Quantitative Aptitude PYQs

Calculation:

Given,

The function is\(y = \left( 1 - \cos(x) \right)^{-1} \), where \(x \neq 2n\pi \)and n  is an integer.

The cosine function has a range of  [-1, 1] , so cos(x) can take values from ( -1) to ( 1 ).

The expression \(1 - \cos(x) \) will take values from:

\( 1 - 1 = 0 \quad \text{to} \quad 1 - (-1) = 2 \)

So, (1 - cos(x)) takes values in the range \((0, 2] \), but \(x \neq 2n\pi \) excludes the value 0.

Since \(y = \frac{1}{1 - \cos(x)} \), the reciprocal function will take values in the range \(\left[ \frac{1}{2}, \infty \right) \)

∴ The range of the function is \([0.5, \infty) \).

The correct answer is Option (b)

Calculation:

Given,

The function is \(y = \left(1 - \cos(x)\right)^{-1} \), where\(x \neq 2n\pi \) and n is an integer.

The given function can be written as:

\( y = \frac{1}{1 - \cos(x)} \)

Using the identity \(1 - \cos(x) = 2\sin^2\left(\frac{x}{2}\right) \), we rewrite the function as:

\( y = \frac{1}{2\sin^2\left(\frac{x}{2}\right)} \)

The integral we need to evaluate is:

\( \int \frac{dx}{2\sin^2\left(\frac{x}{2}\right)} \)

\( \frac{1}{2} \int \csc^2\left(\frac{x}{2}\right) dx \)

The integral of \(\csc^2(x) \) is known to be \( -\cot(x) \), so we get:

\( \frac{1}{2} \left( -\cot\left(\frac{x}{2}\right) \right) + C \)

\( -\frac{1}{2} \cot\left(\frac{x}{2}\right) + C \)

The final result is \( -\cot\left(\frac{x}{2}\right) + C \)

Hence, the correct answer is Option 2.

Calculation:

Given,

The function is\(f(x) = \sin(\lfloor x \rfloor) \), where\(\lfloor x \rfloor \)is the greatest integer function, and g(x) = |x| , the absolute value function.

We are tasked with finding:

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

For \(g(x) = |x| \), we know that:

\( \lim_{x \to 0} g(x) = 0 \)

For \(f(x) = \sin(\lfloor x \rfloor) \), we know that:

For \(x \to 0^+ \)), \( \lfloor x \rfloor = 0 \), so f(x) = sin(0) = 0 .

For \( x \to 0^- \), \( \lfloor x \rfloor = -1 \), so \(f(x) = \sin(-1) \), which is a nonzero constant.

Evaluating the limit:

For\(x \to 0^+ \),\(f(x)g(x) = 0 \times x = 0 \)

For\(x \to 0^- \), \(f(x)g(x) = \sin(-1) \times (-x) \), which approaches 0 as \(x \to 0^- .\).

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

The correct answer is Option (2).

Calculation:

Given,

The function is \( f(x) = \sin(\lfloor x \rfloor) \), where \(\lfloor x \rfloor \) is the greatest integer function, and g(x) = |x| , the absolute value function.

We are tasked with finding:

\( \lim_{x \to 0} \frac{f(x)}{g(x)} \)

For \( g(x) = |x| \), we know that:

\( \lim_{x \to 0} g(x) = 0 \)

For \( f(x) = \sin(\lfloor x \rfloor) \), we know that:

For \( x \to 0^+ \), \( \lfloor x \rfloor = 0 \), so f(x) = sin(0) = 0 .

For \( x \to 0^- \), \( \lfloor x \rfloor = -1 \), so \( f(x) = \sin(-1) \), which is a nonzero constant.

Evaluating the limit:

For \( x \to 0^+ \), \( \frac{f(x)}{g(x)} = \frac{0}{x} = 0 \)

For \( x \to 0^- \), \( \frac{f(x)}{g(x)} = \frac{\sin(-1)}{-x} \), which becomes undefined as \( x \to 0^- \)because the denominator approaches 0, but the numerator remains a nonzero constant.

∴ Since the left-hand and right-hand limits do not match, the limit does not exist.

The correct answer is Option (4):

Calculation:

Given,

The function is \(f(x) = |x - 3| \), which is an absolute value function.

The domain of an absolute value function \(f(x) = |x - a| \) is all real numbers, because the absolute value function is defined for all values of x. The function handles both positive and negative inputs for x .

Since there are no restrictions or undefined points for the absolute value function, the domain is all real numbers.

∴ The domain of the function is \((-\infty, \infty) \)

Hence, the correct answer is Option 3.

Calculation:

Given,

The function is f(x) = |x - 3| , and we need to find the area bounded by the curve and the line y = 3.

To find the points of intersection, we set the function equal to 3:

\( |x - 3| = 3 \)

Solving for x :

• For\( x \geq 3 \), \(x - 3 = 3 \), which gives x = 6 .

• For ( x < 3 ), 3 - x = 3 , which gives x = 0 .

Therefore, the points of intersection are x = 0  and x = 6 .

The area can be calculated by integrating the difference between the curve and the line from x = 0 to x = 6. The integral is split into two parts due to the absolute value function:

\( A = \int_{0}^{3} (3 - x) , dx + \int_{3}^{6} (x - 3) , dx \)

For x in [0, 3] , ( f(x) = 3 - x ), and for ( x in [3, 6] ), ( f(x) = x - 3 ).

Compute both integrals:

• For x in [0, 3] :

\( \int_{0}^{3} (3 - x) , dx = \left[ 3x - \frac{x^2}{2} \right]_{0}^{3} = 9 - 4.5 = 4.5 \)

• For x in [3, 6] :

\( \int_{3}^{6} (x - 3) , dx = \left[ \frac{x^2}{2} - 3x \right]_{3}^{6} = 4.5 \)

Step 4: The total area is the sum of the two areas:

\( A = 4.5 + 4.5 = 9 , \text{square units} \)

∴ The total area bounded by the curve and the line is 9 square units.

The correct answer is Option (4):

Calculation:

Given,

The function is \(f = {(1, 1), (2, 4), (3, 7), (4, 10)} \), and we are also given that the function is \( f(x) = px + q \), where p and q  are constants.

Using the points f(1) = 1  and f(2) = 4 , we create the following system of equations:

• From f(1) = 1 , we have \(p(1) + q = 1 \), so p + q = 1 .......(1).
• From f(2) = 4 , we have p(2) + q = 4 , so 2p + q = 4  ..........( 2).

Subtract Equation 1 from Equation 2:

\( (2p + q) - (p + q) = 4 - 1 \)

\( p = 3 \)

Now, substitute p = 3 into Equation 1:

\( 3 + q = 1 \)

\( q = -2 \)

\( p + q = 3 + (-2) = 1 \)

∴ The value of p + q  is 1.

The correct answer is Option (3).

Calculation:

Given,

The function is \(f(x)=px+q\), a linear polynomial.

Since \(f(x)=px+q\) is linear with \(p\neq0\), it is one-one (injective): different \(x\) give different \(f(x)\).

The actual outputs given are \({1,4,7,10}\), so the range of \(f\) is \({1,4,7,10}\).

The codomain is the set of natural numbers \(\mathbb{N}\). For \(f\) to be onto, its range must equal its codomain, but here:

\({1,4,7,10}\neq\mathbb{N}\)

∴ f is one-one but not onto.

Hence, the correct answer is Option 1.

Calculation:

Given,

The function is \( f(x) = x^2 - 1 \) and we are tasked with finding \( \lim_{x \to 1} (f \circ f(x)) \).

\( f \circ f(x) = f(f(x)) \)

Given \( f(x) = x^2 - 1 \), we substitute \( f(x) \) into itself:

\( f(f(x)) = f(x^2 - 1) \)

Now, apply the function \( f \) to \( (x^2 - 1) \):

\( f(x^2 - 1) = (x^2 - 1)^2 - 1 \)

This simplifies to:

\( f(x^2 - 1) = x^4 - 2x^2 + 1 - 1 = x^4 - 2x^2 \)

Thus, \( f \circ f(x) = x^4 - 2x^2 \).

Now, we need to find \( \lim_{x \to 1} (x^4 - 2x^2) \).

Substitute x = 1 into the function:

\( f \circ f(1) = 1^4 - 2(1^2) = 1 - 2 = -1 \)

∴ The value of \( \lim_{x \to 1} (f \circ f(x)) \) is -1.

The correct answer is Option (1).

Calculation:

 

 

The function is \( y = x^2 - 1 \), and we need to find the area bounded by the curve and the x-axis between x = -1  and x = 1 .

The required area is given by the definite integral of the function from  -1  to 1 :

\( \text{Area} = \int_{-1}^{1} (x^2 - 1) , dx \)

\( \int (x^2 - 1) , dx = \frac{x^3}{3} - x \)

Evaluate the integral from  -1 to  1 :

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

\( = \left(\frac{1}{3} - 1\right) - \left(\frac{-1}{3} + 1\right) = \left(\frac{1}{3} - \frac{3}{3}\right) - \left(\frac{-1}{3} + \frac{3}{3}\right) = \frac{2}{3} - \frac{2}{3} = \frac{4}{3} \)

∴ The area is \( \frac{4}{3} \) square units.

Hence, the correct answer is Option 3.

Calculation:

Given,

The function is \( y = \sin^{-1} \left( x - \frac{4x^3}{27} \right) \), and we are tasked with simplifying it.

\( y = \sin^{-1} \left( x - \frac{4x^3}{27} \right) \)

We recognize that we can factor the expression as:

\( x - \frac{4x^3}{27} = \left( \frac{x}{3} \right) \left( 3 - 4 \left( \frac{x}{3} \right)^3 \right) \)

Thus, the function becomes:

\( y = \sin^{-1} \left( \frac{x}{3} \left( 3 - 4 \left( \frac{x}{3} \right)^3 \right) \right) \)

Recognizing that this fits a known identity for inverse trigonometric functions:

\( \sin^{-1}(x) = 3 \sin^{-1} \left( \frac{x}{3} \right) \)

Thus, we can simplify the expression to:

\( y = 3 \sin^{-1} \left( \frac{x}{3} \right) \)

Hence, the correct answer is Option 4.

Calculation:

Given,

The function is \( y = 3 \sin^{-1} \left( \frac{x}{3} \right) \), and we are tasked with finding the derivative of y , i.e., \( \frac{dy}{dx} \).

The general derivative of \( \sin^{-1}(u) \) is:

\( \frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \), where\( u = \frac{x}{3} \)

\( \frac{du}{dx} = \frac{1}{3} \)

Apply the chain rule to find the derivative of y :

\( \frac{dy}{dx} = 3 \times \frac{1}{\sqrt{1 - \left( \frac{x}{3} \right)^2}} \times \frac{1}{3} \)

\( \frac{dy}{dx} = \frac{1}{\sqrt{1 - \frac{x^2}{9}}} = \frac{3}{\sqrt{9 - x^2}} \)

Hence, the correct answer is Option 3.

Calculation:

Given,

The function is \( f(x) = \sqrt{x^2 + 9} - 3 \) and \( g(x) = \sqrt{x^2 + 16} - 4 \).

We are tasked with finding:

\( \lim_{x \to 0} \frac{f(x)}{g(x)} \)

Multiply both the numerator and denominator by their respective conjugates:

\( \frac{\sqrt{x^2 + 9} - 3}{\sqrt{x^2 + 16} - 4} \times \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 16} + 4} \)

Simplify the numerator:

\( (\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3) = x^2 \)

Simplify the denominator:

\( (\sqrt{x^2 + 16} - 4)(\sqrt{x^2 + 16} + 4) = x^2 \)

Now, the expression becomes:

\( \frac{x^2}{x^2} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} \)

Simplify and evaluate the limit:

\( \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} \) becomes:

\( \frac{\sqrt{16} + 4}{\sqrt{9} + 3} = \frac{4 + 4}{3 + 3} = \frac{8}{6} = \frac{4}{3} \)

Hence, the correct answer is Option 3.

Calculation:

Given,

The function is \( f(x) = x^2 + 9 \).

 

Statement I: f(x) is an increasing function.

The derivative of f(x) is:

\( f'(x) = \frac{d}{dx}(x^2 + 9) = 2x \)

When \(( x > 0 )\), \(( f'(x) > 0 )\), so (f(x) is increasing.

When (x < 0), f'(x) < 0 ), so f(x)  is decreasing.

At (x = 0), ( f'(x) = 0 ), meaning the function is neither increasing nor decreasing at this point.

Hence, f(x)  is not entirely increasing. It is increasing for (x > 0) and decreasing for ( x < 0).

Statement II: f(x) has local maximum at x = 0

Since the function\( f(x) = x^2 + 9 \) is a parabola opening upwards (because the coefficient of x² is positive), it has a global minimum at x = 0, not a local maximum.

Conclusion:

• Statement I is incorrect because the function is not entirely increasing. It is increasing for x > 0  and decreasing for  x < 0 .
• Statement II is incorrect because the function has a global minimum at x = 0, not a local maximum.

Hence, the correct answer is Option 4.

Calculation:

Given,

The function satisfies the equation \( \frac{f(x)}{f(y)} = \frac{x}{y} \) for all positive real values of x and y, and \( f(2) = 3 \).

We are tasked with finding \( f(16) \).

Using the functional equation, for \( f(4) \), we have:

\( \frac{f(4)}{f(2)} = f\left( \frac{4}{2} \right) = f(2) \)

Since \( f(2) = 3 \), we can calculate \( f(4) \)

\( \frac{f(4)}{3} = 3 \quad \Rightarrow \quad f(4) = 9 \)

Next, to find \( f(16) \), we use the functional equation again:

\( \frac{f(16)}{f(4)} = f\left( \frac{16}{4} \right) = f(4) \)

Since \( f(4) = 9 \), we can calculate \( f(16) \)

\( \frac{f(16)}{9} = 9 \quad \Rightarrow \quad f(16) = 81 \)

Hence, the correct answer is Option 4.

Calculation:

Given,

The function satisfies the functional equation:

\( f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)} \)

Also, we are given that:

\( f(2) = 3 \)

We use the functional equation to calculate f(4) . Substituting x = 4 and y = 2 , we get:

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

\( \frac{f(4)}{3} = 3 \quad \Rightarrow f(4) = 3 \times 3 = 9 \)

\( \frac{f(2)}{f(2)} = f(1) \quad \Rightarrow f(1) = 1 \)

\( f(1) \times f(4) = 1 \times 9 = 9 \)

Hence, the correct answer is Option 3.

Calculation:

Given,

The function is such that: \( f(xy) = f(x + y) \) for all real values of x  and y , and \( f(5) = 10 \).

We need to find the value of \( f(0) \).

Substitute \( x = 5 \) and \( y = 0 \) in the given functional equation \( f(xy) = f(x + y) \):

\( f(5 \cdot 0) = f(5 + 0) \)

\( f(0) = f(5) \)

Since \( f(5) = 10 \), we conclude:

\( f(0) = 10 \)

Hence, the correct answer is Option 4.

Calculation:

Given,

The function is \( f(xy) = f(x + y) \) for all real values of x  and y, and f(5) = 10 .

We are tasked with finding:

\( f(20) + f(-20) \)

Using the given functional equation, we have:

For \( f(0 \cdot 5) = f(0 + 5) \), we get:

\( f(0) = f(5) = 10 \)

For \( f(0 \cdot 20) = f(0 + 20) \), we get:

\( f(0) = f(20) = 10 \)

For \( f(0 \cdot -20) = f(0 + (-20)) \), we get:

\( f(0) = f(-20) = 10 \)

Thus,

\( f(20) + f(-20) = 10 + 10 = 20 \)

Hence, the correct answert is Option 3.

Calculation:

Given,

The function is \( f(x) = \left\lfloor x^2 \right\rfloor \), where \( \left\lfloor x^2 \right\rfloor \) is the greatest integer function.

We are tasked with finding:

\( \int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}} \left\lfloor x^2 \right\rfloor , dx \)

Decomposing the integral based on the function:

For \(\frac{\sqrt{3}}{2} \leq x < 1 \), x² lies between \(\frac{3}{4} \) and 1, so \(\left\lfloor x^2 \right\rfloor = 0 \). Therefore,

\( \int_{\frac{\sqrt{3}}{2}}^1 0 , dx = 0 \)

For\( 1 \leq x < \sqrt{2} \), x²  lies between 1 and 2, so \(\left\lfloor x^2 \right\rfloor = 1 .\) Therefore,

\( \int_1^{\sqrt{2}} 1 , dx = \sqrt{2} - 1 \)

For \(\sqrt{2} \leq x < \sqrt{3} \), x²  lies between 2 and 3, so \(\left\lfloor x^2 \right\rfloor = 2\) . Therefore,

\( \int_{\sqrt{2}}^{\sqrt{3}} 2 , dx = 2(\sqrt{3} - \sqrt{2}) \)

Summing up all the results:

\( \int_{\frac{\sqrt{3}}{2}}^{\sqrt{3}} \left\lfloor x^2 \right\rfloor , dx = 0 + (\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) \)

= \( 2(\sqrt{3} - \sqrt{2}) \).

Hence, the correct answer is Option 2.

Calculation:

Given,

The function is \( f(x) = \lfloor x^2 \rfloor \).

We are tasked with finding the value of \( \int_{\sqrt{2}}^{2} f(x) dx \).

We can break the integral into two parts as follows:

\( \int_{\sqrt{2}}^{2} f(x) dx = \int_{\sqrt{2}}^{\sqrt{3}} 2 dx + \int_{\sqrt{3}}^{2} 3 dx \)

For the range \( \sqrt{2} \leq x \leq \sqrt{3} \), \( \lfloor x^2 \rfloor = 2 \). So the first part of the integral is:

\( \int_{\sqrt{2}}^{\sqrt{3}} 2 dx = 2 \times (\sqrt{3} - \sqrt{2}) \)

For the range \( \sqrt{3} \leq x \leq 2 \), \( \lfloor x^2 \rfloor = 3 \). So the second part of the integral is:

\( \int_{\sqrt{3}}^{2} 3 dx = 3 \times (2 - \sqrt{3}) \)

Now, we calculate the values:

\( 2 \times (\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2} \)

\( 3 \times (2 - \sqrt{3}) = 6 - 3\sqrt{3} \)

Combining the two parts:

\( 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} = 6 - 2\sqrt{2} - \sqrt{3} \)

Hence, the correct answer is Option 1.