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.

• Answer All the 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 + A^T and AA^T.
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)

II. Short Answer Type Questions.

• Answer any Five questions.
• 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)

III. Long Answer Type Questions.

• Answer any Five questions.
• 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 = I_B and gof = I_A 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)
⇒ I_A (x) = I_A (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)
= I_B (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) = I_A (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: