TS Inter 1st Year Maths 1A Question Paper March 2019

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TS Inter 1st Year Maths 1A Question Paper March 2019

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 questions.
• Each question carries two marks.

Question 1.
If f(x) = 2x – 1, g(x) = \(\frac{x+1}{2}\) for all x ∈ R, are two functions, then find:
(i) (gof) (x)
(ii) (fog) (x)
Solution:
Given f(x) = 2x – 1 and g(x) = \(\frac{x+1}{2}\)
(i) (gof) (x) = g[f(x)]
= g[2x – 1]
= \(\frac{2 x-1+1}{2}\)
= x
∴ (gof) (x) = x.

(ii) (fog) (x) = f[g(x)]
= f[latex]\frac{x+1}{2}[/latex]
= 2[latex]\frac{x+1}{2}[/latex] – 1
= x + 1 – 1
= x
∴ (fog) (x) = x

Question 2.
Find the domain of the real valued function f(x) = \(\frac{1}{6 x-x^2-5}\).
Solution:
Given f(x) = \(\frac{1}{6 x-x^2-5}\) ∈ R
∴ 6x – x² – 5 ≠ 0
⇒ x² – 6x + 5 ≠ 0
⇒ x² – 5x – x + 5 ≠ 0
⇒ x(x – 5) – 1(x – 5) ≠ 0
⇒ (x – 1) (x – 5) ≠ 0
⇒ x ≠ 1 and x ≠ 5
∴ Domain of f = R – {1, 5}

Question 3.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\) then find 3B – 2A.
Solution:

Question 4.
If A = \(\left[\begin{array}{rrr}
2 & 0 & 1 \\
-1 & 1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
-1 & 1 & 0 \\
0 & 1 & -2
\end{array}\right]\) then find (AB^T)^T.
Solution:

Question 5.
If \(\overline{\mathrm{OA}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{AB}}=3 \overline{\mathrm{i}}-2 \overline{\mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{BC}}=\overline{\mathrm{i}}+2 \overline{\mathrm{j}}-2 \overline{\mathrm{k}}\), \(\overline{C D}=2 \bar{i}+\bar{j}+3 \bar{k}\) then find the vector \(\bar{OD}\).
Solution:

Question 6.
Let \(\bar{a}=2 \bar{i}+4 \bar{j}-5 \bar{k}, \bar{b}=\bar{i}+\bar{j}+\bar{k}\) be three vectors. Find the unit vector in the opposite direction of \(\bar{a}+\bar{b}+\bar{c}\).
Solution:

Question 7.
Find the equation of the plane passing through the point (3, -2, 1) and perpendicular to the vector (4, 7, -4).
Solution:
The equation of the plane passing through the point (3, -2, 1) and perpendicular to the vector (4, 7, -4) is 4(x – 3) + 7(y + 2) – 4(z – 1) = 0
⇒ 4x – 12 + 7y + 14 – 4z + 4 = 0
⇒ 4x + 7y – 4z + 6 = 0

Question 8.
If sin θ = \(-\frac{1}{3}\) and θ do not lie in the third quadrant, then find the values of:
(i) cos θ
(ii) cot θ
Solution:

Question 9.
Evaluate \(\sin ^2 82 \frac{1}{2}^{\circ}-\sin ^2 22 \frac{1}{2}^{\circ}\).
Solution:
\(\sin ^2 82 \frac{1}{2}^{\circ}-\sin ^2 22 \frac{1}{2}^{\circ}\)
= \(\sin \left(82 \frac{1}{2}^{\circ}+22 \frac{1}{2}^{\circ}\right) \cdot \sin \left(82 \frac{1}{2}^{\circ}-22 \frac{1}{2}^{\circ}\right)\)
= sin 105° . sin 60°
= sin (90° + 15°) . sin 60°
= cos 15° . sin 60°
= \(\frac{\sqrt{3}+1}{2 \sqrt{2}} \cdot \frac{\sqrt{3}}{2}\)
= \(\frac{1}{4 \sqrt{2}}(3+\sqrt{3})\)

Question 10.
If cosh x = \(\frac{5}{2}\), then find cosh (2x) and sinh (2x).
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 = \(\left[\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & -1 \\
3 & -1 & 1
\end{array}\right]\), then find A³ – 3A² – A – 3I, where I is unit matrix of order 3.
Solution:
Given A = \(\left[\begin{array}{ccc}
1 & -2 & 1 \\
0 & 1 & -1 \\
3 & -1 & 1
\end{array}\right]\)

Question 12.
If \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar vectors, then find the point of intersection of the line passing through the points \(2 \bar{a}+3 \bar{b}-\bar{c}\), \(3 \bar{a}+4 \bar{b}-2 \bar{c}\) with the line joining the points \(\bar{a}-2 \bar{b}+3 \bar{c}\), \(\overline{\mathrm{a}}-6 \overline{\mathrm{b}}+6 \overline{\mathrm{c}}\).
Solution:
The vector equation of the line through the points

Question 13.
If \(\overline{\mathrm{a}}=2 \overline{\mathrm{i}}+3 \overline{\mathrm{j}}+4 \overline{\mathrm{k}}, \overline{\mathrm{b}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}-\overline{\mathrm{k}}\) and \(\overline{\mathrm{c}}=\overline{\mathrm{i}}-\overline{\mathrm{j}}+\overline{\mathrm{k}}\), then compute compute \(\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})\) and verify that it is perpendicular to a.
Solution:

Question 14.
Prove that \(\left(1+\cos \frac{\pi}{10}\right)\left(1+\cos \frac{3 \pi}{10}\right)\left(1+\cos \frac{7 \pi}{10}\right)\left(1+\cos \frac{9 \pi}{10}\right)=\frac{1}{16}\).
Solution:
L.H.S = \(\left(1+\cos \frac{\pi}{10}\right)\left(1+\cos \frac{3 \pi}{10}\right)\left(1+\cos \frac{7 \pi}{10}\right)\left(1+\cos \frac{9 \pi}{10}\right)\)

Question 15.
If θ₁, θ₂ are solutions of the equation a cos 2θ + b sin 2θ = c, tan θ₁ ≠ tan θ₂ and a + c ≠ 0, then find the values of
(i) tan θ₁ + tan θ₂
(ii) tan θ₁ . tan θ₂
Solution:
Given a cos 2θ + b sin 2θ = c
⇒ \(a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\) = c
⇒ a(1 – tan²θ) + 2b tan θ = c(1 + tan²θ)
⇒ a – a tan²θ + 2b tan θ = c + c tan²θ
⇒ (a + c) tan²θ – 2b tan θ + (c – a) = 0
This is a second-degree quadratic equation in tan θ
Its solutions be tan θ₁, tan θ₂.
(i) Sum of the roots = tan θ₁ + tan θ₂
= \(\frac{-(-2 b)}{a+c}\)
= \(\frac{2 b}{a+c}\)
(ii) Product of the roots = tan θ₁ . tan θ₂ = \(\frac{c-a}{c+a}\)

Question 16.
Prove that \(\sin ^{-1}\left(\frac{3}{5}\right)+\sin ^{-1}\left(\frac{8}{17}\right)=\cos ^{-1}\left(\frac{36}{85}\right)\).
Solution:

Question 17.
In a triangle ABC, if \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\), then show that C = 60°.
Solution:
Given \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\)
⇒ \(\frac{b+c+a+c}{(a+c)(b+c)}=\frac{3}{a+b+c}\)
⇒ (a + b + 2c) (a + b + c) = 3(a + c) (b + c)
⇒ a² + ab + ac + ab + b² + bc + 2ac + 2bc + 2c² = 3ab + 3ac + 3bc + 3c²
⇒ a² + b² – c² = ab
⇒ 2ab cos C = ab
⇒ cos C = \(\frac{1}{2}\)
⇒ C = 60°

Section – C
(5 × 7 = 35 Marks)

III. Long Answer Type Questions.

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

Question 18.
If f: A → B is a bijection, I_A and I_B are identity functions on A and B respectively, then show that fof^-1 = I_B and f^-1of = I_A.

Question 19.
Using mathematical induction, prove that \(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \cdots \cdots\left(1+\frac{2 n+1}{n^2}\right)\) = (n + 1)², ∀ n ∈ N.
Solution:
Let S(n) be the statement that
\(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \cdots \cdots\left(1+\frac{2 n+1}{n^2}\right)\) = (n + 1)²
If n = 1 then
LH.S = (1 + \(\frac{3}{1}\))
= 1 + 3
= 4
R.H.S = (1 + 1)² = 4
∴ L.H.S = R.H.S
∴ S(1) is true.
Assume that S(K) is true.

∴ S(K + 1) is true.
By the principle of mathematical induction S(n) is true for all n ∈ N.
∴ \(\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \cdots \cdots\left(1+\frac{2 n+1}{n^2}\right)\) = (n + 1)²

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:

Question 21.
Solve 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20 by using Matrix inversion method.
Solution:
Given 3x + 4y + 5z = 18
2x – y + 8z = 13
5x – 2y + 7z = 20
The given system of equations can be written as AX = B


∴ x = 3, y = 1, z = 1

Question 22.
If A = (1, -2, -1), B = (4, 0, -3), C = (1, 2, -1) and D = (2, -4, – 5) are four points, then find the shortest distance between the skew lines \(\stackrel{\leftrightarrow}{\mathrm{AB}}\) and \(\stackrel{\leftrightarrow}{\mathrm{CD}}\).

Question 23.
If A, B, C are angled in a triangle, then prove that cos A + cos B – cos C = -1 + 4 \(\cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}\).
Solution:

Question 24.
If ΔABC, show that \(\frac{1}{r^2}+\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}=\frac{a^2+b^2+c^2}{\Delta^2}\).
Solution: