Quantitative Aptitude PYQs

Concept:

The perpendicular distance between the parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by:

d = \(\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}\)

Calculation:

Given, L₁ = 2x + y - 3 = 0 and L₂ = 4x + 2y + 5 = 0

Now, L2 = 4x + 2y + 5 = 0

⇒ 2x + y + 2.5 = 0

⇒ L1  and L2 represent the parallel and opposite sides of the square

⇒ Side of square = Perpendicular distance between L1  and L2 

Now, distance between 2x + y - 3 = 0 and 2x + y + 2.5 = 0 is given by:

d = \(\frac{|-3-2.5|}{\sqrt{2^2+1^2}}\)

= \(\frac{5.5}{\sqrt{5}}\)

∴ Area of square = d² = \(\left(\frac{5.5}{\sqrt{5}}\right)^2\) = \(\frac{30.25}{5}\) = 6.05

∴ The area of the square is 6.05 square units.

The correct answer is Option 1.

Concept:

The midpoint P(x, y) of the line joining A(x₁, y₁) andB(x₂, y₂) is given by:

(x, y) = \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

The centroid of the triangle ABC is given by:

Centroid = \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)

Calculation:

Let the coordinates of B = (x₂, y₂) and C = (x₃, y₃)

Now, A = (3, 5) 

According to the question, mid-point of sides AB is (-1, 2)

⇒ \(\frac{3+x_2}{2}=-1\) and \(\frac{5+y_2}{2}=2\)

⇒ x₂ = - 5 and y₂ = - 1

Also, mid-point of sides AC is (6, 4)

⇒ \(\frac{3+x_3}{2}=6\) and \(\frac{5+y_3}{2}=4\)

⇒ x₃ = 9 and y₃ = 3

∴ The coordinates of the ΔABC are A(3, 5), B(- 5, - 1) and C(9, 3)

∴ Centroid = \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)

= \(\left(\frac{3-5+9}{3}, \frac{5-1+3}{3}\right)\)

= \(\left(\frac{7}{3}, \frac{7}{3}\right)\)

∴ The coordinates of centroid of the triangle ABC is \(\left(\frac{7}{3}, \frac{7}{3}\right)\).

The correct answer is Option 2.

Explanation:

7x - y - 3 = 0 ⇒ y = 7x - 3

x + y - 5 = 0 ⇒ y = - x + 5

Slope of AB is m₁ = 7 and slope of AC is m₂ = -1

Then tan A = \(\left|m_1+m_2\over 1-m_1m_2\right|\) = \(\left|7+1\over 1-7\right|\) = 4/3

 tan A = \(4\over3\)

So, sin A = \(4\over \sqrt{4^2+3^2}\) = \(4\over5\)

tan(\(A\over2\)) = cosec A - cot A = \(5\over4\) - \(3\over4\) = \(1\over2\)

Now, for triangle ABC

θ + θ + A = π

2θ = π - A

θ = \({\pi\over2}-{A\over2}\)

cot θ = cot(\({\pi\over2}-{A\over2}\))

cot θ = tan \(A\over2\) = \(1\over2\)

Option (2) is true.

Calculation:

Given, y² = 8x and P is a point on the parabola.

For x = 6, y² = 8 × 6

⇒ y² = 48

⇒ y = 4√3

⇒ (6, 4√3) lies on the parabola

⇒ The coordinates of P can be (6, 4√3)

Hence, Statement 1 is correct.

Now, comparing with y2 = 4ax, we get a = 2

⇒ Directrix is x = - 2

∴ Perpendicular Distance = |6 − (− 2)| 

= |6 + 2|

= 8

Hence, Statement 2 is correct.

∴ Both 1 and 2 are correct.

The correct answer is Option 3.

Concept:

The equation of an ellipse is given \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), where a and b are major and minor axes.

b2 = a2(1 − e2), where e is the eccentricity of the ellipse.

Calculation:

Let the ellipse be \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

⇒ Foci are S1(ae, 0) and S2(− ae, 0) and the extremity of the minor axis is B (0, b)

Now, angle between the straight lines joining the foci to an extremity of the minor axis is 90° 

⇒ Product of slopes, m_S1B × m_S2B = − 1

⇒ \(\left(\frac{b-0}{0-ae}\right)\)(\left(\frac{b-0}{0+ae}\right)\) = − 1

⇒ b² = a²e²

⇒ a²(1 − e²) = a²e²

⇒ 1 − e² = e²

⇒ e² = \(\frac{1}{2}\)

⇒ e = \(\frac{1}{\sqrt{2}} \)

∴ The eccentricity of the ellipse is \(\frac{1}{\sqrt{2}} \).

The correct answer is Option 4.

Concept:

The cross product or vector product of the vectors \(\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\) and \(\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\) is given by:

\(\vec{a}\times\vec{b}\) = \(\begin{vmatrix} \widehat{i} &\widehat{j} & \widehat{k}\ a_1& a_2 & a_3 \b_1 & b_2 & b_3 \end{vmatrix}\)

Calculation:

Given, \(\vec{a}=\hat{i}-\hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\)

∴ \(\vec{b}\times\vec{a}\) = \(\begin{vmatrix} \widehat{i} &\widehat{j} & \widehat{k}\ 1& 2 & -1 \1 & -1 & 1 \end{vmatrix}\) 

= \(\hat{i}(2-1)-\hat{j}(1+1)+\hat{k}(-1-2) \)

= \(\hat{i}-2\hat{j}-3\hat{k} \)

∴ \(\vec{a} \times(\vec{b} \times \vec{a})\) = \(\begin{vmatrix} \widehat{i} &\widehat{j} & \widehat{k}\ 1& -1 & 1 \1 & -2 & -3 \end{vmatrix}\)

= \(\hat{i}(3+2)-\hat{j}(-3-1)+\hat{k}(-2+1) \)

= \(5\hat{i}+4\hat{j}-\hat{k} \) = \(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\)

⇒ α = 5, β = 4 and γ = - 1

⇒ α + β + γ = 5 + 4 - 1 = 8

∴ The value of α + β + γ is 8.

The correct answer is Option 1.

Explanation:

The vector makes an angle \(\frac{\pi}{3}\) with \(2 \hat{i}\) i.e., x-axis, \(\frac{\pi}{4}\) with \(3 \hat{j}\) i.e., y-axis and an acute angle θ with \(4 \hat{k}\) i.e., with z axis

So, direction cosines are cos\(\frac{\pi}{3}\), cos \(\frac{\pi}{4}\), cos θ = \(\frac12\), \(\frac1{\sqrt2}\), cos θ

Then l² + m² + n² = 1

⇒ \(\frac14\) + \(\frac12\) + cos²θ = 1

⇒ cos2θ = \(\frac12\)

So the vector is (\(\frac12\), \(\frac1{\sqrt2}\), \(\frac12\)) = (1, √2, 1) 

Option (1) is true.

CONCEPT:

Moment of a Force

• The moment of a force (torque) about a point is the rotational effect of the force about that point, calculated as the cross product of the position vector and the force vector.
• The moment of a force about a line involves the projection of the moment about a point onto a direction parallel to the line, making it a vector quantity.
• Key properties to consider are:
• Dependence on the position vector for moments about a point.
• Vector nature of moments about a line.

Explanation:-

• Statement 1: The moment of force about a point is independent of the point of application of force.
• Incorrect: The moment of a force about a point depends on the position vector from the point to the point of application of the force.
• Statement 2: The moment of a force about a line is a vector quantity.
• Correct: The moment of a force about a line is considered a vector quantity as it involves direction and magnitude.

CONCLUSION:

The correct answer is 2 only.

Explanation:

Let \(\vec{r}=\hat i\)

Then \(\vec{r} \cdot \hat{i}=\hat{i} \cdot \hat{i}=1\)

\(\vec{r} \times \hat{i}=\hat{i} \times \hat{i}\) = 0

\(\vec{r} \cdot \hat{j}=\hat{i} \cdot \hat{j}\) = 0

\(\vec{r} \cdot \hat{k}=\hat{i} \cdot \hat{k}\) = 0

So,

\((\vec{r} \cdot \hat{i})(\vec{r} \times \hat{i})+(\vec{r} \cdot \hat{j})\)((\vec{r} \times \hat{j})+(\vec{r} \cdot \hat{k})(\vec{r} \times \hat{k})\)

= 0 + 0 + 0 = 0

Option (1) is true.

Concept:

\(\vec{a}\cdot\vec{b}\) = \(|\vec{a}||\vec{b}|\cos θ\), where θ is the angle between \(\vec{a}\) and \(\vec{b}\)

Calculation:

Given, |\(\vec{a}\)| = 4 and |\(\vec{b}\)| = 4

Also, θ = \(\frac{\pi}{3}\)

Let the angle between \(\vec{a}\) and \(\vec{a}-\vec{b}\) be ϕ.

∴ \(\vec{a}\cdot(\vec{a}-\vec{b})\) = \(|\vec{a}||\vec{a}-\vec{b}|\cos ϕ\)

⇒ \(\vec{a}\cdot\vec{a}-\vec{a}\cdot\vec{b}\) = \(|\vec{a}|\sqrt{|\vec{a}|^2+|\vec{b}|^2-2\cdot|\vec{a}||\vec{b}|\cos θ}\cosϕ\)

⇒ \(|\vec{a}|^2-|\vec{a}||\vec{b}|\cosθ\) = \(|\vec{a}|\sqrt{|\vec{a}|^2+|\vec{b}|^2-2\cdot|\vec{a}||\vec{b}|\cos θ}\cosϕ\)

⇒ \(4^2-4×4\cos\frac{\pi}{3}\) = \(4\sqrt{4^2+4^2-2\cdot4\cdot4\frac{1}{2}}\cosϕ\)

⇒ 16 - 16 × \(\frac{1}{2}\) = 16 cosϕ 

⇒ 16 - 8 = 16 cosϕ 

⇒ 8 = 16 cosϕ

⇒ ϕ = \(\frac{\pi}{3}\)

∴ The angle between \(\vec{a}\) and \(\vec{a}-\vec{b}\) is \(\frac{\pi}{3}\).

The correct answer is Option 2.

Explanation:

\(\frac{d y}{d x}=x\)

dy = x dx

Integrating both sides

y = \(x^2\over2\) + c

y1(0) = 0 ⇒ c = 0

So, y₁(x) = \(x^2\over2\)

and y2(0) = 4 ⇒ c = 4

So, y2(x) = \(x^2\over2\)+ 4

For points of intersections,

y1(x) = y2(x)

⇒ \(x^2\over2\) = \(x^2\over2\) + 4

⇒ 0 = 4 which is not possible.

Therefore no point of intersection.

Option (1) is true.

Explanation:

y = ex(a cos x + b sin x)....(i)

Differentiating both sides

\({dy\over dx}\) = ex(a cos x + b sin x) + ex(-a sin x + b cos x)

\({dy\over dx}\) = y + ex(-a sin x + b cos x)  (using (i))

\({dy\over dx}\) - y = ex(-a sin x + b cos x) ...(ii)

Again differentiating

\({d^2y\over dx^2}\) = \({dy\over dx}\) + ex(-a sin x + b cos x) + ex(-a cos x - b sin x) 

\({d^2y\over dx^2}\) = \({dy\over dx}\) + \({dy\over dx}\) - y - y (using (i) and (ii))

\(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+2 y=0\)

Option (3) is true.

Calculation:

Given, f(x) = ax - b and g(x) = cx + d

According to the question, f(g(x)) = g(f(x))

⇒ f(cx + d) = g(ax - b)

⇒ a(cx + d) - b = c(ax - b) + d

⇒ acx + ad - b = acx - bc + d

⇒ ad - b = -bc + d

⇒ ad - b = -bc - d + 2d

⇒ ad - b = -(bc + d) + 2d

⇒ f(d) = - g(b) + 2d

∴ f(d) + g(b) = 2d

The correct answer is Option 4.

Concept:

\(\int_{-a}^af(x)dx\) = 0 if f(x) is odd function i.e., if f(-x) = -f(x)

sin(-x) = - sin x and cos(-x) = cos x

Explanation:

I = \(\rm \int_{-1}^1(3 \sin x-\sin 3 x) \cos ^2 x d x\)

f(x) = (3sin x - sin 3x)cos²x

f(-x) = (3sin (-x) - sin (-3x))cos2(-x)

        = (- 3sin x + sin 3x)cos2x

       = - (3sin x - sin 3x)cos2x 

       = -f(x)

So. f(x) is odd function.

Hence 

\(\rm \int_{-1}^1(3 \sin x-\sin 3 x) \cos ^2 x d x\) = 0

Option (2) is true.

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

Given, \(\left{2-\left(\frac{d y}{d x}\right)^2\right}^{0.6}=\frac{d^2 y}{d x^2}\)

⇒ \(\left{2-\left(\frac{d y}{d x}\right)^2\right}^6=\left(\frac{d^2 y}{d x^2}\right)^{10}\)

⇒ \(\left{2-\left(\frac{d y}{d x}\right)^2\right}^3=\left(\frac{d^2 y}{d x^2}\right)^5\)

⇒ Order = 2 and Degree= 5.

∴ The order and degree of the differential equation are 2, 5 repectively.

The correct answer is Option 4.

Explanation:

\(\frac{d y}{d x}=2 e^x y^3\)

y^-3dy = 2e^xdx

Integrating both sides

\({y^{-2}\over -2}\) = 2e^x + c

- \({1\over 2y^2}\) = 2ex + c

\(y(0)=\frac{1}{2}\) implies 

• 2 = 2 + c ⇒ c = -4

∴ - \({1\over 2y^2}\) = 2ex - 4

⇒ 2y²(2e^x - 4) = - 1 

⇒ 4y2(2 - ex) = 1

Option (1) is true.

Explanation:

f(x) = e-x

So, \(\rm p=\int_a^b f(x) d x \) = \(\int_a^b e^{-x} d x \)

and \(\rm q=\int_a^b|f(x)| d x\) = \(\rm q=\int_a^b|e^{-x}| d x\) = \(\int_a^b e^{-x} d x \)

So, p = q

Option (4) is true.

Concept:

\(\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\)

Explanation:

I = \(\rm \int_0^{\frac{\pi}{2}} \frac{a+\sin x}{2 a+\sin x+\cos x} d x\)....(i)

I = \(\rm \int_0^{\frac{\pi}{2}} \frac{a+\sin (\frac{\pi}{2}-x)}{2 a+\sin (\frac{\pi}{2}-x)+\cos (\frac{\pi}{2}-x)} d x\)

I = \(\rm \int_0^{\frac{\pi}{2}} \frac{a+\cos x}{2 a+\cos x+\sin x} d x\)....(ii)

Adding (i) and (ii) we get

2I = \(\rm \int_0^{\frac{\pi}{2}} \frac{a+\sin x+a+\cos x}{2 a+\sin x+\cos x} d x\)

2I = \(\rm \int_0^{\frac{\pi}{2}} \frac{2a+\sin x+\cos x}{2 a+\sin x+\cos x} d x\)

2I = \(\rm \int_0^{\frac{\pi}{2}} d x\)

2I = \(\frac{\pi}{2}\)

I = \(\frac{\pi}{4}\)

Option (1) is true.

Explanation:

Let f(x) = \(\frac{16 x^3}{3}-4 b x^2+x\)

f'(x) = 16x² - 8bx + 1

f(x) has neither maximum nor minimum in the range x > 0.

So, f'(x) = 16x2 - 8bx + 1 ≠ 0

⇒ (-8b)² - 4 × 16 × 1 < 0

⇒ 64b² - 64 < 0

⇒ b2 - 1 < 0

⇒ b2 < 1

⇒ -1 < b < 1

Since b is non-negative so 

0 ≤ b < 1

Option (4) is true.

Explanation:

\(​f(x)=\frac{1}{\sqrt{|x|-x}}\)

If x = 0 then |x| - x = 0 so f(x) does not exists

If x is positive real number then |x| - x = 0 so f(x) does not exist.

But when x is negative real number then f(x) exist.

So, domain of f(x) is \(\mathbb R^-\)

\(g(x)=\frac{1}{\sqrt{x-|x|}}\)

If x = 0 then x - |x| = 0 so g(x) does not exists

If x is positive real number then x - |x| = 0 so g(x) does not exist.

If x is negative real number then x - |x| = 0 so g(x) does not exist.

So, g(x) has no domain.

Option (1) is true.