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

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

Time : 3 Hours
Max. Marks : 75

Section – A
(10 × 2 = 20)

I. Very Short Answer Type Questions.

1. Answer all questions.
2. Each question carries two marks.

Time : 3 Hours
Max. 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 :

1. Answer all the questions.
2. Each question carries two marks.

Question 1.
Compute : \(\lim _{x \rightarrow 0} \frac{a^x-1}{b^x-1}\).
Solution:

Question 2.
Find the value of p, if the straight lines 3x + py – 1 = 0, 7x – 3y + 3 = 0 are mutually perpendicular.
Solution:
Given straight line equations are
3x + py – 1 = 0 …….. (1)
7x – 3y + 3 = 0 …..(2)
Since (1), (2) are mutually perpendicular
∴ 3(7) + p(-3) = 0
⇒ 21 – 3p = 0
⇒ 3p = 21
⇒ p = 7

Question 3.
If f(x) = log(tan e^x), then find f(x).
Solution:
Given f(x) = log(tan e^x)
differentiating w.r.to ‘x’ on bothsides, we have

Question 4.
Find the ratio in which the xz-plane divides the line joining A(-2, 3, 4) and B(1, 2, 3).
Solution:
The ratio in which the XZ-plane divides the line joining A(-2, 3, 4) and B(1, 2, 3) is -y₁ : y₂
= 3 : 2
= 3 : 2 externally.

Question 5.
Reduce the equation of the plane x + 2y – 3z – 6 = 0 to the normal form.
Solution:
Given plane equation is x + 2y – 3z – 6 = 0

Question 6.
Evaluate: \(\lim _{x \rightarrow 0} \frac{\log _e(1+5 x)}{x}\)
Solution:

Question 7.
If f(x) = 1 + x + x² + ……… + x¹⁰⁰, then find f'(1).
Solution:
Given f(x) = 1 + x + x² + ……. + x¹⁰⁰
f'(x) = 0 + 1 + 2x + …….. + 100 x⁹⁹
f'(1) = 0 + 1 + 2(1) + ……. + 100(1)⁹⁹
f'(1) = 1 + 2 + 3 + …… + 100
= Σ100
= \(\frac{100(100+1)}{2}\)
= 50(101)
∴ f'(1) = 5050.

Question 8.
Find the angle which the straight line y = \(\sqrt{3}\)x – 4 makes with the y-axis.
Solution:
Given straight line is y = \(\sqrt{3}\)x – 4
m = \(\sqrt{3}\) ⇒ tan θ = \(\sqrt{3}\)
⇒ θ = 60°
∴ The straight line y = \(\sqrt{3}\)x – 4 makes angle with y-axis is 90° – 60° = 30°.

Question 9.
Verify Rolle’s theorem for the function y = f(x) = x² + 4 in [-3, 3],
Solution:
Given f(x) = x² + 4
Since f is a second degree polynomial
∴ f is continuous on [-3, 3] and f is derivable on (-3, 3)
Also f(-3) = (-3)² + 4
= 9 + 4 = 13
f(3) = 3² + 4
= 9 + 4 = 13
∴ f(-3) = f(3)
∴ f satisfies all the conditions of RoHe’s theorem.
∴ There exists c ∈ (-3, 3) such that f(c) = 0
f(x) = x² + 4
⇒ f(x) = 2x
⇒ f'(c) = 2c
⇒ 0 = 2c
⇒ c = 0 ∈ (-3, 3)
Hence Rolle’s theorem is verified.

Question 10.
Find ∆y and dy for the function y = cos x at x = 60° with ∆x = i.
(cos 6i = 0.4848, i = 0.0174 radians)
Solution:
Given y = f(x) cos x, x = 60°, ∆x = 1 °
∆y = f(x + ∆x) – f(x)
= f(60° + 1°) – f(60°)
= f(61 °) – f(60°)
= cos 61°- cos 60°
= 0.4848 – 0.5000
= -0.0152
dy = f'(x) ∆x
= -sin x. ∆x
= -sin 60°. (1°)
= \(\frac{\sqrt{3}}{2}\) .(0.0174)
= -(0.866) (0.0174)
= – 0.01506.

Section – B

II. Short answer type questions :

1. Attempt Any Five questions.
2. Each question carries Four marks.

Question 11.
Check the continuity of the following function at ‘2’:

Solution:

Question 12.
A(1, 2), B(2, -3) and C(-2, 3) are three points. If a point P moves such that PA² + PB² = 2PC², then show that the equation to the locus of P is 7x – 7y + 4 = 0.
Solution:
Given A = (1, 2)
B = (2, -3)
C = (-2, 3)
Let P (x₁, y₁) be a point on locus.
Given geometric condition is PA² + PB² = 2 PC²

Question 13.
A straight line through Q(\(\sqrt{3}\), 2) makes an angle of \(\frac{\pi}{6}\) with the positive direction of the X-axis. If the straight line intersects the line \(\sqrt{3}\)x – 4y + 8 = 0 at P, then find the distance of PQ.
Solution:
Given straight line equation is \(\sqrt{3}\)x – 4y + 8 = 0 …. (1)
Slope of \(\overline{\mathrm{PQ}}\) = tan \(\frac{\pi}{6}\) = tan 30° = \(\frac{1}{\sqrt{3}}\)
Also Q = (\(\sqrt{3}\), 2)
Equation of \(\overline{P Q}\) is

Question 14.
When the axes are rotated through an angle a, find the transformed equation of x cos α + y sin α = p.
Solution:
The given equation x cos α + y sin α = P
The axes are rotated through an angle ‘α’
x = X cos α – Y sin α
y = X sin α + Y cos α
∴ given equation transformed to
(X cos α – Y sin α) cos α + (X sin α + Y cos α) sin α = P
⇒ X cos² α – Y sin α cos α + X sin² α + Y cos α sin α = P
⇒ X (cos² α + sin² α) = p
⇒ X = P.

Question 15.
Show that the tangent at any point θ on the curve x = c sec θ, y = c tan θ is y sin θ = x – c cos θ.
Solution:
Given x = c sec θ
\(\frac{d x}{d \theta}\) = c sec θ tan θ
y = c tan θ
\(\frac{d y}{d \theta}\) = c sec²θ
slope of the tangent at any point ’θ‘ is

∴ The equation of the tangent is
y – c tan θ = cosec θ (x – c sec θ)

Question 16.
Find the derivative of cos²x from the first principle.
Solution:
Let f(x) = cos² x
f(x+ h) = cos²(x + h)
By first principle

Question 17.
A container is in the shape of an inverted cone has height 8 m and radius 6 m at the top. If it is filled with water at the rate of 2 m³/minute, how fast is the height of water changing when the level is 4 m.
Solution:
Let OC = h and CD = r

Given AB = 6, OA = 8 and \(\frac{d V}{d t}\) = 2m³/min
From fig ∆OAB, ∆OCD are similar triangles

Section – C
(5 × 7 = 35 Marks)

III. Long answer type questions :

1. Attempt Any Five questions.
2. Each question carries Seven marks.

Question 18.
Find the orthocentre of the triangle whose vertices are (5, -2), (-1, 2) and (1, 4).
Solution:
Let A = (5, -2)
B = (-1, 2)
C = (1, 4)

Slope of \(\overline{\mathrm{BC}}\) = \(\frac{4-2}{1-(-1)}\) = \(\frac{2}{2}\) = 1
Since \(\overline{\mathrm{AD}}\) ⊥ \(\overline{\mathrm{BC}}\)
∴ Slope of \(\overline{\mathrm{AD}}\) = \(\frac{-1}{1}\) = -1
Equation of \(\overline{\mathrm{AD}}\) is
y – (-2) = -1 (x – 5)
⇒ y + 2 = -x + 5
⇒ x + y – 3 = 0 ……. (1)
Slope of \(\overline{\mathrm{AC}}\) = \(\frac{4+2}{1-5}\)
= \(\frac{6}{-4}\) = \(\frac{-3}{2}\)
Since \(\overline{\mathrm{BE}}\) ⊥ \(\overline{\mathrm{AC}}\)
∴ Slope of \(\overline{\mathrm{BE}}\) = \(\frac{-1}{-3 / 2}\) = \(\frac{2}{3}\)
∴ Equation of \(\overline{\mathrm{BE}}\)
⇒ y – 2 = \(\frac{2}{3}\)(x + 1)
⇒ 3y – 6 = 2x + 2
⇒ 2x – 3y + 8 = 0 ……… (2)
Solving (1) and (2)

Question 19.
Show that the area of the triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\left|\frac{\mathrm{n}^2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{am^{2 } – 2 \mathrm { h } l \mathrm { m } + \mathrm { b } l ^ { 2 }}}\right|\).
Solution:
Let \(\overleftrightarrow{\mathrm{OA}}\) and \(\overleftrightarrow{\mathrm{OB}}\) be the pair of straight lines represented by the equation

ax² + 2hxy + by² = 0 (see figure)
and \(\stackrel{\leftrightarrow}{\mathrm{AB}}\) be the line lx + my + n = 0
Let ax² + 2hxy + by² ≡ (l₁x + m₁y)(l₂x + m₂y),
and \(\overleftrightarrow{O A}\) and \(\overleftrightarrow{O B}\) be the lines.
l₁x + m₁y = 0 and
l₂x + m₂y = 0 respectively.
Let A = (x₁, y₁) and B = (x₂, y₂).
Then l₁x₁ + m₁y₁ = 0 and lx₁ + my₁ + n = 0.
So, by the rule of cross—multiphcation, we obtain

Question 20.
Find the angle between the lines whose direction cosines satisfy the equations :
l + m + n = 0, l² + m² – n² = 0.
Solution:
Given l + m + n = 0 …… (1)
l² + m² – n² = 0 ……. (2)
From (1) l = – m – n …… (3)
Substituting in (2)
(- m – n)² + (m² – n²) = 0.
⇒ (m + n)² + (m + n) (m – n) = 0.
⇒ (m + n) [m + n + m – n] = 0
⇒ 2m (m + n) = 0
⇒ m = 0, m + n = 0
When m = 0
From (3), l = -n
∴ \(\frac{l}{l}\) = \(\frac{m}{0}\) = \(\frac{n}{-l}\)
When m = – n
From (3) l = -(-n) – n
= n – n
= 0.
∴ \(\frac{l}{0}\) = \(\frac{m}{l}\) = \(\frac{n}{-l}\)
∴ The d.rs of the two lines are (1, 0, -1) and (0, 1, -1) let ‘θ’ be the angle between the lines.

Question 21.
If x^{log y} = log x, then show that: \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{y}{x}\left(\frac{1-\log x \log y}{(\log x)^2}\right)\)
Solution:
Given x^{log y} = log x
Taking logarithms on bothsides, we have
log x^{log y} = log(log x)
⇒ (log y) (log x) = log (log x)
differentiating w.r. to ‘x’ on bothsides, we have

Question 22.
If the tangent at any point on the curve x^2/3 + y^2/3 = a^2/3 intersects the co-ordinate axes in A and B, then show that the length AB is a constant.
Solution:
Equation of the curve is x^2/3 + y^2/3 = a^2/3

Differentiate w.r.to x

Question 23.
Find the values of k, if the lines joining the origin to the points of intersection of the curve 2x² – 2xy + 3y² + 2x – y – 1 = 0 and the line x + 2y = k are mutually perpendicular.
Solution:
The given circle equation
S ≡ 2x² + 2xy + 3y² + 2x – y – 1 = 0 …… (1)

AB equation x + 2y = k
\(\frac{x+2 y}{k}\) = 1 ……… (2)
Homogenising (1) with the help of (2) combined equation of OA, OB is
2x² – 2xy + 3y² + 2x.1 – y.1 – 1² = 0
2x² – 2xy + 3y² + 2x\(\frac{(x+2 y)}{k}\) – y\(\frac{(x+2 y)}{k}\) – \(\frac{(x+2 y)^2}{k}\) = 0
Multiplaing with k²
2k²x² – 2k²xy + 3k²y² + 2kx (x + 2y) – ky (x + 2y) – (x + 2y)² = 0
2k²x² – 2k²xy + 3k²y² + 2kx² + 4kxy – kxy – 2ky² – x² – 4xy – 4y² = 0
(2k² + 2k – 1) x² + (-2k² + 3k – 4) xy + (3k² – 2k – 4) y² = 0
since OA, OB are perpendicular.
x² + y² = 0
2k² + 2k – 1 + 3k² – 2k – 4 = 0
5k² = 5 ⇒ k² = 1
∴ k = ± 1.

Question 24.
Find the maximum area of the rectangle that can be formed with fixed perimeter 20.
Solution:
Let x and y denote the length and breadth of a rectangle.
Given 2(x + y) = 20
⇒ x + y = 10 ⇒ y = 10 – x ……… (1)
Let A be the area of the rectangle.
∴ A = xy
= x(10 – x)
= 10x – x²
\(\frac{\mathrm{dA}}{\mathrm{dx}}\) = 10 – 2x
\(\frac{\mathrm{dA}}{\mathrm{dx}}\) = 0 ⇒ 10 – 2x = 0 dx
⇒ 2x = 10.
⇒ x = 5
\(\frac{\mathrm{d}^2 \dot{A}}{\mathrm{dx}^2}\) = 0 – 2
= -2 < 0
\(\left(\frac{d^2 A}{d x^2}\right)_{x=5}\) = -2 < 0
Rectangle area is maximum at x = 5
∴ y = 10 – x
= 10 – 5
= 5
Maximum are A = xy
= 5(5) = 25