Quantitative Aptitude PYQs

Calculation:

We know that \( \cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1 \) ...

(i) And \( \cos(\alpha + \beta) \cdot \cos(\alpha - \beta) = \cos^{2} \alpha - \sin^{2} \beta \)

Substitute and simplify:

\( = \cos^{2} \alpha + \cos^{2} \beta - 1 = 1 - \cos^{2} \gamma \quad \Rightarrow \quad - \cos^{2} \gamma \)

Hence, the correct answer is Option 2.

Calculation:

Given,

Points A(1, 2, -1), B(2, 5, -2), and C(4, 4, -3) are three vertices of the rectangle.

We need to find the area of the rectangle formed by the vectors \( AB \) and \( BC \).

Length of vector \( AB \):

\( \text{Length of AB} = \sqrt{(2 - 1)^2 + (5 - 2)^2 + (-2 - (-1))^2} \)

\(= \sqrt{1^2 + 3^2 + 1^2} = \sqrt{11} \)

Length of vector \( BC \):

\( \text{Length of BC} = \sqrt{(4 - 2)^2 + (4 - 5)^2 + (-3 - (-2))^2} \)

\( = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{6} \)

Area of the rectangle is the product of the lengths of vectors \( AB \) and \( BC \):

\( \text{Area} = \sqrt{11} \times \sqrt{6} = \sqrt{66} \)

∴ The area of the rectangle is \( \sqrt{66} \) square units.

Hence, the correct answer is Option 3.

Calculation:

Given,

Points A(k, 1, -1), B(2k, 0, 2), and C(2 + 2k, k, 1) are the vertices of the triangle.

The vectors AB and BC are given by:

\( \overrightarrow{AB} = B - A = (2k - k, 0 - 1, 2 - (-1)) = (k, -1, 3) \)

\( \overrightarrow{BC} = C - B = (2 + 2k - 2k, k - 0, 1 - 2) = (2, k, -1) \)

The dot product of \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) is:

\( \overrightarrow{AB} \cdot \overrightarrow{BC} = (k)(2) + (-1)(k) + (3)(-1) = 2k - k - 3 = k - 3 \)

Since the vectors are perpendicular, their dot product must be zero:

\( k - 3 = 0 \)

Thus, \( k = 3 \).

∴ The value of k is \(3\).

Hence, the correct answer is Option 4.

Calculation:

Given,

The line equation is: \( \frac{x+1}{p} = \frac{y-1}{q} = \frac{z-2}{r} \),  where p = 2q = 3r .

The line makes an angle θ  with the positive direction of the y-axis.

Using the relation between the direction cosines

\( l = \frac{p}{\sqrt{p^2 + q^2 + r^2}}, m = \frac{q}{\sqrt{p^2 + q^2 + r^2}}, n = \frac{r}{\sqrt{p^2 + q^2 + r^2}} \),

we substitute p = 3r  and  q = \(\frac{3r}{2} \)

Direction cosine with y -axis is given by \( m = \frac{q}{\sqrt{p^2 + q^2 + r^2}} \).

\( m = \frac{\frac{3r}{2}}{\sqrt{(3r)^2 + \left(\frac{3r}{2}\right)^2 + r^2}} = \frac{\frac{3r}{2}}{\sqrt{9r^2 + \frac{9r^2}{4} + r^2}} = \frac{\frac{3r}{2}}{\sqrt{\frac{49r^2}{4}}} = \frac{3}{7} \).

Now, using the double angle identity for cosine \( \cos 2θ = 2 \cos^2 θ - 1 \):

\( \cos 2θ = 2 \left( \frac{3}{7} \right)^2 - 1 = 2 \times \frac{9}{49} - 1 = \frac{18}{49} - 1 = \frac{-31}{49} \).

∴ The value of \( \cos 2θ \) is \( -\frac{31}{49} \).

Hence, the correct answer is Option 1.

Calculation:

Given,

The plane passes through the point \(P(1, 1, 1)\) and is perpendicular to the line with direction ratios \( (3, 2, 1) \).

The general equation of the plane is:

\(Ax + By + Cz = D\), where \( (A, B, C) \) are the direction ratios of the normal to the plane.

Since the plane is perpendicular to the line, the normal vector to the plane is \( (3, 2, 1) \).

Thus, the equation of the plane becomes:

\( 3x + 2y + z = D \).

To find \( D \), substitute the point \( (1, 1, 1) \) into the equation of the plane:

\( 3(1) + 2(1) + 1 = D \), which simplifies to:

\( 3 + 2 + 1 = D \),

\( D = 6 \).

∴ The equation of the plane is \( 3x + 2y + z = 6 \).

Hence, the correct answer is Option 2.

Calculation:

Given,

\( \cos^2(\alpha) + \cos^2(\beta) + \cos^2(\gamma) = 1 \)

Using the identity \( \cos^2(x) = 1 - \sin^2(x) \), we substitute:

\( (1 - \sin^2(\alpha)) + (1 - \sin^2(\beta)) + (1 - \sin^2(\gamma)) = 1 \)

Simplifying the equation:

\( 3 - (\sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma)) = 1 \)

Rearrange to isolate the sine terms:

\( \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) = 2 \)

Now, calculate the dot product:

\( \vec{a} \cdot \vec{b} = \sin^2(\alpha) + \sin^2(\beta) + \sin^2(\gamma) = 2 \)

∴ The value of \( \vec{a} \cdot \vec{b} \)is 2.

Hence, the correct answer is Option 4.

Calculation:

Given,

The vector \( \vec{d} = (\vec{a} \times \vec{b}) \times \vec{c} \)

Statement I: \( \vec{d} \) is coplanar with \( \vec{a} \) and \( \vec{b} \).

We use the vector triple product identity: \( (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} \).

This shows that \( \vec{d} \) is a linear combination of \( \vec{a} \) and \( \vec{b} \), hence \( \vec{d} \) is coplanar with \( \vec{a} \) and \( \vec{b} \).

Therefore, Statement I is correct.

Statement II: \( \vec{d} \) is perpendicular to \( \vec{c} \).

To check this, compute the dot product \( \vec{d} \cdot \vec{c} \). Using the vector triple product identity, we find:

\( \vec{d} \cdot \vec{c} = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{c}) - (\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{c}) = 0 \),

which means \( \vec{d} \) is perpendicular to \( \vec{c} \).

Therefore, Statement II is correct.

∴ Both Statement I and Statement II are correct.

Hence, the correct answer is option 3.

Calculation:

Given,

\( 3\vec{a} - 4\vec{b} + \vec{c} = 0 \)

\( \vec{c} = 4\vec{b} - 3\vec{a} \)

The vector\(\overrightarrow{AB} \) is:

\( \overrightarrow{AB} = \vec{b} - \vec{a} \)

The vector \(\overrightarrow{BC} \) is:

\( \overrightarrow{BC} = \vec{c} - \vec{b} \)

Substituting \(\vec{c} = 4\vec{b} - 3\vec{a} \):

\( \overrightarrow{BC} = (4\vec{b} - 3\vec{a}) - \vec{b} \)

\( \overrightarrow{BC} = 3\vec{b} - 3\vec{a} \)

Step 4: Now, \(\overrightarrow{BC} = 3(\vec{b} - \vec{a}) \), which gives:

\( AB : BC = 1 : 3 \)

∴ The correct ratio is AB : BC = 1 : 3 , 

Hence, the correct answer is Option 2.

Calculation:

Given,

The position vectors of points A, B, and C are \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) respectively, and \( \vec{c} = \cos^2 \theta , \vec{a} + \sin^2 \theta , \vec{b} \).

The expression to evaluate is: \( (\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a}) \).

First, substitute \( \vec{c} \) into the equation:

\( (\vec{a} \times \vec{b}) + (\vec{b} \times (\cos^2 \theta , \vec{a} + \sin^2 \theta , \vec{b})) + (\cos^2 \theta , \vec{a} + \sin^2 \theta , \vec{b}) \times \vec{a} \).

Using the distributive property of the cross product:

\( (\vec{a} \times \vec{b}) + \left[ (\vec{b} \times \cos^2 \theta , \vec{a}) + (\vec{b} \times \sin^2 \theta , \vec{b}) \right] + \left[ (\cos^2 \theta , \vec{a} \times \vec{a}) + (\sin^2 \theta , \vec{b} \times \vec{a}) \right] \).

Since \( \vec{b} \times \vec{b} = 0 \) and \( \vec{a} \times \vec{a} = 0 \), we are left with:

\( (\vec{a} \times \vec{b}) + \cos^2 \theta , (\vec{b} \times \vec{a}) + \sin^2 \theta , (- \vec{a} \times \vec{b}) \).

Substitute \( \vec{b} \times \vec{a} = - (\vec{a} \times \vec{b}) \) into the expression:

\( (\vec{a} \times \vec{b}) + \cos^2 \theta , (- \vec{a} \times \vec{b}) + \sin^2 \theta , (- \vec{a} \times \vec{b}) \).

Factor out \( \vec{a} \times \vec{b} \):

\( \vec{a} \times \vec{b} \left[ 1 - \cos^2 \theta - \sin^2 \theta \right] \).

Since \( \cos^2 \theta + \sin^2 \theta = 1 \), the expression becomes:

\( \vec{a} \times \vec{b} [1 - 1] = 0 \).

∴ The final result is \( \vec{0} \).

Hence, the correct answer is option 1.

Calculation:

Given,

The vectors \(\vec{a},\vec{b} ,(\vec{a}\times\vec{b})\) are unit vectors.

Since \(\vec a\) and  \(\vec{b}\) are unit vectors, we know:

\( |\vec{a}| = 1 \quad \text{and} \quad |\vec{b}| = 1 \).

The magnitude of the cross product \(( \vec{a} \times \vec{b} )\) is given by:

\( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 1 \times 1 \times \sin \theta = \sin \theta \).

Since \( ( |\vec{a} \times \vec{b}| = 1 \), we have:

\( \sin \theta = 1 \), so \(\theta = 90^\circ \), meaning \(\vec a \) and \( \vec{b} \) are perpendicular.

The dot product \( ( \vec{a} \cdot \vec{b} )\) is:

\( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 1 \times 1 \times \cos 90^\circ = 0 \).

∴ The value of\(( \vec{a} \cdot \vec{b} ) \) is 0.

Hence, the correct answer is Option 1.

Calculation:

Given,

\(x = \sec\theta - \cos\theta \)

\(y = \sec^4\theta - \cos^4\theta\)

Compute derivatives w.r.t. θθθ

:

\(\dfrac{dx}{d\theta} = \sec\theta,\tan\theta + \sin\theta\)

\(\dfrac{dy}{d\theta} = 4\sec^4\theta,\tan\theta ;+; 4\cos^3\theta,\sin\theta\)

From the ratio \(\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)

\(\dfrac{dy}{dx} = \frac{4\bigl(\sec^4\theta,\tan\theta + \cos^3\theta,\sin\theta\bigr)} {\sec\theta,\tan\theta + \sin\theta}\)

Using identities \(x^2 = \sec^2\theta + \cos^2\theta - 2\), \(\sec^3\theta + \cos^5\theta)^2 = y^2 + 4\), and \((1 + \cos^2\theta)^2 = x^2 + 4,\)

\(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16 ,\frac{,y^{2} + 4,}{,x^{2} + 4,} \)

∴  \(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = 16,\frac{y^{2} + 4}{x^{2} + 4} \)

Hence, the correct answer is Option 3.

Calculation:

Given,

\(x = \sec\theta - \cos\theta\)

\(y = \sec^4\theta - \cos^4\theta\)

We know from before:

\(\displaystyle \Bigl(\frac{dy}{dx}\Bigr)^{2} = \frac{16,(y^{2}+4)}{,x^{2}+4,} \)

Differentiate both sides w.r.t. x:

\(\displaystyle \frac{d}{dx}\Bigl[(x^{2}+4)\bigl(\tfrac{dy}{dx}\bigr)^{2}\Bigr] = \frac{d}{dx}\bigl[16,(y^{2}+4)\bigr] \)

This yields the differential relation:

\(\displaystyle (x^{2}+4),\frac{d^{2}y}{dx^{2}} ;+;x,\frac{dy}{dx} ;-;16,y ;=;0\)

The required expression is

\(\displaystyle \frac{x^{2}+4}{y^{2}+4},\frac{dy}{dx} \Bigl[(x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y\Bigr] \)

From the differential relation,

\(\displaystyle (x^{2}+4)\frac{d^{2}y}{dx^{2}}-16y = -,x,\frac{dy}{dx} \)

So the expression becomes

\(\displaystyle \frac{x^{2}+4}{y^{2}+4},\frac{dy}{dx}, \bigl(-x,\tfrac{dy}{dx}\bigr) = -,x,\frac{x^{2}+4}{y^{2}+4} \Bigl(\frac{dy}{dx}\Bigr)^{2} \)

Using \(\bigl(\tfrac{dy}{dx}\bigr)^{2} = \tfrac{16,(y^{2}+4)}{x^{2}+4}\), all factors cancel except -16x.

∴  The value of the given expression is  \(-16x\).

Hence, the correct answer is Option 3.

Calculation:

Given,

\(AB + AC = 3\)

Let \(AB = x\) and \(AC = 3 - x\).

Then,

\(BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} \)

The area of the triangle is,

\(A = \tfrac12,x,\sqrt{9 - 6x} \)

To maximize, set up

\(A^2 = \tfrac14,x^2,(9 - 6x)\)

\(\displaystyle \frac{d(A^2)}{dx} = \tfrac14\bigl(2x(9 - 6x) + x^2(-6)\bigr) = \frac{18x(1 - x)}{4} = 0 \)

Hence \(x = 1\) (discarding x=0, so

\(BC = \sqrt{9 - 6}= \sqrt{3},\quad AC = 3 - 1 = 2\)

Therefore,

\(\sin A = \frac{BC}{AC} = \frac{\sqrt{3}}{2} \implies A = \frac{\pi}{3}\)

∴ \(\angle A = \frac{\pi}{3}\).

Hence, the correct answer is Option 3.

Calculation:

Given,

\(AB + AC = 3\)

Let \(AB = x\) and \(AC = 3 - x \).

Then,

\(BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} \)

The area of the triangle is,

\(A = \tfrac12,x,BC = \tfrac12,x,\sqrt{9 - 6x}\)

To maximize, differentiate w.r.t. \(x\) and set to zero:

\(\displaystyle \frac{d}{dx}\bigl(x\sqrt{9-6x}\bigr) = \sqrt{9-6x} ;-;\frac{6x}{2\sqrt{9-6x}} = 0 ;\Longrightarrow; x = 1 \)

At \(x = 1\), we get \(BC = \sqrt{9 - 6} = \sqrt{3}\), so

\(A_{\max} = \tfrac12 \times 1 \times \sqrt{3} = \frac{\sqrt{3}}{2}\)

∴ The maximum area is \(\frac{\sqrt{3}}{2}\) square unit.

Hence, the correct answer is Option  1.

Calculation:

Given,

\((x+y)^{p+q} = x^p,y^q\)

Differentiate implicitly w.r.t. \(x\):

\((p+q)(x+y)^{p+q-1}\bigl(1+\tfrac{dy}{dx}\bigr) = p,x^{p-1}y^q ;+; q,x^p,y^{q-1}\tfrac{dy}{dx}\)

Rearrange to collect \(\tfrac{dy}{dx}\):

\(\tfrac{dy}{dx}\bigl[(p+q)(x+y)^{p+q-1} - q,x^p,y^{q-1}\bigr] = p,x^{p-1}y^q - (p+q)(x+y)^{p+q-1}\)

Use \((x+y)^{p+q-1}=\frac{x^p,y^q}{x+y}\) to simplify:

\(\tfrac{dy}{dx} = \tfrac{y}{x}\)

∴ \(\displaystyle \frac{dy}{dx} = \frac{y}{x}\), independent of \(p\) and \(q\).

Hence, the correct answer is Option 4.

Calculation:

Given,

\( (x + y)^{p+q} = x^p,y^q \) and \( p + q = 10 \).

Differentiate both sides with respect to \(x\) implicitly:

\(\frac{d}{dx}\bigl((x+y)^{p+q}\bigr) = \frac{d}{dx}\bigl(x^p y^q\bigr) \)

Left side:

\((p+q),(x+y)^{p+q-1}\bigl(1 + \tfrac{dy}{dx}\bigr) \)

Right side (product rule):

\(p,x^{p-1}y^q ;+; q,x^p y^{q-1},\tfrac{dy}{dx} \)

Rearrange to collect \( \tfrac{dy}{dx} \) terms and use

\( (x+y)^{p+q} = x^p y^q \implies (x+y)^{p+q-1} = \tfrac{x^p y^q}{x+y} \).

After cancellation of the common factor \( p,y - q,x \), you obtain:

∴ \( \frac{dy}{dx} = \frac{y}{x} \)

Hence, the correct answer is Option 1.

Calculation:

Given,

The slope of the tangent to the curve y = f(x) at (x, f(x)) is 4 for every real number x , and the curve passes through the origin.

The slope of the tangent is the derivative of the function, so we have:

\( f'(x) = 4 \)

Integrating f'(x) = 4  with respect to  x :

\( f(x) = 4x + C \)

The curve passes through the origin, so when x = 0 , y = 0 . Substituting these values into the equation f(x) = 4x + C :

\( 0 = 4(0) + C \quad \Rightarrow \quad C = 0 \)

Therefore, the equation of the curve is:

\( f(x) = 4x \)

This is the equation of a straight line with a slope of 4, passing through the origin.

∴ The curve is a straight line with a slope of 4, passing through the origin.

Hence, the correct answer is option 1.

Calculation: 

 

Given,

The equation of the curve is y = 4x , and the line x = 4 intersects the curve at the point (4, 16) . We need to find the area bounded by the curve, the x-axis, and the line x = 4 .

The region of interest is a right triangle with a base along the x-axis from x = 0 to x = 4  and a height of 16 units, corresponding to the point (4, 16) .

The area of the triangle is given by the formula:

\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

Substituting the values of the base (4 units) and the height (16 units):

\( \text{Area} = \frac{1}{2} \times 4 \times 16 = 32 , \text{square units} \)

∴ The area is 32 square units.

Hence, the correct answer is option 3.

Calculation:

Given,

The function is defined as:

\( f(x) = \begin{cases} x^3, & \text{if} , |x| < 1 \ x^2, & \text{if} , |x| \geq 1 \end{cases} \)

We are tasked with finding:

\( \lim_{x \to 0} f'(x) \)

For  |x| < 1 , the function is  f(x) = x³, so the derivative is:

\( f'(x) = 3x^2 \)

Now, compute the limit of the derivative as x to 0:

\( \lim_{x \to 0} f'(x) = \lim_{x \to 0} 3x^2 = 0 \)

∴ The value of  \(\lim_{x \to 0} f'(x) \)is 0.

The correct answer is Option (c)

Calculation:

Given,

The function is defined as:

\( f(x) = \begin{cases} x^3, & \text{if} , |x| < 1 \ x^2, & \text{if} , |x| \geq 1 \end{cases} \)

We are tasked with finding:

\( \lim_{x \to -1} f'(x) \)

Check the left-hand limit for continuity at x = -1 :

\( \lim_{x \to -1^-} f(x) = (-1)^2 = 1 \)

Check the right-hand limit for continuity at x = -1 :

\( \lim_{x \to -1^+} f(x) = (-1)^3 = -1 \)

Since the left-hand limit (L.H.S) and right-hand limit (R.H.S) are not equal, the function is discontinuous at x = -1 .

Check the differentiability at x = 1 :

The left-hand derivative at x = 1  is \(\text{L.H.D} = 3 \) and the right-hand derivative at  x = 1  is \(\text{R.H.D} = 2 \) which means the function is not differentiable at x = 1 

∴ The function is neither continuous at x = -1  nor differentiable at x = 1 . 

Hence, the correct answer is Option 4.