Quantitative Aptitude PYQs

Explanation:

⇒ BC = \(√{(6+8)^2+ (8-6)^2} =√200 = 10√2\)

Let O is centre of circle

x² +y² = 100,

Radius = OB = OC = 10 units

OB² + OC² = \(√{100+100} = 10√2\)

 In ∆BOC, OB² + OC² = BC² = (10√2 )²

= 200

So ∠ BOC = 90° (by converse of Pythagoras theorem)

Then, ∠ BAC = 45° or 180° – 45° = \(\frac{\pi}{4} or \frac{3\pi}{4}\) .. ((Angle subtended by a chord at centre is double the angle at any point subtended by it on the remaining circle.)

∴ Option (c) is correct.

Explanation:

Point A can be taken at any point on the circumference of a circle.

So coordinates of A cannot be determined due to insufficient data. 

∴ Option (d) is correct

Explanation:

In isosceles trapezium, diagonals bisect each other.

O is the midpoint of AC and BD

Then, \(\frac{5+2}{2} = \frac{4+a}{2}, \frac{3+b}{2} = \frac{3+1}{2}\)

⇒ a = 3, b = 1 

⇒ D(a, b) = (3, 1)

∴ Option (d) is correct

Explanation:

Since diagonals of isosceles trapezium bisect each other, then the intersecting point of diagonal is O

Coordinates of O is \(\frac{2 + 5}{2}, \frac{3 + 1}{2} = \frac{7}{2}, 2\)

∴ Option (c) is correct

Explanation:

Given:

Sphere: 2x² + 2y² + 2z² + 3x + 3y + 3z – 6 = 0

⇒ \(x^2 + y^2+z^2 + \frac{3}{2}x +\frac{3}{2}y + \frac{3}{2}z\) -3 = 0

Now Radius = \(\sqrt{(\frac{3}{4})^2+ (\frac{3}{4})^2+ (\frac{3}{4})^2 + 3}\)

= \(\sqrt{\frac{27}{16} +3} = 5 \frac{\sqrt3}{4}\)

Diamter = \(5 \frac{\sqrt3}{2}\)

∴ Option (b) is correct

Explanation:

The Centre of the sphere is C \((\frac{-3}{4}, \frac{-3}{4}, \frac{-3}{4})\)

4x + 8y + 8z + 15 = 0,

Plane satisfy the centre of sphere – 3 –6 – 6 + 15 = 0 

⇒ 0 = 0

∴ Option (d) is correct

Explanation:

Given:

⇒ Two planes are x + y + z = 1 and 2x + 3y – 4z = 8.

Let a, b, c are direction ratios of lines of intersection of planes.

Then the lines is perpendicular to normal of both planes

⇒ a + b + c = 0 ...(i)

⇒ 2a + 3b – 4c = 0 ....(ii) 

⇒ \(\frac{a}{-4-3} = \frac{b}{2+4} = \frac{c}{3-2} = λ \)

⇒  a = – 7λ, b = 6λ  c = λ 

Thus, Direction ratios are (–7, 6, 1).  

∴ Option (b) is correct

Explanation:

We know: Direction ratios are (–7, 6, 1).

⇒Direction ratios of S are l = \(\frac{-7}{\sqrt86}\)

m = \(\frac{6}{\sqrt86 }, n =\frac{1}{\sqrt86}\)

Thus, 43 (l ² –m² –n² ) = 43 \((\frac{49}{86}- \frac{36}{86} - \frac{1}{86})\)

= 43× 12/86 = 6

∴ Option (a) is correct.

Explanation:

Given:

L = x + y + z + 4 = 0 = 2x – y – z + 8 and

P = x + 2y + 3z + 1 = 0 is a plane. 

⇒ Let direction ratios of line (a, b, c)

Since line is obtained by the intersection of planes.

⇒ x + y + z + 4 = 0 and 2x – y – z + 8 = 0

So a + b + c = 0....(i)

⇒ 2a – b – c = 0...(ii)

Solve (i) and (ii)

⇒\(\frac{a}{-1+1} = \frac{b}{2+1} = \frac{c}{-1-2} =λ \)

⇒ a = 0, b = 3λ  and c = –3λ 

⇒ Direction ratios of line is (0, 3, –3) or (0, 1, –1).

∴ Option (c) is correct.

Explanation:

Given:

L = x + y + z + 4 = 0 = 2x – y – z + 8 and

P = x + 2y + 3z + 1 = 0 is a plane.

To find the point of intersection of L and P,

The point must satisfy both line L and plane P and Option (d) (–4, –3, 3) will satisfy both line L and plane P.

So option (d) is the correct answer.

Explanation:

Since, {x}+[x] = x

⇒x – [x] = {x}

⇒ 0≤ x – [x] < 1 (0 ≤ {x} < 1)

⇒ –1 ≤ [x] –x ≤ 0

But x is positive and non-integer; then

⇒ –1 < [x]–x < 0

⇒ –1 < y < 0

⇒ [y] = –1 

∴ Option (a) is correct.

Explanation:

Given

f(x) = 4x + 1, g(x) = kx + 2

⇒ fog(x) = 4(kx + 2) + 1 = 4kx + 9

⇒ go f(x) = k(4x + 1) + 2 = 4kx + k + 2

Also fog(x) = go f(x) 

⇒ 9 = k + 2 

⇒ k = 7

∴ Option (a) is correct

Explanation:

f(x) = log₁₀ (x² + 2x + 11) for minimum f'(x) = 0

⇒ \(\frac{2x + 2 }{x^2+2x+11} = 0\) .. (Since x² +2x + 11 > 0)

At x = –1, we get minumum value of f(x).

⇒ f(–1) = log₁₀10 = 1

Explanation:

Given:

\(\rm \lim_{x\rightarrow 3}\frac{|x-3|}{x-3}\)

LHL = \(\rm \lim_{x\rightarrow 3}\frac{|x-3|}{x-3} = \frac{-(x-3)}{x-3} = -1\)

RHL = \(\rm \lim_{x\rightarrow 3}\frac{|x-3|}{x-3} = \frac{(x-3)}{x-3} =1\)

LHL ≠ RHL; limit does not exist

∴ Option (d) is correct.

Explanation:

Since, maximum value of acosx + bsinx is \(\sqrt{a^2+b^2}\)

So maximum value of acosx + bsinx + c is \(\sqrt{a^2+b^2} +c\)

∴ Option (b) is correct.

Explanation:

f(2x) = 4x² + 1

⇒f(x) = x² + 1

⇒ f(4x) = 16x² + 1

⇒ (4x² + 1)² = (x² + 1) × (16x² + 1)      [f(2x) is GM of f(x) & f(4x)]

⇒ x² = 0

⇒x = 0

Thus One real value of x is possible

∴ Option (c) is correct.

Explanation:

f(x) = [x]² – 30[x] + 221 = 0 

⇒ [x] = \(\frac{30± \sqrt{900-884}}{2} = \frac{30±4}{2}\)

⇒ [x] = \(\frac{30+ 4}{2 }, \frac{30 -4 }{2}\)

⇒ [x] = 17 , 13

Thus, Sum of integral solution of f(x) = 30.

∴ Option (d) is correct

Explanation:

Given:

f(x) = 9x - 8√x 

g(x) = f(x) - 1

Let x = t²

⇒ g(t² ) = 9t² – 8t – 1 = 0

⇒ t = \(\frac{8± 10}{18} = 1, \frac{-1}{9}\)

⇒ x = 1 , 1/81

 g(x) = 0 has only one real root which is an integer.

∴ Option (b) is correct.

Explanation:

\(\rm \lim_{x\rightarrow \frac{\pi}{2}}(\sec \theta-\tan \theta)\)

= \(\rm \lim_{x\rightarrow \frac{\pi}{2}} \frac{1-sin\theta }{cos\theta} [\frac{0}{0}\) Form]

= \(\rm \lim_{x\rightarrow \frac{\pi}{2}}( \frac{-cos\theta}{-sin\theta}) =0\)  (using L' hospital rule)

∴ Option (b) is correct.

Explanation:

Given:

f(x)f(y) = f(xy)

Let f(x) = xⁿ and f(y) = yⁿ . 

⇒f(xy) = (xy)ⁿ

f(2) = 4

⇒ 4 = (2)ⁿ

⇒ n= 2

\(f(\frac{1}{2}) =( \frac{1}{2})^2 =\frac{1}{4}\)

∴ Option (a) is correct.