AP Inter 1st Year Maths 1A Question Paper March 2018

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AP 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.
Find the domain of the real-valued function f(x) = \(\sqrt{x^2-25}\).
Solution:
Given f(x) = \(\sqrt{x^2-25}\)
f(x) ∈ R ⇔ x² – 25 ≥ 0
⇔ (x + 5) (x – 5) ≥ 0
⇔ x ∈ (-∞, -5) ∪ (5, ∞)
∴ Domain of f = (-∞, -5) ∪ (5, ∞).

Question 2.
If f : R → R, g : R → R are defined by f(x) = 3x – 1, g(x) = x² + 1 then find (fog) (2).
Solution:
Given f(x) = 3x – 1 and g(x) = x² + 1
(fog) (2) = f[g(2)]
= f[2² + 1]
= f[4 + 1]
= f(5)
= 3(5) – 1
= 15 – 1
= 14
∴ (fog) (2) = 14

Question 3.
Define a symmetric matrix. Give one example of order 3 × 3.
Solution:
Symmetric Matrix: A square matrix ‘A’ is said to be a symmetric matrix if A^T = A.
Example: A = \(\left[\begin{array}{ccc}
1 & 2 & 0 \\
2 & -3 & -1 \\
0 & -1 & 4
\end{array}\right]\)

Question 4.
Find the inverse of the matrix \(\left[\begin{array}{cc}
1 & 2 \\
3 & -5
\end{array}\right]\).
Solution:

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

Question 6.
Find the vector equation of the plane passing through the points (0, 0, 0), (0, 5, 0) and (2, 0, 1).
Solution:
Let A = (0, 0, 0)
B = (0, 5, 0)
C = (2, 0, 1)
The vector equation of the plane passing through the points A(\(\bar{a}\)), B(\(\bar{b}\)), C(\(\bar{c}\)) is

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

Question 8.
Find the value of sin 330° cos 120° + cos 210° sin 300°.
Solution:
sin 330° . cos 120° + cos 210° . sin 300°
= sin (360° – 30°) . cos (180° – 60°) + cos(180° + 30°) . sin(360° – 60°)
= (-sin 30°) . (-cos 60°) + (-cos 30°) . (-sin 60°)
= \(\left(\frac{-1}{2}\right)\left(\frac{-1}{2}\right)+\left(\frac{-\sqrt{3}}{2}\right)\left(\frac{-\sqrt{3}}{2}\right)\)
= \(\frac{1}{4}+\frac{3}{4}\)
= 1
∴ sin 330° . cos 120° + cos 210° . sin 300° = 1

Question 9.
Find the extreme values of cos 2x + cos²x.
Solution:
Let f(x) = cos 2x + cos²x
= cos 2x + \(\frac{1+\cos 2 x}{2}\)
= cos 2x + \(\frac{1}{2}\) + \(\frac{1}{2}\) cos 2x
= \(\frac{1}{2}\) + (1 + \(\frac{1}{2}\)) cos 2x
= \(\frac{1}{2}\) + \(\frac{3}{2}\) cos 2x
We know -1 ≤ cos 2x ≤ 1
⇒ \(\frac{-3}{2} \leq \frac{3}{2} \cos 2 x \leq \frac{3}{2}\)
⇒ \(\frac{1}{2}-\frac{3}{2} \leq \frac{1}{2}+\frac{3}{2} \cos 2 x \leq \frac{1}{2}+\frac{3}{2}\)
⇒ -1 ≤ f(x) ≤ 2
∴ Minimum value = -1, Maximum value = 2

Question 10.
For any x ∈ R show that cosh 2x = 2 cosh²x – 1.
Solution:
L.H.S = cosh 2x
= cosh²x + sinh²x
= cosh²x + cosh²x – 1 {∵ cosh²x – sinh²x = 1}
= 2 cosh²x – 1
= R.H.S
∴ L.H.S = R.H.S
Hence cosh 2x = 2 cosh²x – 1

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

• Answer any five questions.
• Each question carries four marks.

Question 11.
If A = \(\left[\begin{array}{rr}
7 & -2 \\
-1 & 2 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{rr}
-2 & -1 \\
4 & 2 \\
-1 & 0
\end{array}\right]\) then find AB’ and BA’.
Solution:

Question 12.
If \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar vectors, prove that the following four points are co-planar.
\(-\overline{\mathrm{a}}+4 \overline{\mathrm{b}}-3 \overline{\mathrm{c}}, 3 \overline{\mathrm{a}},+2 \overline{\mathrm{b}}-5 \overline{\mathrm{c}},-3 \overline{\mathrm{a}}+8 \overline{\mathrm{b}}-5 \overline{\mathrm{c}} \text { and }-3 \overline{\mathrm{a}}+2 \overline{\mathrm{b}}+\overline{\mathrm{c}}\)
Solution:

Question 13.
Let \(\bar{a}\) and \(\bar{b}\) be vectors, satisfying |\(\bar{a}\)| = |\(\bar{b}\)| = 5 and (\(\bar{a}\), \(\bar{b}\)) = 45°. Find the area of the triangle having \(\bar{a}-2 \bar{b}\) and \(3 \bar{a}+2 \bar{b}\) as two of its sides.
Solution:
Given |\(\bar{a}\)| = |\(\bar{b}\)| = 5 and (\(\bar{a}\), \(\bar{b}\)) = 45°
Let \(\overline{\mathrm{AB}}=\overline{\mathrm{a}}-2 \overline{\mathrm{b}}\)

Question 14.
If A is not an integral multiple of \(\frac{\pi}{2}\) then prove that
(i) tan A + cot A = 2 cosec 2A
(ii) cot A – tan A = 2 cot 2A
Solution:

Question 15.
Solve the equation √3 sin θ – cos θ = √2.
Solution:

Question 16.
Prove that \(\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)+\sin ^{-1}\left(\frac{16}{65}\right)=\frac{\pi}{2}\).
Solution:
Let \(\sin ^{-1}\left(\frac{4}{5}\right)\) = A and \(\sin ^{-1}\left(\frac{5}{13}\right)\) = B
Then sin A = \(\frac{4}{5}\) and sin B = \(\frac{5}{13}\)

Question 17.
If a = (b – c) sec θ, prove that tan θ = \(\frac{2 \sqrt{b c}}{b-c} \sin \frac{A}{2}\).
Solution:
Given a = (b – c) sec θ
⇒ sec θ = \(\frac{a}{b-c}\)
We know tan2θ = sec²θ – 1

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: A → C is a bijection.
Solution:
Given f: A → B, g: B → C be bijections.
(i) To prove gof: A → C, is one-one
Let a₁, a₂ ∈ a Then f(a₁), f(a₂) ∈ B
(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 onto
∴ There exists b ∈ B such that g(b) = c
Since f: A → B is onto
∴ 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
Hence gof: A → C is a bijective function.

Question 19.
Using mathematical induction, prove that \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots .+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\) for all n ∈ N.
Solution:
Let p(n) be the statement that


∴ p(k + 1) is true.
∴ By the principle of finite mathematical induction p(n) is true for all n ∈ N.
∴ \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\ldots .+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1}\) for all n ∈ N.

Question 20.
Show that \(\left|\begin{array}{ccc}
a+b+2 c & a & b \\
c & b+c+2 a & b \\
c & a & c+a+2 b
\end{array}\right|\) = 2(a + b + c)³.
Solution:

= 2(a + b + c) (1) (b + c + a) (c + a + b)
= 2(a + b + c)³
= R.H.S.
∴ L.H.S = R.H.S
Hence \(\left|\begin{array}{ccc}
a+b+2 c & a & b \\
c & b+c+2 a & b \\
c & a & c+a+2 b
\end{array}\right|\) = 2(a + b + c)³

Question 21.
Solve the following system of equations by Gauss-Jordan Method:
2x – y + 3z = 9, x + y + z = 6 and x – y + z = 2
Solution:
Given that the system of equations are
2x – y + 3z = 9
x + y + z = 6
x – y + z = 2
The given system of equations can be expressed as AX = B


∴ x = 1, y = 2, z = 3

Question 22.
For any four vectors \(\bar{a}, \bar{b}, \bar{c}\) and \(\bar{d}\), prove that
(i) \((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{c}} \times \overline{\mathrm{d}})=[\overline{\mathrm{a}} \overline{\mathrm{c}} \overline{\mathrm{d}}] \overline{\mathrm{b}}-[\overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{d}}] \overline{\mathrm{a}}\)
(ii) \((\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(\overline{\mathrm{c}} \times \overline{\mathrm{d}})=[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{d}}] \overline{\mathrm{c}}-[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}] \overline{\mathrm{d}}\)
Solution:

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

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

Question 24.
Show that in any triangle ABC, r + r₃ + r₁ – r₂ = 4R cos B.
Solution: