[Download PDF] TS ICET 2018 Shift 1 Paper with Solutions

Instructions

For the following questions answer them individually

Question 111

If p and q are two statements then $$\sim (p \wedge q)$$ is equivalent to

Video Solution

Question 112

$$\sim (p \vee (\sim p \wedge q))$$ is equivalent to

Video Solution

Question 113

If $$f(x) = \frac{x}{\sqrt{1 - x^2}}$$ and $$g(x) = \frac{x}{\sqrt{1 + x^2}}$$ then $$(fog)x =$$

Video Solution

Question 114

If $$\mid x \mid < 1$$ and $$f(x) = 1 + x + x^2 + .......$$ then $$f^{-1}(x) =$$

Video Solution

Question 115

The domain of the function $$f(x) = \frac{1}{x^2 + 1}$$ is

Video Solution

Question 116

The area of the region (in sq. units) bounded by $$y = \mid x \mid - 5$$ with the coordinates axes is

Video Solution

Question 117

The equation of the line passing through the point of intersection of the lines $$3x + 4y + 5 = 0$$ and $$4x + y - 2 = 0$$ and perpendicular to the line $$\frac{x}{5} + \frac{y}{2} = 1$$ is

Video Solution

Question 118

If $$2 \cot \theta = 3$$, then $$\frac{2 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta} =$$

Video Solution

Question 119

$$(1 + \tan 17^\circ)(1 + \tan 28^\circ) =$$

Video Solution

Question 120

$$\frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} =$$

Video Solution

CAT Content

TS ICET 2018 Shift 1

TS ICET 2018 Shift 1

If p and q are two statements then $$\sim (p \wedge q)$$ is equivalent to

$$\sim (p \vee (\sim p \wedge q))$$ is equivalent to

If $$f(x) = \frac{x}{\sqrt{1 - x^2}}$$ and $$g(x) = \frac{x}{\sqrt{1 + x^2}}$$ then $$(fog)x =$$

If $$\mid x \mid < 1$$ and $$f(x) = 1 + x + x^2 + .......$$ then $$f^{-1}(x) =$$

The domain of the function $$f(x) = \frac{1}{x^2 + 1}$$ is

The area of the region (in sq. units) bounded by $$y = \mid x \mid - 5$$ with the coordinates axes is

The equation of the line passing through the point of intersection of the lines $$3x + 4y + 5 = 0$$ and $$4x + y - 2 = 0$$ and perpendicular to the line $$\frac{x}{5} + \frac{y}{2} = 1$$ is

If $$2 \cot \theta = 3$$, then $$\frac{2 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta} =$$

$$(1 + \tan 17^\circ)(1 + \tan 28^\circ) =$$

$$\frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A} =$$

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