Quantitative Aptitude PYQs

Calculation:

We are given the line equation: \( \sqrt{3}x + 2y = 7 \)

The general perpendicular form is: \( x\cos\alpha + y\sin\alpha = p \)

Compare with \( Ax + By = C \) where \( A = \sqrt{3}, B = 2, C = 7 \)

\( \sqrt{A^2 + B^2} = \sqrt{(\sqrt{3})^2 + 2^2} = \sqrt{3 + 4} = \sqrt{7} \)

Normalize coefficients

\( \cos\alpha = \frac{\sqrt{3}}{\sqrt{7}} \)

\( \sin\alpha = \frac{2}{\sqrt{7}} \)

\( p = \frac{7}{\sqrt{7}} = \sqrt{7} \)

Final perpendicular form

\( \frac{\sqrt{3}}{\sqrt{7}}x + \frac{2}{\sqrt{7}}y = \sqrt{7} \)

Therefore, the correct answer is Option 3.

Calculation:

\( d = BD = \sqrt{(4 - 2)^2 + (1 - 3)^2} \)

\( = \sqrt{4 + 4} = 2\sqrt{2} \)

Area of the square \( = \frac{1}{2} d^2 \)

\( = \frac{1}{2} (2\sqrt{2})^2 = \frac{1}{2} \times 8 \)

\( = 4 \) square units

Calculation:

\( px + qy = p + q \Rightarrow m_1 = -\frac{p}{q} \)

\( p(x - y) + q(x + y) = 2q \)

\( x(p + q) - y(p - q) = 2q \Rightarrow m_2 = \frac{p + q}{p - q} \)

\( \tan \theta = \frac{m_2 - m_1}{1 + m_1 \cdot m_2} \)

\( = \left| \frac{(p + q)}{(p - q)} + \frac{p}{q} \right| \Big/ \left| 1 - \frac{p}{q} \cdot \frac{(p + q)}{(p - q)} \right| \)

\( = \left| \frac{pq + q^2 + p^2 - pq}{pq - q^2 - p^2 + pq} \right| = 1 \)

\( \tan \theta = \tan\left(\frac{\pi}{4}\right) \Rightarrow \theta = \frac{\pi}{4} \)

\( \therefore \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \)

Hence, the correct answer is Option 4.

Calculation:

\( x^2 + y^2 - 2kx - 2ky + k^2 = 0 \)

\( \Rightarrow (x - k)^2 + (y - k)^2 = k^2 \)

\(\therefore\) Centre = \( (k, k) \) and radius = \( k \)

So circle touches the x-axis at \( P(k, 0) \) and y-axis at \( Q(0, k) \)

\(\therefore PQ = \sqrt{k^2 + k^2} = \sqrt{2k^2} = \sqrt{2}k \)

Hence, the correct answer is Option 1.

Calculation:

The given equation of the hyperbola is:

\( \dfrac{x^2}{1} - \dfrac{y^2}{\tfrac{1}{4}} = 1 \)

Comparing this with the standard form of a hyperbola

\( \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \)

we get \( a^2 = 1 \) and \( b^2 = \tfrac{1}{4} \).

For a hyperbola, the focal distance satisfies:

\( c^2 = a^2 + b^2 \)

Substituting the values,

\( c^2 = 1 + \tfrac{1}{4} = \tfrac{5}{4} \)

Thus,

\( c = \dfrac{\sqrt{5}}{2} \)

The distance between the two foci of the hyperbola is:

\( 2c = \sqrt{5} \)

Hence, the correct answer is option 2

Calculation:

\( |\vec{p}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 4 + 4 - 4 = 4 \)

\( \Rightarrow |\vec{p}| = \sqrt{4} = 2 \)

⇒ \( |\vec{q}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 4 + 4 + 4 = 12 \)

\( \Rightarrow |\vec{q}| = \sqrt{12} = 2\sqrt{3} \)

⇒ \( \vec{p} \cdot \vec{q} = |\vec{a}|^2 - |\vec{b}|^2 = 4 - 4 = 0 \)

\( \therefore \vec{p} \perp \vec{q} \Rightarrow \theta = 90^\circ \)

\( |\vec{p} \times \vec{q}| = |\vec{p}||\vec{q}| \sin \theta = 2 \cdot 2\sqrt{3} \cdot \sin 90^\circ = 4\sqrt{3} \)

Hence, the correct answer is Option 4.

Calculation:

A vector perpendicular to both the given vectors is obtained using the cross product.

Let the given vectors be represented in determinant form:

\( \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 1 \ 1 & 1 & 3 \end{vmatrix} \)

Expanding the determinant, we get:

\( \vec{n} = \hat{i}(-3 - 1) - \hat{j}(6 - 1) + \hat{k}(2 + 1) \)

\( \vec{n} = -4\hat{i} - 5\hat{j} + 3\hat{k} \)

Multiplying the vector by 2,

\( 2\vec{n} = -8\hat{i} - 10\hat{j} + 6\hat{k} \)

Dividing by 50 to obtain the required form,

\( \frac{1}{50}\vec{n} = \frac{1}{50}(-4\hat{i} - 5\hat{j} + 3\hat{k}) \)

Thus, Given all three vectors perpendicualr two given vectors

Hence, the correct answer is Option 4.

Calculation:

A vector perpendicular to both the given vectors is obtained using the cross product.

Let the given vectors be represented in determinant form:

⇒ \( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 3 \ 2 & 1 & 2 \end{vmatrix} \)

Expanding the determinant, we get:

⇒ \( \vec{b}_1 \times \vec{b}_2 = \hat{i}(4 - 3) - \hat{j}(2 - 6) + \hat{k}(1 - 4) \)

⇒ \( \vec{b}_1 \times \vec{b}_2 = \hat{i} + 4\hat{j} - 3\hat{k} \)

The magnitude of the cross product (and hence the area of the parallelogram formed by the two vectors) is:

⇒ \( \left\lvert \vec{b}_1 \times \vec{b}_2 \right\rvert = \sqrt{1^2 + 4^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26} \)

Hence, the correct answer is option 3.

Calculation:

We calculate vector \(\vec{AC}\) by subtracting the position vectors of points C and A:

\( \vec{AC} = (2\hat{i} - 3\hat{j} + 2\hat{k}) - (3\hat{i} + 4\hat{j} - 2\hat{k}) \)

\( \vec{AC} = -\hat{i} - 7\hat{j} + 4\hat{k} \)

\( |\vec{AC}| = \sqrt{(-1)^2 + (-7)^2 + 4^2} = \sqrt{1 + 49 + 16} = \sqrt{66} \)

Next, compute vector \(\vec{BD}\) using point subtraction:

\( \vec{BD} = (6\hat{i} - 2\hat{j} + \hat{k}) - (4\hat{i} - 4\hat{j} - 3\hat{k}) = 2\hat{i} + 2\hat{j} + 4\hat{k} \)

Now calculate the magnitude of \(\vec{BD}\):

\( |\vec{BD}| = \sqrt{(2)^2 + (2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} \)

Compute the dot product \(\vec{AC} \cdot \vec{BD}\):

\( \vec{AC} \cdot \vec{BD} = (-1)(2) + (-7)(2) + (4)(4) = -2 - 14 + 16 = 0 \)

Use the dot product formula to find the angle between the vectors:

\( \cos \theta = \frac{\vec{AC} \cdot \vec{BD}}{|\vec{AC}| |\vec{BD}|} = \frac{0}{\sqrt{66} \cdot \sqrt{24}} = 0 \Rightarrow \theta = 90^\circ \)

Hence, the correct answer is Option 1.

Calculation:

We begin by calculating the position vector \(\vec{r}\) from point B to point A:

⇒ \( \vec{r} = \vec{A} - \vec{B} = (1 + 1)\hat{i} + (2 + 2)\hat{j} + (5 - 3)\hat{k} = 2\hat{i} + 4\hat{j} + 2\hat{k} \)

Next, compute the moment vector \(\vec{M}\) using the cross product:

⇒ \( \vec{M} = \vec{r} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 4 & 2 \ 2\lambda & 2 & 5 \end{vmatrix} \)

Expanding the determinant:

⇒ \( \vec{M} = \hat{i}(4 \cdot 5 - 2 \cdot 2) - \hat{j}(2 \cdot 5 - 2 \cdot 2\lambda) + \hat{k}(2 \cdot 2 - 4 \cdot 2\lambda) \)

⇒ \( \vec{M} = \hat{i}(20 - 4) - \hat{j}(10 - 4\lambda) + \hat{k}(4 - 8\lambda) = (16 + 0)\hat{i} - 6\hat{j} + 2\hat{k} \)

Equating components to solve for \(\lambda\):

⇒ \( 16 = 20 + 2\lambda \Rightarrow \lambda = -2 \)

⇒ \( 2\lambda = -2\lambda - 8 \Rightarrow \lambda = -2 \)

Thus, the value of \(\lambda\) is \(-2\)

Hence, the correct answer is Option 1.

Calculation:

The function \( f(x) \) is defined piecewise as:

\( f(x) = \begin{cases} \frac{1 - \cos 2x}{x^2}, & x < 0 \ 9, & x = 0 \ \frac{\sqrt{x}}{\sqrt{(16 + \sqrt{-4})}}, & x > 0 \end{cases} \)

To evaluate the limit \(\lim_{x \to 0} f(x)\), we consider the left-hand and right-hand limits separately.

For \(x \to 0^-\):

⇒ \( \lim_{x \to 0^-} \frac{1 - \cos 2x}{x^2} = \lim_{x \to 0^-} \frac{2 \sin^2 x}{x^2} = 2 \)

Hence, the correct answer is Option 1.

Calculation:

We evaluate the right-hand limit of the function \( f(x) \) as \( x \to 0^+ \):

\( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{(16 + \sqrt{x})} - 4} \)

To simplify, multiply numerator and denominator by the conjugate of the denominator:

\( = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{(16 + \sqrt{x})} + 4)}{(16 + \sqrt{x}) - 16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{(16 + \sqrt{x})} + 4)}{\sqrt{x}} \)

Canceling \(\sqrt{x}\) from numerator and denominator:

\( = \lim_{x \to 0^+} \sqrt{(16 + \sqrt{x})} + 4 \)

As \( x \to 0^+ \), \(\sqrt{x} \to 0\), so:

\( \lim_{x \to 0^+} f(x) = \sqrt{16} + 4 = 4 + 4 = 8 \)

Hence, the correct answer is option 3.

Calculation:

The function \( f(x) = x|x| \) is defined as:

\( f(x) = \begin{cases} x^2, & x \geq 0 \ -x^2, & x < 0 \end{cases} \)

Evaluating the limit as \( x \to -1 \):

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

Hence, the correct answer is Option 1.

Calculation:

Required area:

\( \left| \int_{-2}^{0} x^2 , dx \right| + \int_{0}^{1} x^2 , dx \)

\( = -\left[ \frac{x^3}{3} \right]_{-2}^{0} + \left[ \frac{x^3}{3} \right]_{0}^{1} \)

\( = \left[ 0 + \frac{8}{3} \right] + \left[ \frac{1}{3} - 0 \right] = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3 \)

Hence, the correct answer is Option 4.

Calculation:

\(f(x) = \begin{cases} x^2, & x \ge 0 \ - x^2, & x < 0 \end{cases} \)

It is clear from the graph that \( f(x) \) is increasing in the interval \((-\infty, \infty)\).

\( \therefore \text{L.H.D.} = 0 = \text{R.H.D.} \)

\( \Rightarrow f(x) \text{ is differentiable at } x = 0 \)

Hence both I and II are correct.

Calculation:

Given the function:

\( f(x) = \frac{x}{1 - x} \)

Evaluate \( f(x + 1) \):

⇒ \( f(x + 1) = \frac{x + 1}{1 - (x + 1)} = \frac{x + 1}{-x} \)

Now compute \( \frac{f(x)}{f(x + 1)} \):

⇒ \( \frac{f(x)}{f(x + 1)} = \frac{\frac{x}{1 - x}}{\frac{x + 1}{-x}} = \frac{1 - x}{x + 1} \cdot \frac{x}{-x} \)

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

Hence, the correct answer is option 1.

Calculation:

Given the function:

\( f(x) = \frac{x}{1 - x} \Rightarrow f(\sqrt{x}) = \frac{\sqrt{x}}{1 - \sqrt{x}} \)

Evaluate \( f(\sqrt{x} + 1) \):

⇒ \( f(\sqrt{x} + 1) = \frac{\sqrt{x} + 1}{1 - \sqrt{x} - 1} = \frac{\sqrt{x} + 1}{-\sqrt{x}} \)

Now compute the expression:

⇒ \( (1 - x) \cdot \frac{\sqrt{x}}{1 - \sqrt{x}} - x \cdot \frac{\sqrt{x} + 1}{\sqrt{x}} \)

Simplifying the numerator:

\( = \frac{(1 - x)\sqrt{x}(1 - \sqrt{x}) - x(1 - \sqrt{x})(\sqrt{x} + 1)}{\sqrt{x}(1 - \sqrt{x})} \)

\( = \frac{x - x^2 - x + x^2}{\sqrt{x}(1 - \sqrt{x})} = 0 \)

Hence, the correct answer is Option 4.

Calculation:

Given the function:

\( y = f(x) = x \cdot \frac{\sin^{-1}x}{\sqrt{1 - x^2}} + \ln \sqrt{1 - x^2} \)

Differentiating \( f(x) \) step-by-step:

⇒ \( f'(x) = \left( \sin^{-1}x + \frac{x}{\sqrt{1 - x^2}} \right) \cdot \sqrt{1 - x^2} + x \cdot \sin^{-1}x + \frac{2x}{2\sqrt{1 - x^2}} \)

\( + \frac{1}{\sqrt{1 - x^2}} \cdot \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) \)

Simplifying:

\( = \sqrt{1 - x^2} \cdot \sin^{-1}x + x + \frac{x^2 \cdot \sin^{-1}x}{\sqrt{1 - x^2}} - \frac{x}{1 - x^2} \)

Combining terms:

⇒​ \( f'(x) = \frac{(1 - x^2)\sin^{-1}x + x\sqrt{1 - x^2} + x^2 \sin^{-1}x - x\sqrt{1 - x^2}}{(1 - x^2)\sqrt{1 - x^2}} \)

\( = \frac{\sin^{-1}x}{(1 - x^2)\sqrt{1 - x^2}} \)

Evaluating the derivative at \( x = 0.5 \):

⇒​ \( f'(0.5) = \frac{\sin^{-1}(0.5)}{(1 - 0.25)\sqrt{1 - 0.25}} = \frac{\frac{\pi}{6}}{\frac{3}{4} \cdot \frac{\sqrt{3}}{2}} = \frac{\pi}{6} \cdot \frac{8}{3\sqrt{3}} = \frac{4\pi\sqrt{3}}{27} \)

Thus, the slope of \( y = f(x) \) at \( x = 0.5 \) is \(\frac{4\pi\sqrt{3}}{27}\).

Calculation:

\( \frac{dy}{dx} = \frac{\sin^{-1}x}{(1 - x^2)^{3/2}} \)

⇒ \( \frac{dy}{dx^2} = \frac{\frac{1}{\sqrt{1 - x^2}}(1 - x^2)^{3/2} - \sin^{-1}x \cdot \frac{3}{2}(1 - x^2)^{1/2}(-2x)}{(1 - x^2)^3} \)

\( = \frac{\sqrt{1 - x^2} + 3x\sqrt{1 - x^2} \sin^{-1}x}{(1 - x^2)^3} \)

\( \left( \frac{d^2 y}{dx^2} \right)_{x=0} = \frac{1 + 0}{(1 - 0)^3} = 1 \)

Hence, the correct answer is Option 3.

Calculation:

\( \frac{dy}{dx} = \frac{\sin^{-1}x}{(1 - x^2)^{3/2}} \)

\( \left( \frac{dy}{dx} \right) = \sin \theta = \frac{\sin^{-1}(\sin \theta)}{(1 - \sin^2 \theta)^{3/2}} \)

\( = \frac{\theta}{\cos^3 \theta} = \theta \cdot \sec^3 \theta \)

Hence, the correct answer is Option 3.