AP Inter 1st Year Maths 1A Question Paper March 2020

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AP Inter 1st Year Maths 1A Question Paper March 2020

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(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) ∀ x ∈ R then find (gof) (x).
Solution:
f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) ∀ x ∈ R
(gof) (x) = g(f(x))
= g(2x – 1), {∵ f(x) = 2x – 1}
= \(\frac{(2 x-1)+1}{2}\) {∵ g(x) = \(\frac{x+1}{2}\)}
= x
∴ (g o f) (x) = x

Question 2.
If f = {(1, 2), (2, -3), (3, -1)} then find (i) 2f (ii) f².
Solution:
Given f = {(1, 2), (2, -3), (3, -1)}
(i) 2f = {(1, 2(2)), (2, 2(-3), (3, 2(-1))} = {(1, 4), (2, -6), (3, -2)}
(ii) f² = {(1, 22), (2, (-3)2), (3, (-1)2)} = {(1, 4), (2, 9), (3, 1)}

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

Question 4.
Find the rank of the matrix A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]\).
Solution:
A ~ \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]\) (on interchanging R₁ and R₂)
~ \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 2 \\
0 & -4 & -8
\end{array}\right]\)
R₃ → R₃ – 3R₁
~ \(\left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 0
\end{array}\right]\)
R₁ → R₁ – 2R₂
R₃ → R₃ + 4R₂
The last matrix is singular and \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) is a non-singular submatrix of it. Hence its rank is 2.
∴ Rank (A) = 2.

Question 5.
If \(\overline{\mathrm{a}}=2 \overline{\mathrm{i}}+5 \overline{\mathrm{j}}+\overline{\mathrm{k}}\) and \(\overline{\mathrm{b}}=4 \overline{\mathrm{i}}+\mathrm{m} \overline{\mathrm{j}}+n \overline{\mathrm{k}}\) are collinear vectors then find m, n.
Solution:
Since \(\overline{\mathrm{a}}=2 \overline{\mathrm{i}}+5 \overline{\mathrm{j}}+\overline{\mathrm{k}}\) and \(\overline{\mathrm{b}}=4 \overline{\mathrm{i}}+\mathrm{m} \overline{\mathrm{j}}+n \overline{\mathrm{k}}\) are collinear.
⇒ \(\frac{2}{4}=\frac{5}{m}=\frac{1}{n}\)
⇒ \(\frac{1}{2}=\frac{5}{m}\) and \(\frac{1}{n}=\frac{1}{2}\)
⇒ m = 10 and n = 2

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.
Find the angle between the planes \(\bar{r} \cdot(2 \bar{i}-\bar{j}+2 \bar{k})=3\) and \(\bar{r} \cdot(3 \bar{i}+6 \bar{j}+\bar{k})=4\).
Solution:

Question 8.
If tan 20° = λ then show that \(\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \tan 110^{\circ}}=\frac{1-\lambda^2}{2 \lambda}\).
Solution:

Question 9.
Find the range of 7 cos x – 24 sin x + 5.
Solution:

Question 10.
Prove that (cosh x – sinh x)ⁿ = cosh (nx) – sinh (nx). Such as for any n ∈ Rx.
Solution:

Section – B
(5 × 4 = 20 Marks)

II. Short Answer Type Questions.

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

Question 11.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\) then show that AA’ = A’A = I.
Solution:

Question 12.
If the points whose position vectors are \(3 \bar{i}-2 \bar{j}-\bar{k}, 2 \bar{i}+3 \bar{j}-4 \bar{k},-\bar{i}+\bar{j}+2 \bar{k}\) and \(4 \bar{i}+5 \bar{j}+\lambda \bar{k}\) are coplanar then show that λ = \(\frac{-146}{17}\).
Solution:
Let ‘O’ be the origin and Let A, B, C, and D be the given points. Then

Question 13.
Find the vector area and area of the parallelogram having \(\overline{\mathrm{a}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}-\overline{\mathrm{k}}\) and \(\overline{\mathrm{b}}=2 \overline{\mathrm{i}}-\overline{\mathrm{j}}+2 \overline{\mathrm{k}}\) as adjacent sides.
Solution:

Question 14.
Prove that \(\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}=\frac{3}{2}\).
Solution:

Question 15.
Solve 7 sin²θ + 3 cos²θ = 4.
Solution:
Given that 7 sin²θ + 3 cos²θ = 4
⇒ 7 sin²θ + 3(1 – sin²θ) = 4
⇒ 4 sin²θ = 1
⇒ sin θ = ±\(\frac{1}{2}\)
∴ Principal solutions are θ = ±\(\frac{\pi}{6}\) and the general solution is given by θ = nπ ± \(\frac{\pi}{6}\), n ∈ Z.

Question 16.
Prove that \(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\).
Solution:

Question 17.
If cot \(\frac{A}{2}\), cot \(\frac{B}{2}\), cot \(\frac{C}{2}\) are in A.P. then prove that a, b, c are in A.P.
Solution:
cot \(\frac{A}{2}\), cot \(\frac{B}{2}\), cot \(\frac{C}{2}\) are in A.P
⇒ \(\frac{(s)(s-a)}{\Delta}, \frac{(s)(s-b)}{\Delta}, \frac{(s)(s-c)}{\Delta}\) are in A.P
⇒ s – a, s – b, s – c are in A.P
⇒ -a, -b, -c are in A.P
⇒ a, b, c are in A.P

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

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

Question 18.
(a) If f(x) = \(\frac{x+1}{x-1}\), (x ≠ ±1) then find (fofof) (x).
(b) If f: A → B, g: B → C, h: C → D are functions then show that ho(gof) = (hog)of.
Solution:
(a) f(x) = \(\frac{x+1}{x-1}\), (x ≠ ±1)
(fofof) (x) = (fof) [f(x)]
= (fof) \(\left(\frac{x+1}{x-1}\right)\), {∵ f(x) = \(\frac{x+1}{x-1}\)}

(b) f: A → B and g: B → C, gof: A → C
Now g o f: A → C and h: C → D
⇒ h o (g o f): A → D
Similarly (h o g) o f: A → D
Thus h o (g o f) and (h o g) of both exist and have the same domain A and co-domain D.
Let a be an element of A.
Now [h o (g o f)] (a) = h[(g o f)(a)]
= h[g(f(a))]
= (h o g) [f(a)]
= [(h o g) o f] (a)
∴ h o (g o f) = (h o g) o f
Thus composition of mappings is associative.

Question 19.
Show that 49ⁿ + 16n – 1 is divisible by 64 for all positive integers by using Mathematical induction.
Solution:
Let p(n) be the statement.
49ⁿ + 16n – 1 is divisible by 64
Since 49¹ + 16(1) – 1 = 64 is divisible by 64
∴ The statement is true for n = 1
Assume that the statement p(n) is true for n = k
(i.e.,) 49^k + 16k – 1 is divisible by 64
Then (49^k + 16k – 1) = 64t, for some t ∈ N ……….(1)
we show that the statement is true for n = k + 1.
(i.e.,) we show that 49^k+1 + 16(k + 1) – 1 is divisible by 64
From (1), we have 49^k + 16k – 1 = 64t
⇒ 49^k = 64t – 16k + 1
⇒ 49^k . 49 = (64t – 16k + 1) . 49
⇒ 49^k+1 + 16(k + 1) – 1 = (64t – 16k + 1)49 + 16(k + 1) – 1
⇒ 49^k+1 + 16(k + 1) -1 = 64 (49t – 12k + 1)
Here (49t – 12k + 1) is an integer.
∴ 49^k+1 + 16(k + 1) – 1 is divisible by 64.
∴ The statements true for n = k + 1
∴ By the principle of mathematical induction,
∴ p(n) is true for all n ∈ N.
(i.e.,) 49ⁿ + 16n – 1 is divisible by 64, ∀ n ∈ N.

Question 20.
Show that \(\left|\begin{array}{ccc}
a^2+2 a & 2 a+1 & 1 \\
2 a+1 & a+2 & 1 \\
3 & 3 & 1
\end{array}\right|\) = (a – 1)³.
Solution:

= (a – 1)² [0(6 – 3) – 0[3(a + 1) – 3)] + 1(a + 1 – 2)]
= (a – 1)² (a – 1)
= (a – 1)³
= R.H.S.

Question 21.
Solve the equations using Cramer’s Rule x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3.
Solution:
Cramer’s rule
∆ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right|\)
= 1(18 – 12) – 1(18 – 3) + 1(8 – 2)
= 6 – 15 + 6
= -3
∆₁ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
6 & 2 & 3 \\
3 & 4 & 9
\end{array}\right|\)
= 1(18 – 12) – 1(54 – 9) + 1(24 – 6)
= 6 – 45 + 18
= -21
∆₂ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 6 & 3 \\
1 & 3 & 9
\end{array}\right|\)
= 1(54 – 9) – 1(18 – 3) + 1(6 – 6)
= 45 – 15
= 30
∆₃ = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 6 \\
1 & 4 & 3
\end{array}\right|\)
= 1(6 – 24) – 1(6 – 6) + 1(8 – 2)
= -18 – 0 + 6
= -12
x = \(\frac{\Delta_1}{\Delta}=\frac{-21}{-3}\) = 7
y = \(\frac{\Delta_2}{\Delta}=\frac{30}{-3}\) = -10
z = \(\frac{\Delta_3}{\Delta}=\frac{-12}{-3}\) = 4
Solution is x = 7, y = -10, z = 4

Question 22.
Find the shortest distance between the skew lines \(\overline{\mathrm{r}}=(6 \overline{\mathrm{i}}+2 \overline{\mathrm{j}}+2 \overline{\mathrm{k}})+\mathbf{t}(\overline{\mathrm{i}}-2 \overline{\mathrm{j}}+2 \overline{\mathrm{k}})\) and \(\overline{\mathbf{r}}=(-4 \bar{i}-\bar{k})+s(3 \bar{i}-2 \bar{j}-2 \bar{k})\).
Solution:
The first line passes through point A(6, 2, 2) and is parallel to the vector b = i – 2j + 2k.
Second line passes through the point C(-4, 0, -1) and is parallel to the vector d = 3i – 2j – 2k
Shortest distance = \(\frac{|[\mathrm{AC} \quad \mathrm{b} d]|}{|\mathrm{b} \times \mathrm{d}|}\)
d = PQ = ST |cos θ| = \(\frac{\left(b_1 \times b_2\right) \cdot\left(a_2-a_1\right)}{\left|b_1 \times b_2\right|}\)
b × d = \(\left|\begin{array}{ccc}
i & j & k \\
1 & -2 & 2 \\
3 & -2 & -2
\end{array}\right|\) = 8i + 8j + 4k and |b × d| = 12
∴ The shortest distance between the skew lines
\(\frac{|[\mathrm{ACbd}]|}{|\mathrm{b} \times \mathrm{d}|}=\frac{108}{12}\) = 9
[A C b d] = \(\left|\begin{array}{ccc}
-10 & -2 & -3 \\
1 & -2 & 2 \\
3 & -2 & -2
\end{array}\right|\) = -108

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:
A + B + C = \(\frac{\pi}{2}\) ………(1)
L H S = cos 2A + cos 2B + cos 2C
= 2 cos\(\left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right)\) cos\(\left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)\) + cos 2C
= 2 cos (A + B) . cos (A – B) + cos 2C
= 2 cos (90° – C) cos (A – B) + cos 2C
= 2 sin C cos (A – B) + (1 – 2 sin2C)
= 1 + 2 sin C [cos (A – B) – sin C]
= 1 + 2 sin C [cos (A – B) – sin (90° – \(\overline{A+B}\))]
= 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
∴ cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C

Question 24.
Prove that r + r₃ + r₁ – r₂ = 4R cos B.
Solution: