Quantitative Aptitude PYQs

Concept:

The general equation of circle is given by: x² + y² + z² + 2ux + 2vy + 2wz + d = 0

• Center = (- u, - v, - w)
• Radius, r = \(\sqrt{u^2+v^2+w^2-d}\)

Calculation:

Given, A(1, -1, 2) and B(2, 1, -1) are the end points of the diameter.

⇒ Centre = Midpoint of AB

= \(\left(\frac{1+2}{2},\frac{-1+1}{2},\frac{2-1}{2} \right )\)

= \(\left(\frac{3}{2},0,\frac{1}{2} \right )\)

Given, x2 + y2 + z2 + 2ux + 2vy + 2wz - 1 = 0

Centre of circle = (- u, - v, - w) = \(\left(\frac{3}{2},0,\frac{1}{2} \right )\)

⇒ u = - 3/2

v = 0

w = - 1/2

∴ u + v + w = - 3/2 - 1/2 = - 4/2 = -2

∴ The value of u + v + w = - 2.

Concept:

The angle at the circumference in a semicircle is 90°.

Calculation:

Given, P(x, y, z) is any point on the sphere.

 

Given, A(1, -1, 2) and B(2, 1, -1) are the end points of the diameter.

∴ AB = \(\sqrt{(1-2)^2+(-1-1)^2+(2+1)^2}\)

= \(\sqrt{1+4+9}\)

= \(\sqrt{14}\)

Now, the angle at circumference is 90^∘.

⇒ ΔPAB is a right angled triangle

⇒ PA2 + PB2 = AB² 

⇒ PA2 + PB2 = (\(\sqrt{14}\))² 

⇒ PA2 + PB2 = 14

∴ The value of PA2 + PB2 is 14.

Concept:

If two lines having direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) inclines at an angle of θ, then:

cosθ = \(\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}\)

Calculation:

The direction ratios of the lines are (2, -1, 2) and (k, 3, 5). 

The angle between the lines is \(\frac{\pi}{4}\).

∴ cos(\(\frac{\pi}{4}\)) = \(\frac{2k-3+10}{\sqrt{2^2+(-1)^2+2^2}\sqrt{k^2+3^2+5^2}}\)

⇒ \(\frac{1}{\sqrt{2}}\) = \(\frac{2k+7}{3\sqrt{k^2+34}}\)

Squaring both sides and cross multiplying, we get:

⇒ 9(k2 + 34) = 2(4k2 + 28k + 49)

⇒ 9k² + 306 = 8k2 + 56k + 98

⇒ k² - 56k + 208 = 0

⇒ k² - 52k - 4k + 208 = 0

⇒ k(k - 52) - 4(k - 52) = 0

⇒ (k - 52)(k - 4) = 0

⇒ k = 4, 52

∴ Required value of k is 4.

Concept:

If \(\vec{a}\) and \(\vec{b}\) are two vectors having direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂), then the direction ration of a line perpendicular to both the lines is given by:

\(\vec{c}\) =λ \(\begin{vmatrix} ̂{i} & ̂{j} & ̂{k}\ a_1 & a_2 & a_3\ b_1 & b_2 & b_3 \end{vmatrix}\), where λ is a real number.

Calculation:

The direction ratios are (2, -1, 2) and (k, 3, 5).

k = 4

⇒ Direction ratio is (2, -1, 2) and (4, 3, 5).

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∴ \(\vec{a}=2̂{i} -̂{j}+2̂{k}\) and \(\vec{b}=4̂{i}+ 3̂{j}+ 5̂{k}\)

∴ \(\vec{c}\) = λ\(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\ a_1 & a_2 & a_3\ b_1 & b_2 & b_3 \end{vmatrix}\) = λ\(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\ 2 & -1 & 2\ 4 & 3 & 5 \end{vmatrix}\)

= λ[î(-5 - 6) - ĵ(10 - 8) + k̂(6 + 4)]

= λ[-11î - 2ĵ + 10k̂]

Putting λ = - 1, we get:

\(\vec{c}\) = 11î + 2ĵ - 10k̂

∴ The direction ratios of a line which is perpendicular to both the lines is (11, 2, - 10).

Concept:

If \(\vec{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}\) and \(\vec{b}=x_2\hat{i}+y_2\hat{j}+z_2\hat{k}\) are two vectors, then:

\(\vec{a} \cdot \vec{b}=x_1x_2+y_1y_2+z_1z_2\)

\( \vec{a} \times \vec{b}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\ x_1 & y_1 & z_1\ x_2 & y_2 & z_2 \end{vmatrix}\)

Calculation:

Given, \(\vec{a}=3 \hat{i}+3 \hat{j}+3 \hat{k} \) and \( \vec{c}=\hat{j}-\hat{k}\) 

Also, \(\vec{a} \cdot \vec{b}=27 \) and \( \vec{a} \times \vec{b}=\overrightarrow{9 c}\)

Let \(\vec{b}=x \hat{i}+y \hat{j}+z \hat{k} \)

⇒ \(\vec{a} \cdot \vec{b}\) = 3x + 3y + 3z = 27

⇒ x + y + z = 9 ...(i)

Also, \( \vec{a} \times \vec{b}\)

= \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\ 3 & 3 & 3\ x & y & z \end{vmatrix}\)

= \(3(z-y)\hat{i}-3(z-x)\hat{j}+3(y-x)\hat{k}\)

= \(9\vec{c}\) = \(9\hat{j}-9\hat{k}\)

Since, \( \vec{a} \times \vec{b}\)  = \(9\vec{c}\)

\(3(z-y)\hat{i}-3(z-x)\hat{j}+3(y-x)\hat{k}\) = \(9\hat{j}-9\hat{k}\)

On comparing both side

⇒ z - y = 0 ⇒ y = z          ...(ii)

• 3(z - x) = 9 ⇒ z - x = - 3 ...(iii)

3(y - x) = - 9 ⇒ y - x = - 3 ...(iv)

(ii) + (iii) + (iv) 

⇒ 2z - 2x = - 6 

⇒ z - x = - 3 

⇒ x = z + 3

Putting the values of x and y in (i), we get:

z + 3 + z + z = 9

⇒ 3z = 6

⇒ z = 2

∴ y = 2 and x = 5

∴ \(\vec{b}=x \hat{i}+y \hat{j}+z \hat{k} \)

= \(5 \hat{i}+2 \hat{j}+2 \hat{k}\)

∴ \(\vec{b}\) = \(5 \hat{i}+2 \hat{j}+2 \hat{k}\)

Concept:

If \(\vec{a}\) and \(\vec{b}\) are two vectors having angle θ between them, then

θ = \(\frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}\)

Calculation:

Given, \(\vec{a}=3 \hat{i}+3 \hat{j}+3 \hat{k} \) and \( \vec{c}=\hat{j}-\hat{k}\) 

Also, \(\vec{b}=5\hat{i}+2 \hat{j}+2 \hat{k} \)

∴ \((\vec{a}+\vec{b})\) = \(8\hat{i}+5\hat{j}+5 \hat{k} \)

Let angle between \((\vec{a}+\vec{b})\) and \(\vec{c}\) be θ.

∴ \((\vec{a}+\vec{b})\cdot\vec{c}\) = \(|\vec{a}+\vec{b}||\vec{c}|\cos θ\)

⇒ 0 + 5 - 5 = \(\sqrt{8^2+5^2+5^2}\)(\sqrt{0^2+1^2+(-1)^2}\)cosθ 

⇒ cosθ = 0

⇒ θ = \(\frac{\pi}{2}\)

∴ Angle between \((\vec{a}+\vec{b})\) and \(\vec{c}\) is \(\frac{\pi}{2}\).

Concept:

If a vector is given by \(\vec{a}=x \hat{i}+y \hat{j}+z\hat{k}\), then the direction cosines are given by:

cosα = \(\frac{x}{\sqrt{x^2+y^2+z^2}}\)

cosβ = \(\frac{y}{\sqrt{x^2+y^2+z^2}}\)

cosγ = \(\frac{z}{\sqrt{x^2+y^2+z^2}}\)

Calculation:

Given,  \(\vec{a}=4 \hat{i}-8 \hat{j}+\hat{k}\) and it make angles α, β, γ with the positive directions of x, y, z axes respectively.

∴ cosα = \(\frac{4}{\sqrt{4^2+(-8)^2+1^2}}\)

= \(\frac{4}{\sqrt{16+64+1}}\)

= \(\frac{4}{\sqrt{81}}\)

= \(\frac{4}{9}\)

∴ The value of cosα is \(\frac{4}{9}\).

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Concept:

If a vector is given by \(\vec{a}=x \hat{i}+y \hat{j}+z\hat{k}\), then the direction cosines are given by:

l = cosα = \(\frac{x}{\sqrt{x^2+y^2+z^2}}\)

m = cosβ = \(\frac{y}{\sqrt{x^2+y^2+z^2}}\)

n = cosγ = \(\frac{z}{\sqrt{x^2+y^2+z^2}}\)

Also, cos²α + cos²β + cos²γ = 1

Calculation:

Given,  \(\vec{a}=4 \hat{i}-8 \hat{j}+\hat{k}\) and it make angles α, β, γ with the positive directions of x, y, z axis respectively.

Also, cosα = \(\frac{4}{9}\)

∴ cos2β + cos2γ

= 2cos²β - 1 + 2cos²γ - 1 [∵ cos2θ = 2cos²θ - 1]

= 2[cos2β + cos2γ] - 2

= 2[1 - cos2α] - 2 [∵  cos2α + cos2β + cos2γ = 1]

= 2(1 - \(\frac{16}{81}\)) - 2

= 2 - \(\frac{32}{81}\) - 2

= \(-\frac{32}{81}\)

∴ The value of cos2β + cos2γ is equal to \(-\frac{32}{81}\).

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Concept:

Equation of line joining A((x1, y1, z1) and B(x2, y2, z2) is given by:

\(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\)

Calculation:

The given vectors are \(\vec{A}=\hat{i}-\hat{j}\) and \(\vec{B}=\hat{j}+\hat{k}\)

∴ A passes through (1, -1, 0) = (x₁, y₁, z₁) and B passes through (0, 1, 1) = (x₂, y₂, z₂)

∴ Equation of line joining A and B is given by:

\(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\)

⇒ \(\frac{x-1}{0-1}=\frac{y+1}{2}=\frac{z-0}{1-0}\)

⇒ \(\frac{x-1}{-1}=\frac{y+1}{2}=\frac{z}{1}\) ...(i)

Putting (-1, -3, 1) in (i), we get 

\(\frac{-1-1}{-1}=\frac{-3+1}{2}=\frac{1}{1}\)

⇒ 2 = - 1 = 1, which is not true

⇒ (-1, -3, 1) does not lie on AB

Putting (-1, 3, 2) in (i), we get 

\(\frac{-1-1}{-1}=\frac{3+1}{2}=\frac{2}{1}\)

⇒ 2 = 2 = 2, which is true

⇒ (-1, 3, 2) lie on AB

Putting (-1, 3, 2) in (i), we get 

\(\frac{-2-1}{-1}=\frac{5+1}{2}=\frac{3}{1}\)

⇒ 3 = 3 = 3, which is true

⇒ (-2, 5, 3) lie on AB

∴  Point 2 and 3 lie on AB.

Concept:

If any vector is given by \(\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}\), then the magnitude is given by:

|\(\vec{a}\)| = \(\sqrt{a^2+b^2+c^2}\)

Calculation:

The position vectors of two points A and B are î - ĵ and ĵ + k̂ respectively. 

∴ \(\overrightarrow{OA}\) = \(\hat{i}-\hat{j}\) and \(\overrightarrow{OB}\) = \(\hat{j}+\hat{k}\)

∴ \(\overrightarrow{A B}\) = \(\overrightarrow{OB}\) - \(\overrightarrow{OA}\)

= \(\hat{j}+\hat{k}\) - (\(\hat{i}-\hat{j}\))

= \(-\hat{i}+2\hat{j}+\hat{k}\)

∴ |\(\overrightarrow{A B}\)| = \(√{(-1)^2+2^2+1^2}\) = \(√{1+4+1}\) = √6

∴ The magnitude of \(\overrightarrow{A B}\) is √6.

Concept:

\(\frac{d}{dx}e^{mx}\) = me^mx 

\(\int e^{mx}dx\) = \(\frac{e^{mx}}{m}\) + C

Calculation:

Given f(x) = Pe^x + Qe^2x + Re^3x 

∴ f(0) = P + Q + R = 6(given) ...(i)

∴ f'(x) = Pex + 2Qe2x + 3Re3x 

⇒ f'(ln 3) = Pe^{ln 3} + 2Qe^{2(ln 3)} + 3Re^{3(ln 3)} 

= Pe^{ln 3} + 2Qe^{ln 9} + 3Re^{ln 27} 

= 3P + 18Q + 81R = 282 (given)

∴ 3P + 18Q + 81R = 282

⇒ P + 6Q + 27R = 94 ...(ii)

Now, \(\int_0^{\ln 2}f(x)dx\)

= \(\int_0^{\ln 2}(Pe^x+Qe^{2x}+Re^{3x})dx\)

= \(\left[Pe^x+\frac{Qe^{2x}}{2}+\frac{Re^{3x}}{3} \right ]_0^{\ln 2}\)

= P[2 - 1] + \(\frac{Q}{2}\)[4 - 1] + \(\frac{R}{3}\)[8 - 1]

= P + \(\frac{3Q}{2}\) + \(\frac{7R}{3}\) = 11(given)

⇒ 6P + 9Q + 14R = 66 ...(iii)

∴ The system of equation is:

• P + Q + R = 6
• P + 6Q + 27R = 94
• 6P + 9Q + 14R = 66​

​∴ Δ = \(\begin{vmatrix} 1 & 1 & 1\ 1 & 6 & 27\ 6 & 9 & 14 \end{vmatrix} \) = - 38

Δ₁ = \(\begin{vmatrix} 6 & 1 & 1\ 94 & 6 & 27\ 66 & 9 & 14 \end{vmatrix} \) = - 38

Δ₂ = \(\begin{vmatrix} 1 & 6 & 1\ 1 & 94 & 27\ 6 & 66 & 14 \end{vmatrix}\) = - 76

Δ₃ = \(\begin{vmatrix} 1 & 1 & 6\ 1 & 6 & 94\ 6 & 9 & 66\end{vmatrix} \) = - 114

∴ P = \(\frac{\Delta_1}{\Delta}\) = \(\frac{-38}{-38}\) = 1

Q = \(\frac{\Delta_2}{\Delta}\) = \(\frac{-76}{-38}\) = 2

R = \(\frac{-114}{-38}\) = 3

∴ The value of Q is 2.

Concept:

\(\frac{d}{dx}e^{mx}\) = memx 

\(\int e^{mx}dx\) = \(\frac{e^{mx}}{m}\) + C

Calculation:

Given f(x) = Pex + Qe2x + Re3x 

∴ f(0) = P + Q + R = 6(given) ...(i)

∴ f'(x) = Pex + 2Qe2x + 3Re3x 

⇒ f'(ln 3) = Peln 3 + 2Qe2(ln 3) + 3Re3(ln 3) 

= Peln 3 + 2Qeln 9 + 3Reln 27 

= 3P + 18Q + 81R = 282 (given)

∴ 3P + 18Q + 81R = 282

⇒ P + 6Q + 27R = 94 ...(ii)

Now, \(\int_0^{\ln 2}f(x)dx\)

= \(\int_0^{\ln 2}(Pe^x+Qe^{2x}+Re^{3x})dx\)

= \(\left[Pe^x+\frac{Qe^{2x}}{2}+\frac{Re^{3x}}{3} \right ]_0^{\ln 2}\)

= P[2 - 1] + \(\frac{Q}{2}\)[4 - 1] + \(\frac{R}{3}\)[8 - 1]

= P + \(\frac{3Q}{2}\) + \(\frac{7R}{3}\) = 11(given)

⇒ 6P + 9Q + 14R = 66 ...(iii)

∴ The system of equation is:

• P + Q + R = 6
• P + 6Q + 27R = 94
• 6P + 9Q + 14R = 66​

​∴ Δ = \(\begin{vmatrix} 1 & 1 & 1\ 1 & 6 & 27\ 6 & 9 & 14 \end{vmatrix} \) = - 38

Δ1 = \(\begin{vmatrix} 6 & 1 & 1\ 94 & 6 & 27\ 66 & 9 & 14 \end{vmatrix} \) = - 38

Δ2 = \(\begin{vmatrix} 1 & 6 & 1\ 1 & 94 & 27\ 6 & 66 & 14 \end{vmatrix}\) = - 76

Δ3 = \(\begin{vmatrix} 1 & 1 & 6\ 1 & 6 & 94\ 6 & 9 & 66\end{vmatrix} \) = - 114

∴ P = \(\frac{\Delta_1}{\Delta}\) = \(\frac{-38}{-38}\) = 1

Q = \(\frac{\Delta_2}{\Delta}\) = \(\frac{-76}{-38}\) = 2

R = \(\frac{-114}{-38}\) = 3

∴ The value of R is 3.

Concept:

\(\frac{d}{dx}e^{mx}\) = memx 

Calculation:

∴ P = 1

Q = 2

R = 3

∴ f(x) = Pex + Qe2x + Re3x  = ex + 2e2x + 3e3x 

⇒ f'(x) = ex + 4e2x + 9e3x 

⇒ f'(0) = e⁰ + 4e⁰ + 9e⁰ = 1 + 4 + 9 = 14

∴ The value of f'(0) = 14

Concept:

Order: The order of a differential equation is defined by the highest order derivative.

Calculation:

Given, y² = 2cx + 2c√c

Differentiating both sides w.r.t. x,

2yy' = 2c 

⇒ c = yy'

∴ Equation becomes, y² = 2(yy')x + 2(yy')^3/2 

⇒ [y² - 2(yy')x] = 2(yy')^3/2 

⇒ [y² - 2(yy')x]² = 4(yy')³ 

The highest order derivative is 1

⇒ Order = 1

∴ Order of the differential equation is 1.

Concept:

Degree: The degree of a differential equation is defined by the highest power on a variable.

Calculation:

The differential equation is given by:

Given, y2 = 2cx + 2c√c

Differentiating both sides w.r.t. x,

2yy' = 2c 

⇒ c = yy'

∴ Equation becomes, y2 = 2(yy')x + 2(yy')3/2 

⇒ [y2 - 2(yy')x] = 2(yy')3/2 

[y2 - 2(yy')x]2 = 4(yy')3 

The degree of highest order derivative is 3

∴ Degree of the differential equation is 3.

Concept:

If A = \(\begin{bmatrix} a & λ b & c\ d & λ e & f\ g & λ h & i \end{bmatrix}\), then the determinant of A given by:

|A| = \(\begin{vmatrix} a & λ b & c\ d & λ e & f\ g & λ h & i \end{vmatrix}\) = λ\(\begin{vmatrix} a & b & c\ d & e & f\ g & h & i \end{vmatrix}\)

Calculation:

Given, f(x) = \(\left|\begin{array}{ccc} \cos x & x & 1 \ 2 \sin x & x^2 & 2 x \ \tan x & x & 1 \end{array}\right|\)

∴  f(x) = - x2(cos x - tan x) 

∴ f(0) = - 02(cos 0 - tan 0) 

= 0

∴ f(0) = 0

Concept:

If A = \(\begin{bmatrix} a & λ b & c\ d & λ e & f\ g & λ h & i \end{bmatrix}\), then the determinant of A given by:

|A| = \(\begin{vmatrix} a & λ b & c\ d & λ e & f\ g & λ h & i \end{vmatrix}\) = λ\(\begin{vmatrix} a & b & c\ d & e & f\ g & h & i \end{vmatrix}\)

Calculation:

Given, f(x) = \(\left|\begin{array}{ccc} \cos x & x & 1 \ 2 \sin x & x^2 & 2 x \ \tan x & x & 1 \end{array}\right|\)

∴  f(x) = - x2(cos x - tan x) 

∴ \(\lim _{x \rightarrow 0} \frac{f(x)}{x}\)

= \(\lim _{x \rightarrow 0} -x(\cos x -\tan x)\)

= 0

∴ \(\lim _{x \rightarrow 0} \frac{f(x)}{x}\) = 0

Concept:

If A = \(\begin{bmatrix} a & λ b & c\ d & λ e & f\ g & λ h & i \end{bmatrix}\), then the determinant of A given by:

|A| = \(\begin{vmatrix} a & λ b & c\ d & λ e & f\ g & λ h & i \end{vmatrix}\) = λ\(\begin{vmatrix} a & b & c\ d & e & f\ g & h & i \end{vmatrix}\)

Calculation:

Given, f(x) = \(\left|\begin{array}{ccc} \cos x & x & 1 \ 2 \sin x & x^2 & 2 x \ \tan x & x & 1 \end{array}\right|\)

⇒ f(x) = \(x\left|\begin{array}{ccc} \cos x & 1 & 1 \ 2 \sin x & x & 2 x \ \tan x & 1 & 1 \end{array}\right|\)

C₃ → C₃ - C₂  

⇒ f(x) = \(x\left|\begin{array}{ccc} \cos x & 1 & 0 \ 2 \sin x & x & x \ \tan x & 1 & 0\end{array}\right|\)

⇒ f(x) = x[ -x(cos x - tan x)] [Expanding by C₃]

∴  f(x) = - x²(cos x - tan x) 

∴ \(\lim _{x \rightarrow 0} \frac{f(x)}{x^2}\)

= \(\lim _{x \rightarrow 0} -(\cos x -\tan x)\)

= - (1 - 0)

= - 1

∴ \(\lim _{x \rightarrow 0} \frac{f(x)}{x^2}\) = - 1

Concept:

Greatest Integer Function: (Floor function)

The function f (x) = [x] is calleld the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x. Domain of [x] is R and range is I

Calculation:

Given, f(x) = sin[π2]x + cos[-π2]x 

Now, π² = 9.869

⇒ [π²] = [9.869] = 9

⇒ [- π2] = [- 9.869] = - 10

∴ f(x) = sin 9x + cos (-10x) = sin 9x + cos 10x

∴ \(f\left(\frac{\pi}{2}\right)\) = sin \(\frac{9\pi}{2}\) + cos \(\frac{10\pi}{2}\)

= sin \(\frac{9\pi}{2}\) + cos 5x

= 1 + (- 1) 

= 0

∴ The value of \(f\left(\frac{\pi}{2}\right)\) is 0.

Concept:

Greatest Integer Function: (Floor function)

The function f (x) = [x] is calleld the greatest integer function and means greatest integer less than or equal to x i.e [x] ≤ x. Domain of [x] is R and range is I

Calculation:

Given, f(x) = sin[π2]x + cos[-π2]x 

Now, π2 = 9.869

⇒ [π2] = [9.869] = 9

⇒ [- π2] = [- 9.869] = - 10

∴ f(x) = sin 9x + cos (-10x) = sin 9x + cos 10x

∴ \(f\left(\frac{\pi}{4}\right)\) = sin \(\frac{9\pi}{4}\) + cos \(\frac{10\pi}{4}\)

= sin \(\frac{9\pi}{4}\) + cos \(\frac{5\pi}{2}\)

= \(\frac{1}{\sqrt{2}}\) + 0

= \(\frac{1}{\sqrt{2}}\)

∴ The value of \(f\left(\frac{\pi}{4}\right)\) is \(\frac{1}{\sqrt{2}}\).