Thoroughly analyzing AP Inter 1st Year Maths 1A Model Papers and AP Inter 1st Year Maths 1A Question Paper March 2020 helps students identify their strengths and weaknesses.

## 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) f^{2}.

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) f^{2} = {(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 R_{1} and R_{2})

~ \(\left[\begin{array}{ccc}

1 & 2 & 3 \\

0 & 1 & 2 \\

0 & -4 & -8

\end{array}\right]\)

R_{3} → R_{3} – 3R_{1}

~ \(\left[\begin{array}{ccc}

1 & 0 & -1 \\

0 & 1 & 2 \\

0 & 0 & 0

\end{array}\right]\)

R_{1} → R_{1} – 2R_{2}

R_{3} → R_{3} + 4R_{2}

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 sin^{2}θ + 3 cos^{2}θ = 4.

Solution:

Given that 7 sin^{2}θ + 3 cos^{2}θ = 4

⇒ 7 sin^{2}θ + 3(1 – sin^{2}θ) = 4

⇒ 4 sin^{2}θ = 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 49^{n} + 16n – 1 is divisible by 64 for all positive integers by using Mathematical induction.

Solution:

Let p(n) be the statement.

49^{n} + 16n – 1 is divisible by 64

Since 49^{1} + 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.,) 49^{n} + 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 + r_{3} + r_{1} – r_{2} = 4R cos B.

Solution: