[Download PDF] AP ICET 2016 Question Paper (16th May) Paper with Solutions

Instructions

For the following questions answer them individually

Question 131

The sum to infinite terms of the progression $$6, 4\frac{4}{5}, 3\frac{21}{25}, ...........$$ is

Video Solution

Question 132

If A is an $$n \times n$$ matrix such that $$A^2 = I$$ and $$A$$ ≠ $$A^2$$, then det(I + A) =

Video Solution

Question 133

If $$A = (a_{ij})$$ is a $$3 \times 3$$ matrix and $$a_{ij} = 4i + 7j$$ for $$1 \leq i, j \leq 3$$, then trace of 3A =

Video Solution

Question 134

$$\lim_{x \rightarrow 0} \frac{\tan^3 3x}{5x^3} =$$

Video Solution

Question 135

$$\lim_{x \rightarrow 0} \frac{e^{3x} - e^{4x}}{\sin 3x \cos 4x} =$$

Video Solution

Question 136

If $$x = \sin \theta + \theta \cos \theta$$ and $$y = \cos \theta + \theta \sin \theta$$, then $$\frac{dy}{dx}$$ at $$\theta = \frac{\pi}{2}$$ is

Video Solution

Solution

$$\frac{dy}{d\theta\ }=-\sin\theta\ +\sin\theta\ +\theta\ \cos\theta\ =\theta\ \cos\theta\ $$

$$\frac{dx}{d\theta\ }=\cos\theta\ +\cos\theta\ -\theta\ \sin\theta\ =2\cos\theta\ -\theta\ \sin\theta\ $$

$$\frac{dy}{dx}=\frac{dy}{d\theta\ }\times\ \frac{d\theta}{dx}=\frac{\theta\ \cos\theta}{2\cos\theta\ -\theta\ \sin\theta\ }$$

Substituting $$\theta\ =\frac{\pi}{2}$$ we get $$\frac{4}{\pi\ }$$

Question 137

A, B are end points of the longest chord of a circle of radius 3 units. If O is the centre of the circle and C is a point on the circle such that AC = AO,then the perimeter of the triangle ABC is

Video Solution

Question 138

If the diagonals of a rhombus are 14 cm and 48 cm, then its area in square centimetres is

Video Solution

Solution

Given, diagonals of the rhombus are 14 cm and 48 cm

$$\therefore\ $$Area of the rhombus = $$\frac{1}{2}\times d_1\times d_2$$ = $$\frac{1}{2}\times14\times48$$ = 336 cm$$^2$$

Hence, the correct answer is Option C

Question 139

If the y-intercept of the straight line (2 + 3k) x + (7 - 2k) y + (4k + 3) = 0 is 1, then k =

Video Solution

Solution

Since y-intercept is 1, so at x = 0 y should be equal to 1.

Substituting x = 0 and y = 1 in the equation,

(2 + 3k)0 + (7 - 2k)1 + (4k + 3) = 0

$$\Rightarrow$$ 7 - 2k + 4k + 3 = 0

$$\Rightarrow$$ 2k = -10

$$\Rightarrow$$ k = -5

Hence, the correct answer is Option D

Question 140

If (0, 0), (0, 4), (3, 0) are vertices of a triangle, then the distance between the circumcentre and the orthocentre of the triangle is

Video Solution

Solution

As we can see the given triangle is a right-angled triangle. Hence its orthocentre(0,0). Circumcentre will be at the midpoint of hypotenuse i.e. (1.5,2)

Distance between them $$\sqrt{\ \left(1.5\right)^2+\left(2\right)^2}=\sqrt{\ 6.25}=2.5$$

CAT Content

AP ICET 2016 Question Paper (16th May)

AP ICET 2016 Question Paper (16th May)

The sum to infinite terms of the progression $$6, 4\frac{4}{5}, 3\frac{21}{25}, ...........$$ is

If A is an $$n \times n$$ matrix such that $$A^2 = I$$ and $$A$$ ≠ $$A^2$$, then det(I + A) =

If $$A = (a_{ij})$$ is a $$3 \times 3$$ matrix and $$a_{ij} = 4i + 7j$$ for $$1 \leq i, j \leq 3$$, then trace of 3A =

$$\lim_{x \rightarrow 0} \frac{\tan^3 3x}{5x^3} =$$

$$\lim_{x \rightarrow 0} \frac{e^{3x} - e^{4x}}{\sin 3x \cos 4x} =$$

If $$x = \sin \theta + \theta \cos \theta$$ and $$y = \cos \theta + \theta \sin \theta$$, then $$\frac{dy}{dx}$$ at $$\theta = \frac{\pi}{2}$$ is

A, B are end points of the longest chord of a circle of radius 3 units. If O is the centre of the circle and C is a point on the circle such that AC = AO,then the perimeter of the triangle ABC is

If the diagonals of a rhombus are 14 cm and 48 cm, then its area in square centimetres is

If the y-intercept of the straight line (2 + 3k) x + (7 - 2k) y + (4k + 3) = 0 is 1, then k =

If (0, 0), (0, 4), (3, 0) are vertices of a triangle, then the distance between the circumcentre and the orthocentre of the triangle is

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