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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.

- Answer all questions.
- 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 :

- Answer all the questions.
- 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 -y_{1} : y_{2}

= 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 + x^{2} + ……… + x^{100}, then find f'(1).

Solution:

Given f(x) = 1 + x + x^{2} + ……. + x^{100}

f'(x) = 0 + 1 + 2x + …….. + 100 x^{99}

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) = x^{2} + 4 in [-3, 3],

Solution:

Given f(x) = x^{2} + 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) = 3^{2} + 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) = x^{2} + 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 :

- Attempt Any Five questions.
- 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 PA^{2} + PB^{2} = 2PC^{2}, 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 (x_{1}, y_{1}) be a point on locus.

Given geometric condition is PA^{2} + PB^{2} = 2 PC^{2}

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 cos^{2} α – Y sin α cos α + X sin^{2} α + Y cos α sin α = P

⇒ X (cos^{2} α + sin^{2} 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 sec^{2}θ

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 cos^{2}x from the first principle.

Solution:

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

f(x+ h) = cos^{2}(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 m^{3}/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 \(\frac{d V}{d t}\) = 2m^{3}/min

From fig ∆OAB, ∆OCD are similar triangles

Section – C

(5 × 7 = 35 Marks)

III. Long answer type questions :

- Attempt Any Five questions.
- 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 ax^{2} + 2hxy + by^{2} = 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

ax^{2} + 2hxy + by^{2} = 0 (see figure)

and \(\stackrel{\leftrightarrow}{\mathrm{AB}}\) be the line lx + my + n = 0

Let ax^{2} + 2hxy + by^{2} ≡ (l_{1}x + m_{1}y)(l_{2}x + m_{2}y),

and \(\overleftrightarrow{O A}\) and \(\overleftrightarrow{O B}\) be the lines.

l_{1}x + m_{1}y = 0 and

l_{2}x + m_{2}y = 0 respectively.

Let A = (x_{1}, y_{1}) and B = (x_{2}, y_{2}).

Then l_{1}x_{1} + m_{1}y_{1} = 0 and lx_{1} + my_{1} + 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, l^{2} + m^{2} – n^{2} = 0.

Solution:

Given l + m + n = 0 …… (1)

l^{2} + m^{2} – n^{2} = 0 ……. (2)

From (1) l = – m – n …… (3)

Substituting in (2)

(- m – n)^{2} + (m^{2} – n^{2}) = 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 2x^{2} – 2xy + 3y^{2} + 2x – y – 1 = 0 and the line x + 2y = k are mutually perpendicular.

Solution:

The given circle equation

S ≡ 2x^{2} + 2xy + 3y^{2} + 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

2x^{2} – 2xy + 3y^{2} + 2x.1 – y.1 – 1^{2} = 0

2x^{2} – 2xy + 3y^{2} + 2x\(\frac{(x+2 y)}{k}\) – y\(\frac{(x+2 y)}{k}\) – \(\frac{(x+2 y)^2}{k}\) = 0

Multiplaing with k^{2}

2k^{2}x^{2} – 2k^{2}xy + 3k^{2}y^{2} + 2kx (x + 2y) – ky (x + 2y) – (x + 2y)^{2} = 0

2k^{2}x^{2} – 2k^{2}xy + 3k^{2}y^{2} + 2kx^{2} + 4kxy – kxy – 2ky^{2} – x^{2} – 4xy – 4y^{2} = 0

(2k^{2} + 2k – 1) x^{2} + (-2k^{2} + 3k – 4) xy + (3k^{2} – 2k – 4) y^{2} = 0

since OA, OB are perpendicular.

x^{2} + y^{2} = 0

2k^{2} + 2k – 1 + 3k^{2} – 2k – 4 = 0

5k^{2} = 5 ⇒ k^{2} = 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 – x^{2}

\(\frac{\mathrm{dA}}{\mathrm{dx}}\) = 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