[Download PDF] AP ICET 2018 Question Paper Shift-1 (2nd May) Paper with Solutions

Instructions

For the following questions answer them individually

AP ICET 2018 Shift-1 (2nd May) - Question 131

The coefficient of $$x^8$$ in $$\left(x - \frac{2}{x}\right)^{20}$$ is

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AP ICET 2018 Shift-1 (2nd May) - Question 132

If the system of equations $$3x - 2y + z = 0, \lambda x - 14y + 15z = 0, x + 2y + 3z = 0$$ has non-trivial solution, then $$\lambda$$ is a root of the equation

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AP ICET 2018 Shift-1 (2nd May) - Question 133

$$A = \begin{bmatrix}4 & 0 & 0 \\0 & 5 & 0 \\0 & 0 & 6 \end{bmatrix} \Rightarrow A^{-1} =$$

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AP ICET 2018 Shift-1 (2nd May) - Question 134

$$\lim_{x \rightarrow 0}\frac{1 - \cos 4x}{1 - \cos 5x} =$$

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AP ICET 2018 Shift-1 (2nd May) - Question 135

If $$y = \sqrt{x + y}$$ then $$\frac{dx}{dy} =$$

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Solution

Given, $$y = \sqrt{x + y}$$

Squaring on both sides, we get

$$y^2=x+y$$

$$\Rightarrow$$ $$x=y^2-y$$

$$\Rightarrow$$ $$\frac{dx}{dy} = 2y^{2-1}-y^{1-1}$$

$$\Rightarrow$$ $$\frac{dx}{dy} = 2y-1$$

Hence, the correct answer is Option D

AP ICET 2018 Shift-1 (2nd May) - Question 136

If each interior angle of a regular polygon is $$135^\circ$$, then the number of sides of the polygon is

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Solution

Each interior angle of the regular polygon = $$\frac{\left(n-2\right)180^{\circ\ }}{n}$$

Given, each interior angle of the regular polygon is $$135^\circ$$

$$\Rightarrow$$ $$\frac{\left(n-2\right)180^{\circ\ }}{n}=135^{\circ\ }$$

$$\Rightarrow$$ $$4\left(n-2\right)\ =3n$$

$$\Rightarrow$$ $$\ 4n-8=3n$$

$$\Rightarrow$$ $$n=8$$

$$\therefore\ $$Number of sides of the polygon = 8

AP ICET 2018 Shift-1 (2nd May) - Question 137

ABCD is a quadrilateral and P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Then PQORS is a

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AP ICET 2018 Shift-1 (2nd May) - Question 138

A parallelogram and a triangle have same base and the third vertex of the triangle lies on the side parallel to base. If the area of the triangle is 36 sq.cms then the area (in square cms) ofthe parallelogram is

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AP ICET 2018 Shift-1 (2nd May) - Question 139

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

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Solution

Area of triangle formed by points (2, 3), (3, 4) and (4, 2) = $$\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$$

$$=\frac{1}{2}\left|2\left(4-2\right)+3\left(2-3\right)+4\left(3-4\right)\right|$$

$$=\frac{1}{2}\left|2\left(2\right)+3\left(-1\right)+4\left(-1\right)\right|$$

$$=\frac{1}{2}\left|4-3-4\right|$$

$$=\frac{1}{2}\left|-3\right|$$

$$=\frac{3}{2}$$

AP ICET 2018 Shift-1 (2nd May) - Question 140

PQR is a right angled triangle where P and R are the points (2, 4) and (6, 10) respectively and $$\angle Q = 90^\circ$$. Then the circumcentre of the triangle is

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Solution

In a right-angled triangle, circumcentre is the centre of the hypotenuse.

Hence circumcentre = $$\frac{\left(6+2\right)}{2},\ \frac{\left(10+4\right)}{2}=\left(4,7\right)$$

CAT Content

AP ICET 2018 Question Paper Shift-1 (2nd May)

AP ICET 2018 Question Paper Shift-1 (2nd May)

AP ICET 2018 Shift-1 (2nd May) - Question 131

AP ICET 2018 Shift-1 (2nd May) - Question 132

AP ICET 2018 Shift-1 (2nd May) - Question 133

AP ICET 2018 Shift-1 (2nd May) - Question 134

AP ICET 2018 Shift-1 (2nd May) - Question 135

AP ICET 2018 Shift-1 (2nd May) - Question 136

AP ICET 2018 Shift-1 (2nd May) - Question 137

AP ICET 2018 Shift-1 (2nd May) - Question 138

AP ICET 2018 Shift-1 (2nd May) - Question 139

AP ICET 2018 Shift-1 (2nd May) - Question 140

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