# AP Inter 1st Year Maths 1A Question Paper May 2018

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## AP 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.

• Each Question carries Two marks.

Question 1.
If f: R – {±1} → R is defined by f(x) = $$\log \left|\frac{1+x}{1-x}\right|$$, then show that $$f\left(\frac{2 x}{1+x^2}\right)$$ = 2f(x).
Solution:

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

Question 3.
If A = $$\left[\begin{array}{cc} 2 & -4 \\ -5 & 3 \end{array}\right]$$ then find A + AT and AAT.
Solution:

Question 4.
Find the rank of the 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 × 2 minors of A is zero.
∴ Rank of (A) ≠ 2
Since A is a non-zero matrix.
∴ Rank of (A) = 1

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

Question 7.
If $$\overline{\mathrm{a}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}-3 \overline{\mathrm{k}}$$ and $$\bar{b}=3 \bar{i}-\bar{j}+2 \bar{k}$$, then show that $$\bar{a}+\bar{b}$$ and $$\bar{a}-\bar{b}$$ are perpendicular to each other.
Solution:

Question 8.
If sec θ + tan θ = $$\frac{2}{3}$$, find the value of sin θ.
Solution:
Given sec θ + tan θ = $$\frac{2}{3}$$

Question 9.
If A is not an integral multiple of $$\frac{\pi}{2}$$, prove that tan A + cot A = 2 cosec 2A.
Solution:

Question 10.
If cosh x = $$\frac{5}{2}$$, find the values of (i) cosh (2x) and (ii) sinh (2x).
Solution:
Given cosh x = $$\frac{5}{2}$$
We know cosh2x – sinh2x = 1
⇒ sinh2x = cosh2x – 1
= $$\frac{25}{4}$$ – 1

Section – B
(5 × 4 = 20 Marks)

• Each question carries Four marks.

Question 11.
If θ – φ = $$\frac{\pi}{2}$$, then show that $$\left[\begin{array}{cc} \cos ^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin ^2 \theta \end{array}\right]\left[\begin{array}{cc} \cos ^2 \phi & \cos \theta \sin \phi \\ \cos \phi \sin \phi & \sin ^2 \phi \end{array}\right]$$ = 0.
Solution:
Given θ – φ = $$\frac{\pi}{2}$$

Question 12.
Let ABCDEF be a regular hexagon with a center ‘O’. Then show that $$\overline{\mathrm{AB}}+\overline{\mathrm{AC}}+\overline{\mathrm{AD}}+\overline{\mathrm{AE}}+\overline{\mathrm{AF}}=3 \overline{\mathrm{AD}}=6 \overline{\mathrm{AO}}$$.
Solution:

Question 13.
Let $$\overline{\mathrm{a}}=4 \overline{\mathrm{i}}+5 \overline{\mathrm{j}}-\overline{\mathrm{k}}, \overline{\mathrm{b}}=\overline{\mathrm{i}}-4 \overline{\mathrm{j}}+5 \overline{\mathrm{k}}$$ and $$\overline{\mathrm{c}}=3 \overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}$$. Find vector $$\bar{\alpha}$$ which is perpendicular to both $$\bar{a}$$ and $$\bar{b}$$ and $$\bar{\alpha} \cdot \bar{c}$$ = 21.
Solution:

Question 14.
Prove that cos2 76° + cos2 16° – cos 76° cos 16° = $$\frac{3}{4}$$.
Solution:
L.H.S = cos2 76° + cos2 16° – cos 76° cos 16°
= cos2 76° + 1 – sin2 16° – $$\frac{1}{2}$$(2 cos 76° cos 16°)
= 1 + cos2 76°- sin2 16° – $$\frac{1}{2}$$[cos(76° + 16°) + cos(76° – 16°)]
= 1 + cos(76° + 16°) cos(76° – 16°) – $$\frac{1}{2}$$[cos 92° – cos 60°]
= 1 + cos 92° . $$\frac{1}{2}$$ – $$\frac{1}{2}$$ cos 92° – $$\frac{1}{2}$$ . $$\frac{1}{2}$$
= 1 – $$\frac{1}{4}$$
= $$\frac{3}{4}$$
= R.H.S
∴ L.H.S = R.H.S
∴ cos2 76° + cos2 16° – cos 76° cos 16° = $$\frac{3}{4}$$

Question 15.
Solve √2(sin x + cos x) = √3.
Solution:
Given that √2 (sin x + cos x) = √3
⇒ sin x + cos x = $$\frac{\sqrt{3}}{\sqrt{2}}$$
By dividing both sides by √2, we get
⇒ cos x . $$\frac{1}{\sqrt{2}}$$ + sin x . $$\frac{1}{\sqrt{2}}$$ = $$\frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}$$
⇒ cos x cos 45° + sin x sin 45° = $$\frac{\sqrt{3}}{2}$$
⇒ cos (x – 45°) = $$\frac{\sqrt{3}}{2}$$
The principal value of cos(x – 45°) = $$\frac{\sqrt{3}}{2}$$ is 30°
∴ x – 45° = 2nπ ≠ 30°, n ∈ Z
x – 45° = 2nπ + 30° (or) x – 45° = 2nπ – 30°, n ∈ Z
⇒ x = 2nπ + 75° (or) x = 2nπ + 15°, n ∈ Z
⇒ x = 2nπ + $$\frac{5 \pi}{12}$$ (or) 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 $$\cos \left(2 \tan ^{-1} \frac{1}{7}\right)=\sin \left(2 \tan ^{-1} \frac{3}{4}\right)$$.
Solution:

Question 17.
Show that a2 cot A + b2 cot B + c2 cot C = $$\frac{abc}{R}$$ in a triangle.
Solution:
L.H.S = a2 cot A + b2 cot B + c2 cot C
= $$4 R^2 \sin ^2 A \cdot \frac{\cos A}{\sin A}+4 R^2 \sin ^2 B \cdot \frac{\cos B}{\sin B}+4 R^2 \sin ^2 C \cdot \frac{\cos C}{\sin C}$$
= 2R2 (2 sin A cos A + 2 sin B cos B + 2 sin C cos C)
= 2R2 (sin 2A + sin 2B + sin 2C)
= 2R2 (4 sin A sin B sin C)
Since sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
= $$\frac{1}{R}$$ (2R sin A) (2R sin B) (2R sin C)
= $$\frac{1}{R}$$ a.b.c
= $$\frac{abc}{R}$$
= R.H.S.
Hence a2 cot A + b2 cot B + c2 cot C = $$\frac{abc}{R}$$

Section – C
(5 × 7 = 35 Marks)

• Each question carries Seven marks.

Question 18.
Let f: A → B be a function. Then f is a bijection if and only if there exists a function g: B → A such that fog = IB and gof = IA and, in this case, g = f-1.
Solution:
To show that g = f-1
first, we have to show that f-1 exists,
i.e., we have to show that f is a bijection.
(i) To prove f is one-one.
Let x, y ∈ A and f(x) = f(y)
⇒ g[f(x)l = g[f(y)]
⇒ (gof) (x) = (gof) (y)
⇒ IA (x) = IA (y)
⇒ x = y
∴ f is one-one.
(ii) To prove f is onto
Let b ∈ B
Since g: B → A
⇒ g(b) ∈ A
let g(b) = a Then f(a) = f[g(b)]
= (fog) (b)
= IB (b)
= b
∴ f is onto
∴ f is one-one and onto
∴ f is a bijection
∴ f-1 exists.
(iii) To prove g = f-1
f:A → B ⇒ f-1: B → A
∴ g and f-1 have the same domain B.
Let b ∈ B, since f-1: B → A
⇒ f-1(b) ∈ A
Let f-1(b) = a then f(a) = b
g(b) = g[f(a)] = (gof) (a) = IA (a) = a = f-1(b)
Hence g = f-1

Question 19.
Using mathematical induction, prove that $$3.5^{2 n+1}+2^{3 n+1}$$ is divisible by 17, for all n ∈ N.
Solution:
Let p(n) be the statement that $$3.5^{2 n+1}+2^{3 n+1}$$ is divisible by 17.
If n = 1 then
$$3.5^{2+1}+2^{3+1}$$ = 3 . 53 + 24
= 375 + 16
= 391
= 17(23)
∴ p(1) is true
Assume that p(k) is true.

∴ p(k + 1) is true.
∴ By the principle of finite mathematical induction p(n) is true for all n ∈ N.
Hence $$3.5^{2 n+1}+2^{3 n+1}$$ is divisible by 17 for all n ∈ N.

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

= (a + b + c) (-a – b – c) (-a – b – c)
= (a + b + c) (-1) (a + b + c) (-1) (a + b + c)
= (a + b + c)3
= R.H.S
∴ L.H.S = R.H.S
∴ $${det}\left[\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right]$$ = (a + b + c)3

Question 21.
Solve the following equations by Gauss-Jordan Method:
x + y + z = 1, 2x + 2y + 3z = 6 and x + 4y + 9z = 3.
Solution:
Given x + y + z = 1
2x + 2y + 3z = 6
x + 4y + 9z = 3
The given system of equations can be expressed as AX = B

∴ x = 7, y = -10, z = 4

Question 22.
If $$\bar{a}=\bar{i}-2 \bar{j}-3 \bar{k}, \bar{b}=2 \bar{i}+\bar{j}-\bar{k}$$ and $$\bar{c}=\bar{i}+3 \bar{j}-2 \bar{k}$$, verify that $$\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \neq(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}$$.
Solution:

Question 23.
If A, B, C are the angles in a triangle, then prove that $$\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}-\sin ^2 \frac{C}{2}=1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$$.
Solution:
Given A, B, C are the angles of a triangle.
∴ A + B + C = 180°
⇒ $$\frac{A}{2}+\frac{B}{2}+\frac{C}{2}$$ = 90°

∴ $$\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}-\sin ^2 \frac{C}{2}=1-2 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$$

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