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General

Easy

Question

# The area (in square units)bounded by the curves y=a x is, and lying in the first quadrant is

- 36
- 18
- 27/4
- 9

## The correct answer is: 9

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### The area under the curve y= and above x-axis is:

### The area under the curve y= and above x-axis is:

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### The area of the region (in sq. units), in the first quadrant, bounded by the parabola y=and the lines x=0,y=1 and y=4 is:

### The area of the region (in sq. units), in the first quadrant, bounded by the parabola y=and the lines x=0,y=1 and y=4 is:

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### The area bounded by the curve y=ln (x) and the lines y=0, y=ln(3) and x=0 is equal to:

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### Let A= The area (in square units) of the region A is:

### Let A= The area (in square units) of the region A is:

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### The area (in sq units) of the region bounded by the curves and is

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### The area between the parabolas4a(x+a) and

=-4a(x-a)in sQ units….

### The area between the parabolas4a(x+a) and

=-4a(x-a)in sQ units….

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### The area of the region between the curve4 and x = 0; x =1 is….

### The area of the region between the curve4 and x = 0; x =1 is….

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### The area of the region bounded by y=|x-1| and y=1 in sq. units is

### The area of the region bounded by y=|x-1| and y=1 in sq. units is

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### The area of the elliptic quadratic with the semi major axis and semi minor axis as 6 and 4 respectively

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### The area of circle circumscribing ΔABC is

### The area of circle circumscribing ΔABC is

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Statement-I : The statement that circumradius and inradius of a triangle are 12 and 8 respectively can not be correct. Statement-II : Circumradius 2 (inradius)

Statement-I : The statement that circumradius and inradius of a triangle are 12 and 8 respectively can not be correct. Statement-II : Circumradius 2 (inradius)

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### The complete solution set of the equation is

### The complete solution set of the equation is

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### The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0, ] is/are –

### The solution(s) of the equation cos2x sin6x = cos3x sin5x in the interval [0, ] is/are –

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### The equation has

(not possible for real x)

Hence, the solutions are x = 0, p, 2p, 3p.

### The equation has

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(not possible for real x)

Hence, the solutions are x = 0, p, 2p, 3p.

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### if

### if

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