[Download PDF] AP ICET 26th April 2019 Shift-2 Paper with Solutions

Instructions

For the following questions answer them individually

Question 131

If the expansion of $$(x - 3x^2)^{25}$$ is a polynomial of $$n^{th}$$ degree in x, then n =

Video Solution

Solution

$$(x-3x^2)^{25}=^{25}C_0x^{25}+^{25}C_1x^{24}\left(-3x^2\right)+....+^{25}C_{24}x\left(-3x^2\right)^{24}+^{25}C_{25}\left(-3x^2\right)^{25}$$

$$=^{25}C_0x^{25}+^{25}C_1\left(-3\right)x^{26}+....+^{25}C_{24}\left(-3\right)^{24}x^{49}+^{25}C_{25}\left(-3\right)^{25}x^{50}$$

The degree of the polynomial (n) = highest power of $$x$$ = 50

Hence, the correct answer is Option C

Question 132

If $$A = \begin{bmatrix}1 & -1 \\2 & -1 \end{bmatrix}, B = \begin{bmatrix}x & 1 \\y & -1 \end{bmatrix}$$ and $$(A + B)^2 = A^2 + B^2$$, then (x, y) =

Video Solution

Question 133

If $$\begin{bmatrix}2 & -1 \\1 & 1 \end{bmatrix}\begin{bmatrix}x \\y \end{bmatrix} = \begin{bmatrix}4 \\5 \end{bmatrix}$$, then $$x^2 + y^2 =$$

Video Solution

Question 134

$$\lim_{x \rightarrow 0}\frac{\sqrt{1 + x^3} - \sqrt{1 - x^2}}{x^2} =$$

Video Solution

Question 135

$$\frac{d}{dx}\left(\tan^{-1}\left(\frac{1 + \tan x}{1 - \tan x}\right)\right) =$$

Video Solution

Question 136

The ratio of the sides of a triangle whose angles are in the ratio $$1 : 2 : 3$$ is

Video Solution

Solution

The angles will be 30,60,90

We know that, $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{SinC}$$

a:b:c = $$\sin A:\sin B:\sin C$$ = $$\frac{1}{2}:\frac{\sqrt{3\ }}{2}:1$$ = $$1:\sqrt{3\ }:2$$

Question 137

The perimeter (in cms) of a Rhombus whose diagonals are 9 cm and 12 cm is

Video Solution

Solution

We know diagonals bisect each other perpendicularly

Hence side of the rhombus would be $$\sqrt{\ 4.5^2+6^2}=7.5$$

Perimeter = 4*7.5 = 30

Question 138

A tangent is drawn to a circle from a point 13 cm away from its centre. If the length of that tangent is 12 cm, then the radius (in cms) of the circle is

Video Solution

Solution

Let the radius of the circle = r

From the triangle,

r$$^2$$ + 12$$^2$$ = 13$$^2$$

$$\Rightarrow$$ r$$^2$$ + 144 = 169

$$\Rightarrow$$ r$$^2$$ = 25

$$\Rightarrow$$ r = 5 cm

$$\therefore\ $$Radius of the circle = 5 cm

Question 139

The area (in sq. units) of the triangle formed by the points (0, 0), (2, -3) and (4, 5) is

Video Solution

Solution

Area of the triangle when one vertex is origin is $$\frac{1}{2}\left|x_2y_3-x_3y_2\right|$$ = $$\frac{1}{2}\left|2\cdot5\ -\ \left(4\right)\cdot\left(-3\right)\right|=11$$

Question 140

The ratio in which the X-axis divides the line segment joining the points (2, 3) and (3, -5) is

Video Solution

CAT Content

AP ICET 26th April 2019 Shift-2

If the expansion of $$(x - 3x^2)^{25}$$ is a polynomial of $$n^{th}$$ degree in x, then n =

If $$A = \begin{bmatrix}1 & -1 \\2 & -1 \end{bmatrix}, B = \begin{bmatrix}x & 1 \\y & -1 \end{bmatrix}$$ and $$(A + B)^2 = A^2 + B^2$$, then (x, y) =

If $$\begin{bmatrix}2 & -1 \\1 & 1 \end{bmatrix}\begin{bmatrix}x \\y \end{bmatrix} = \begin{bmatrix}4 \\5 \end{bmatrix}$$, then $$x^2 + y^2 =$$

$$\lim_{x \rightarrow 0}\frac{\sqrt{1 + x^3} - \sqrt{1 - x^2}}{x^2} =$$

$$\frac{d}{dx}\left(\tan^{-1}\left(\frac{1 + \tan x}{1 - \tan x}\right)\right) =$$

The ratio of the sides of a triangle whose angles are in the ratio $$1 : 2 : 3$$ is

The perimeter (in cms) of a Rhombus whose diagonals are 9 cm and 12 cm is

A tangent is drawn to a circle from a point 13 cm away from its centre. If the length of that tangent is 12 cm, then the radius (in cms) of the circle is

The area (in sq. units) of the triangle formed by the points (0, 0), (2, -3) and (4, 5) is

The ratio in which the X-axis divides the line segment joining the points (2, 3) and (3, -5) is

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