Thoroughly analyzing AP Inter 1st Year Maths 1A Model Papers and AP Inter 1st Year Maths 1A Question Paper March 2018 helps students identify their strengths and weaknesses.

## 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 ⇔ x^{2} – 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) = x^{2} + 1 then find (fog) (2).

Solution:

Given f(x) = 3x – 1 and g(x) = x^{2} + 1

(fog) (2) = f[g(2)]

= f[2^{2} + 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 + cos^{2}x.

Solution:

Let f(x) = cos 2x + cos^{2}x

= 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 cosh^{2}x – 1.

Solution:

L.H.S = cosh 2x

= cosh^{2}x + sinh^{2}x

= cosh^{2}x + cosh^{2}x – 1 {∵ cosh^{2}x – sinh^{2}x = 1}

= 2 cosh^{2}x – 1

= R.H.S

∴ L.H.S = R.H.S

Hence cosh 2x = 2 cosh^{2}x – 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 = sec^{2}θ – 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 a_{1}, a_{2} ∈ a Then f(a_{1}), f(a_{2}) ∈ B

(gof)(a_{1}) = (gof)(a_{2})

⇒ g[f(a_{1})] = g[f(a_{2})]

⇒ f(a_{1}) = f(a_{2}) {∵ g is one-one}

⇒ a_{1} = a_{2} {∵ 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 + r_{3} + r_{1} – r_{2} = 4R cos B.

Solution: