) (cot x – √3) = 0
∴ cot x – 1 = 0
⇒ cot x = 1
⇒ tan x = 1
⇒ x = nπ + \(\frac{\pi}{4}\), n ∈ z
∴ cot x – √3 = 0
⇒ cot x = √3
⇒ tan x = \(\frac{1}{\sqrt{3}}\)
⇒ x = nπ + \(\frac{\pi}{6}\), n ∈ z
0 < x < \(\frac{\pi}{2}\)
∴ x = \(\frac{\pi}{6}\), \(\frac{\pi}{4}\)

Question 16.
Prove that \(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}-\tan ^{-1} \frac{2}{9}\) = 0.
Solution:

Question 17.
Prove that \(\frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}=\frac{a^2+b^2}{a^2+c^2}\)
Solution:
L.H.S = \(\frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}\)

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

• Attempt Any Five questions.
• Each question carries Seven marks.

Question 18.
Let f: A → B, g: B → C be bijections, then prove that (gof)^-1 = f^-1 o g^-1.
Solution:
f: A → B, g: B → C are bijections
⇒ gof: A → C is a bijection
Also g^-1: C → B and f^-1: B → A are bijections
⇒ f^-1 o g^-1: C → A is a bijection.
Let c be any element of C.
Then ∃ an element b ∈ B such that g(b) = c
⇒ b = g^-1(c)
Also ∃ an element a ∈ A such that f(a) = b
⇒ a = f^-1(b)
Now (gof) (a) = g(f(a) = g(b) = c
⇒ a = (g o f)^-1 (c)
⇒ (g o f)^-1 (c) = a
Also (f^-1 o g^-1) (c) = f^-1(g^-1(c))
= f^-1(b)
= a
From (1) and (2);
(g o f)^-1(c) = f^-1 o g^-1(c)
⇒ (g o f)^-1 = f^-1 o g^-1

Question 19.
Using mathematical induction, prove that:
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{2 n+1}{n^2}\right)=(n+1)^2\)
Solution:
Let p(n) be the statement.
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{2 n+1}{n^2}\right)=(n+1)^2\)
and let S(n) be the product on the LHS
since S(1) = 1 + 3 = 4 = (1 + 1)² = 4
∴ P(a) 4 time for n = 1
Assume that p(n) is true for n = K
S(K) = \(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \ldots\left(1+\frac{2 k+1}{k^2}\right)=(k+1)^2\)
We show that P(n) is true for n = K + 1
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \cdots\left(\frac{k^2+2 K+1}{K^2}\right)\left(1+\frac{2 k+2+1}{(K+1)^2}\right)\)
= (k + 1)² + 2k + 3
= k² + 2k + 1 + 2k + 3
= k² + 4k + 4
= (k + 2)²
= (k + 1 + 1)²
∴ P(n) is true for n = k + 1
By the principle of Mathematical Induction p(n) is true & n ∈ N

Question 20.
Find the value of x if \(\left|\begin{array}{ccc}
x-2 & 2 x-3 & 3 x-4 \\
x-4 & 2 x-9 & 3 x-16 \\
x-8 & 2 x-27 & 3 x-64
\end{array}\right|\) = 0
Solution:

⇒ (x – 2) (30 – 24) – (2x – 3) (10 – 6) + (3x – 4)(4 – 3) = 0
⇒ (x – 2) (6) – (2x – 3) (4) + (3x – 4) (1) = 0
⇒ 6x – 12 – 8x + 12 + 3x – 4 = 0
⇒ x – 4 = 0
⇒ x = 4

Question 21.
Examine whether the equations are consistent or inconsistent and if consistent find the complete solution:
x + y + z = 6, x – y + z = 2, 2x – y + 3z = 9
Solution:
Given system of equations are
x + y + z = 6
x – y + z = 2
2x – y + 3z = 9
Given the system of equations can be represented as a matrix equation AX = B


Here r₁ = 3, r₂ = 3 and n = 3
∴ r₁ = r₂ = n
∴ The given system of equations is consistent and it has unique solution.
∴ Given equations are equivalent to
x + y + z = 6 ………(1)
y = 2
z = 3
From (1), x + 2 + 3 = 6
⇒ x = 1
∴ x = 1, y = 2, z = 3

Question 22.
\(\bar{a}, \bar{b}, \bar{c}\) are three vectors, then prove that \((\bar{a} \times \bar{b}) \times \bar{c}=(\bar{a} \cdot \bar{c}) \bar{b}-(\bar{b} \cdot \bar{c}) \bar{a}\).
Solution:
Equation (1) is evidently true if a and b are parallel.
Now suppose that a and b are non-parallel.
Let O denote the origin. Choose points A and B such that \(\overline{\mathrm{OA}}=\overline{\mathrm{a}}\) and \(\overline{\mathrm{OB}}=\overline{\mathrm{b}}\)
Since \(\bar{a}\) and \(\bar{b}\) are non-parallel, the points O, A, and B are non-collinear. Hence they determine a plane.
Let i denote the unit vector along \(\bar{OA}\).
Let j be a unit vector in the OAB plane perpendicular to i.

Question 23.
If A, B, C are angles in a triangle, then prove that: sin²A + sin²B – sin²C = 2 sin A sin B cos C.
Solution:
Since A, B, C are angles in a triangle.
∴ A + B + C = 180°
L.H.S = sin²A + sin²B – sin²C
= sin²A + sin (B + C) sin (B – C)
= sin²A + sin(180 – A) sin (B – C)
= sin²A + sin A sin (B – C)
= sin A [sin A + sin (B – C)]
= sin A [sin(180 – (B + C)) + sin(B – C)]
= sin A [sin (B + C) + sin (B – C)]
= sin A [2 sin B cos C]
= 2 sin A sin B cos C
= R.H.S.
∴ sin²A + sin²B – sin²C = 2 sin A sin B cos C

Question 24.
In ΔABC, show that: (r₁ + r₂) \(\sec ^2 \frac{C}{2}\) = (r₂ + r₃) \(\sec ^2 \frac{A}{2}\) = (r₃ + r₁) \(\sec ^2 \frac{B}{2}\).
Solution: