Quantitative Aptitude PYQs

Explanation -

Given integral

\(\int \frac{3 \cos x + 4 \sin x}{2 \cos x + 5 \sin x} , dx\)

we use substitution.

Let:  u = 2 cos x + 5 sin x

Then, the derivative du with respect to x is:

du = (-2 sin x + 5 cos x ) dx

Notice that the numerator of the integral (3 cos x + 4 sin x) can be rewritten as a combination of the derivative's terms.

We need to express (3 cos x + 4 sin x) in a form that fits A(5 cos x - 2 sin x) + B (2 cos x + 5 sin x):

Let:

3 cos x + 4 sin x = A (5 cos x - 2 sin x) + B (2 cos x + 5 sin x)

Equate coefficients of (cos x) and (sin x):

3 = 5A + 2B ..... (1)

4 = -2A + 5B .... (2) Solve the system of equations:

Multiplying (1) by 5 and (2) by 2 to align coefficients:

15 = 25A + 10B

8 = -4A + 10B

Subtract the second from the first:

15 - 8 = 25A + 10B - (-4A + 10B)

7 = 29A

A = 7/29

Substitute A = 7/29 back into (1):

3 = 5 x 7/29 + 2B

3 = 35/29 + 2B

2B = 3 - 35/29

2B = 87/29 - 35/29 

2B = 52/29

B = 26/29

Thus, 

\(\frac{3 \cos x + 4 \sin x}{2 \cos x + 5 \sin x} = \frac{\frac{7}{29}(5 \cos x - 2 \sin x) + \frac{26}{29} (2 \cos x + 5 \sin x)}{2 \cos x + 5 \sin x}\)

Therefore:

\(\int \frac{3 \cos x + 4 \sin x}{2 \cos x + 5 \sin x} , dx = \frac{7}{29} \int \frac{5 \cos x - 2 \sin x}{2 \cos x + 5 \sin x} , dx + \frac{26}{29} \int \frac{2 \cos x + 5 \sin x}{2 \cos x + 5 \sin x} , dx \)

The integral becomes:

\(\frac{7}{29} \int \frac{du}{u} + \frac{26}{29} \int dx = \frac{7}{29} ln (u) + \frac{26}{29} x + C\)

Substituting back u = 2 cos x + 5 sin x 

\(\frac{7}{29} ln (2 \cos x + 5 \sin x) + \frac{26}{29} x + C\)

Thus: α = 26, β = 7

Hence the correct option is (4).

Explanation -

Given integral

\(\int \frac{3 \cos x + 4 \sin x}{2 \cos x + 5 \sin x} , dx\)

we use substitution.

Let:  u = 2 cos x + 5 sin x

Then, the derivative du with respect to x is:

du = (-2 sin x + 5 cos x ) dx

Notice that the numerator of the integral (3 cos x + 4 sin x) can be rewritten as a combination of the derivative's terms.

We need to express (3 cos x + 4 sin x) in a form that fits A(5 cos x - 2 sin x) + B (2 cos x + 5 sin x):

Let:

3 cos x + 4 sin x = A (5 cos x - 2 sin x) + B (2 cos x + 5 sin x)

Equate coefficients of (cos x) and (sin x):

3 = 5A + 2B ..... (1)

4 = -2A + 5B .... (2) Solve the system of equations:

Multiplying (1) by 5 and (2) by 2 to align coefficients:

15 = 25A + 10B

8 = -4A + 10B

Subtract the second from the first:

15 - 8 = 25A + 10B - (-4A + 10B)

7 = 29A

A = 7/29

Substitute A = 7/29 back into (1):

3 = 5 x 7/29 + 2B

3 = 35/29 + 2B

2B = 3 - 35/29

2B = 87/29 - 35/29 

2B = 52/29

B = 26/29

Thus, 

\(\frac{3 \cos x + 4 \sin x}{2 \cos x + 5 \sin x} = \frac{\frac{7}{29}(5 \cos x - 2 \sin x) + \frac{26}{29} (2 \cos x + 5 \sin x)}{2 \cos x + 5 \sin x}\)

Therefore:

\(\int \frac{3 \cos x + 4 \sin x}{2 \cos x + 5 \sin x} , dx = \frac{7}{29} \int \frac{5 \cos x - 2 \sin x}{2 \cos x + 5 \sin x} , dx + \frac{26}{29} \int \frac{2 \cos x + 5 \sin x}{2 \cos x + 5 \sin x} , dx \)

The integral becomes:

\(\frac{7}{29} \int \frac{du}{u} + \frac{26}{29} \int dx = \frac{7}{29} ln (u) + \frac{26}{29} x + C\)

Substituting back u = 2 cos x + 5 sin x 

\(\frac{7}{29} ln (2 \cos x + 5 \sin x) + \frac{26}{29} x + C\)

Thus: α = 26, β = 7

Hence the correct option is (1).

Explanation -

Given function:

\(f(x)=\frac{x}{\ln x} ;(x>1)\)

Statement 1: f(x) is increasing in the interval (e, ∞)

To determine whether f(x) is increasing or decreasing, we need to find its derivative and analyze its sign.

Calculate the derivative f'(x) :

\(f'(x) = \left( \frac{x}{\ln x} \right)' = \frac{(x)'(\ln x) - (x)(\ln x)'}{(\ln x)^2} = \frac{\ln x - 1}{(\ln x)^2}\)

We analyze the sign of f'(x) over different intervals:

For x > e , ln x > 1 so ln x - 1 > 0.

\(\Rightarrow f'(x) = \frac{\ln x - 1}{(\ln x)^2} > 0\)

This means f(x) is increasing in the interval (e, ∞).

Statement 2: f(x) is decreasing in the interval (1, e)

For 1 < x < e , 0 < ln x < 1 , so ln x - 1 < 0.

\(\Rightarrow f'(x) = \frac{\ln x - 1}{(\ln x)^2} < 0\)

This means f(x) is decreasing in the interval (1, e).

Statement 3: 9 ln 7 > 7 ln 9

Let's compare 9 ln 7 and 7 ln 9 :

Using the logarithm properties and approximations:

ln 7 ≈ 1.945910

ln 9 ≈ 2.197225

Calculating:

9 ln 7 ≈ 9 × 1.945910 = 17.51319

7 ln 9 ≈ 7 × 2.197225 = 15.380575

Since 17.51319 > 15.380575 :

9 ln 7 > 7 ln 9

Hence all the statements are correct.

So, the correct answer is (4)

Explanation -

Given the function f(x) =\( \frac{x}{\ln x} \) for x > 1,

let's analyze the statements.

Statement 1: f"(e) = \(\frac{1}{e}\)

First, we find the first and second derivatives of f(x) .

Using the quotient rule for f(x) =\( \frac{x}{\ln x} \) 

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

Now, differentiate f'(x) :

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

Using the product rule and chain rule to differentiate:

\(f''(x) = \left[ \frac{d}{dx} (\ln x - 1) \right] (\ln x)^{-2} + (\ln x - 1) \left[ \frac{d}{dx} (\ln x)^{-2} \right]\)

⇒ \(f''(x) = \left( \frac{1}{x} \right) (\ln x)^{-2} + (\ln x - 1) \left( -2 (\ln x)^{-3} \cdot \frac{1}{x} \right)\)

⇒ \(f''(x) = \frac{1}{x (\ln x)^2} - \frac{2 (\ln x - 1)}{x (\ln x)^3}\)

Combine the terms:

\(f''(x) = \frac{1}{x (\ln x)^2} - \frac{2 (\ln x - 1)}{x (\ln x)^3}\)

\(f''(x) = \frac{(\ln x)^3 - 2 (\ln x - 1)}{x (\ln x)^5}\)

Evaluating f''(x) at x = e :

⇒ \(f''(e) = \frac{(\ln e)^3 - 2 (\ln e - 1)}{e (\ln e)^5} = \frac{1 - 2 \cdot 0}{e \cdot 1} = \frac{1}{e}\)

So, f''(e) = 1/e is correct.

Statement 2: f(x) attains a local minimum value at x = e

For the critical points,  f'(x) = 0 :

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

This gives ln x - 1 = 0 , so ln x = 1 , hence x = e .

To confirm it's a minimum, check the second derivative:

f''(e) = 1/e > 0

Since f''(e) > 0 , f(x) has a local minimum at x = e .

So, this statement is correct.

Statement 3: A local minimum value of f(x) is e

Evaluate f(x) at x = e :

\(f(e) = \frac{e}{\ln e} = \frac{e}{1} = e\)

So, the local minimum value is e.

This statement is correct.

Hence Option (4) is correct.

Given -

\(g(x)=x-\frac{1}{x}\) and \(f og(x)=x^3-\frac{1}{x^3}\)

Concept -

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

Explanation -

We have \(f og(x)=x^3-\frac{1}{x^3}\)

⇒ \(f (g(x))=x^3-\frac{1}{x^3}\)

Using the formula, we get -

⇒ \(f (g(x))=(x-\frac{1}{x})^3 +3(x-\frac{1}{x})\)

we have \(g(x)=x-\frac{1}{x}\)

⇒ \(f (x -\frac1x)=(x-\frac{1}{x})^3 +3(x-\frac{1}{x})\)

Replace x - 1/x = y, we get 

⇒ \(f (y)=y^3 + 3y\)

Now f(x) = x³ + 3x

So f(x) - 3x = x3 + 3x - 3x = x³

Now g[f(x) - 3x] = g(x³) = \(x^3-\frac{1}{x^3}\)

Hence Option (1) is correct.

Given -

\(g(x)=x-\frac{1}{x}\) and \(f og(x)=x^3-\frac{1}{x^3}\)

Concept -

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

Explanation -

We have \(f og(x)=x^3-\frac{1}{x^3}\)

⇒ \(f (g(x))=x^3-\frac{1}{x^3}\)

Using the formula, we get -

⇒ \(f (g(x))=(x-\frac{1}{x})^3 +3(x-\frac{1}{x})\)

we have \(g(x)=x-\frac{1}{x}\)

⇒ \(f (x -\frac1x)=(x-\frac{1}{x})^3 +3(x-\frac{1}{x})\)

Replace x - 1/x = y, we get 

⇒ \(f (y)=y^3 + 3y\)

Now f(x) = x3 + 3x

Now differentiate this function w.r.t. x, we get -

⇒ f'(x) = 3x² + 3

⇒ f''(x) = 6x

Hence Option (4) is correct.

Given -

Let f(x) = |x| + 1 and g(x) = [x] - 1, \(h(x)=\frac{f(x)}{g(x)}\)

where [.] is the greatest integer function.

Let \(h(x)=\frac{f(x)}{g(x)}\)

Explanation -

Statement 1: f(x) is differentiable for all x < 0 .

The function f(x) = |x| + 1 can be written as:

\(f(x) = -x + 1 \quad \text{for} ; x < 0\)

The function -x + 1 is differentiable for all x < 0 because it is a linear function.

Therefore, Statement 1 is correct.

Statement 2: g(x) is continuous at x = 0.0001 .

The function g(x) = [x] - 1 :

For \(0 ≤ x < 1 , \) [x] = 0

Therefore, g(x) =[x] - 1 = 0 - 1 = -1

At x = 0.0001 , which is in the interval 0 ≤ x < 1 :

⇒ g(0.0001) = -1

Since the greatest integer function is not continuous at integer points, but 0.0001 is not an integer, g(x) does not have any discontinuity at x = 0.0001 .

Therefore, Statement 2 is correct.

Statement 3: The derivative of g(x) at x = 2.5 is 1.

The function g(x) =[x] - 1 is a piecewise constant function, which means its derivative is zero almost everywhere.

For 2 ≤ x < 3 , [x] = 2

Therefore, g(x) = 2 - 1 = 1

Since g(x) is constant in the interval 2 ≤ x < 3 , its derivative is 0 in this interval.

Thus, Statement 3 is incorrect.

Hence Option (1) is Correct.

Given -

Let f(x) = |x| + 1 and g(x) = [x] - 1, \(h(x)=\frac{f(x)}{g(x)}\)

where [.] is the greatest integer function.

Let \(h(x)=\frac{f(x)}{g(x)}\)

Explanation -

Step 1: Evaluate \(\displaystyle \lim _{x \rightarrow 0^{-}} h(x)\)

For x → 0^- (approaching 0 from the left):

f(x) = |x| + 1 = -x + 1 (since x < 0 )

g(x) = [x] - 1 = -1 - 1 = -2 (since [x] = -1 when -1 < x < 0 )

Thus, 

h(x) = \(\frac{f(x)}{g(x)} = \frac{-x + 1}{-2} = \frac{x - 1}{2}\)

As x → 0^- :

\(lim_{{x → 0^-}} h(x) = \frac{0 - 1}{2} = -\frac{1}{2}\)

Step 2: Evaluate \(\lim_{{x → 0^+}} h(x)\)

For x → 0⁺ (approaching 0 from the right):

f(x) = |x| + 1 = x + 1 (since x > 0 )

g(x) = [x] - 1 = -1 (since [x] = 0 when 0 ≤ x < 1 )

Thus,

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

As x → 0⁺ :

\(\lim_{{x → 0^+}} h(x) = -(0 + 1) = -1\)

Step 3: Sum of the limits

Now, sum the left-hand limit and the right-hand limit:

\(\lim_{{x → 0^-}} h(x) + \lim_{{x → 0^+}} h(x) = -\frac{1}{2} + (-1) = -\frac{1}{2} - 1 = -\frac{3}{2}\)

So, the answer is -3/2.

Explanation -

\(\varphi(a) = \int_{a}^{a + 100π} |\sin x| , dx \)

Understanding the Integral of |sin x|

The sine function, sin x, oscillates between -1 and 1 with a period of 2π.

The absolute value function |sin x| makes the sine function always positive, essentially flipping the negative parts of the sine wave to positive.

The period of |sin x| is π .

Over one period of π, the integral of |sin x| is:

\(\int_{0}^{π} |\sin x| , dx = \int_{0}^{π} \sin x , dx\)  

= \(\left[ -\cos x \right]_{0}^{π}\)

= \(-\cos(π) - (-\cos(0))\)

= -(-1) - (-1) = 1 + 1 = 2

So, \(\int_{0}^{π} |\sin x| , dx = 2\)  

The interval given is from a to a + 100π.

Since |sin x| is periodic with a period of π:

The interval a to a + 100π covers exactly 100 periods of |sin x|.

Thus,

\(\int_{a}^{a + 100π} |\sin x| , dx = 100 \times \int_{0}^{π} |\sin x| , dx = 100 \times 2 = 200 \)

⇒ \(\varphi(a) = 200\)

The correct answer is 200.

Concept -

The Leibniz rule states that if we have an integral of the form:

\(\int_{u(a)}^{v(a)} f(x, a) , dx\)

The derivative with respect to a is given by:

\(\frac{d}{da} \left( \int_{u(a)}^{v(a)} f(x, a) , dx \right) = f(v(a), a) \cdot v'(a) - f(u(a), a) \cdot u'(a) + \int_{u(a)}^{v(a)} \frac{\partial f(x, a)}{\partial a} , dx\)

Explanation -

\(\varphi(a) = \int_{a}^{a+100π} |\sin x| , dx\)

We need to find the derivative of \(\varphi(a)\) with respect to a .

Using the Leibniz Rule for Differentiation Under the Integral Sign -

f(x, a) = |sin x| , which does not explicitly depend on a .

u(a) = a and v(a) = a + 100π .

Thus, the partial derivative \(\frac{\partial f(x, a)}{\partial a}\) is zero.

Applying the Leibniz Rule

\(\varphi'(a) = |\sin(v(a))| \cdot v'(a) - |\sin(u(a))| \cdot u'(a)\)

Where:

v(a) = a + 100π and v'(a) = 1 .

u(a) = a and u'(a) = 1 .

Substituting these into the equation:

⇒ \(\varphi'(a) = |\sin(a + 100π)| \cdot 1 - |\sin(a)| \cdot 1 \)

⇒ \(\varphi'(a) = |\sin(a + 100π)| - |\sin(a)|\)

Simplifying the Sine Function with Periodicity

Since sin(x) has a period of 2π : sin(a + 100π) = sin(a)

Therefore \(\varphi'(a) = |\sin(a)| - |\sin(a)| = 0\)

Explanation -

Given Conditions

f(x) has a local maximum at x = 0  and y = 2f(x) + ax - b .

Understanding the Conditions for a Local Maximum

For f(x) to have a local maximum at x = 0 :

1. The first derivative f'(0) = 0 .

2. The second derivative f''(0) < 0 .

3. Statement 1: f'(0) = 0

Since f(x) has a local maximum at x = 0 , by definition, the first derivative at that point must be zero.

Therefore, Statement 1 is correct. 2. Statement 2: f''(0) < 0

For x = 0 to be a local maximum, the second derivative must be negative at that point, which ensures the concavity of the function is downwards at x = 0 .

Therefore, Statement 2 is correct.

Hence Option (3) is true.

Given -

A differentiable function f(x) has a local maximum at x = 0 and y = f(x) + ax - b .

Concept -

For y to have a relative maximum at x = 0 :

1. The first derivative of y with respect to x should be zero at x = 0 .

2. The second derivative of y with respect to x should be negative at x = 0 .

Explanation -

We have y = 2f(x) + ax - b

y' = 2f'(x) + a

At x = 0 : y'(0) = 2f'(0) + a

Since f(x) has a local maximum at x = 0 , f'(0) = 0 .

Therefore: y'(0) = 2 x 0 + a = a

For y to have a relative maximum at x = 0 ,

we need y'(0) = 0 : a = 0

y'' = 2f''(x)

At x = 0 :  y''(0) = 2f''(0)

Since f(x) has a local maximum at x = 0 , f''(0) < 0 .

Therefore: y''(0) = 2 f''(0)

For y to have a relative maximum at x = 0 , y''(0) must be negative.

Given that f''(0) < 0 :

⇒ 2f''(0) < 0

This condition is satisfied as f''(0) < 0 .

Hence Option (2) is correct.

Given - 

Let f(x) = |x - 1|, g(x) = [x] and h(x) = f(x)g(x) where [.] is greatest integer function.

Concept -

\(f(x) = |x-1| = \begin{cases} -(x-1) & x < 1 \ x -1 & 1\leq x \ \end{cases}\)

Explanation -

Now h(x) = (|x -1|)([x])

For -1 < x < 0, h(x) = -(x - 1) x -1 = (x - 1)

So \(\int_{-1}^0 h(x) d x\)

= \(\int_{-1}^0 (x - 1) d x\)

= \([ \frac{x^2}{2} - x]_{-1}^{0} \)

= - [ 1/2 + 1] = -3/2

Hence Option (1) is correct.

Given - 

Let f(x) = |x - 1|, g(x) = [x] and h(x) = f(x)g(x) where [.] is greatest integer function.

Concept -

\(f(x) = |x-1| = \begin{cases} -(x-1) & x < 1 \ x -1 & 1\leq x \ \end{cases}\)

Explanation -

Now h(x) = (|x -1|)([x])

For 0 < x < 2,

Case - I

For 0 < x < 1 then h(x) = -(x - 1) x 0 = 0

Case - II

For 1 < x < 2 then h(x) = (x - 1) x 1 = x - 1

So \(\int_0^2 h(x) d x\)

= \(\int_{0}^1 0 d x\) + \(\int_1^2 (x - 1) d x\)

= \([ \frac{x^2}{2} - x]_{1}^{2} \)

= - [ 1/2 - 1] = 1/2

Hence Option (4) is correct.

Concept:

\(\int x^n \ dx = \frac{x^{n+1}}{n+1}+C\), where C is the constant of integration.

Calculation:

Given, \(\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}\) = \(\alpha(x+1)^{\frac{3}{2}}+ \beta(x-1)^{\frac{3}{2}}+c\)

Let I = \(\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}\) = \(\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}\times\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)

= \(\frac{1}{2}\int \sqrt{x+1}+\sqrt{x-1} \ dx\)

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

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

= \(\alpha(x+1)^{\frac{3}{2}}+ \beta(x-1)^{\frac{3}{2}}+c\)

∴ The value of α is \(\frac{1}{3}\).

The correct answer is Option 1.

Concept:

\(\int x^n \ dx = \frac{x^{n+1}}{n+1}+C\), where C is the constant of integration.

Calculation:

Given, \(\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}\) = \(\alpha(x+1)^{\frac{3}{2}}+ \beta(x-1)^{\frac{3}{2}}+c\)

Let I = \(\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}\) = \(\int \frac{d x}{\sqrt{x+1}-\sqrt{x-1}}\times\frac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\)

= \(\frac{1}{2}\int \sqrt{x+1}+\sqrt{x-1} \ dx\)

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

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

= \(\alpha(x+1)^{\frac{3}{2}}+ \beta(x-1)^{\frac{3}{2}}+c\)

∴ The value of β is \(\frac{1}{3}\).

The correct answer is Option 3.

Calculation:

Given, x2 + y2 - 2x = 0

⇒ x2 + y2 - 2x + 1 = 1

⇒ (x - 1)2 + y2 = 1, which is a circle with centre at (- 1, 0) and radius = 1

A₁ = ar(ABDEA) 

= ar(ΔABF) + ar(AFBDEA)

= \(\frac{1}{2}\times 1\times 1\) + \(\frac{3}{4}\pi(1)^2\)

= \(\frac{3 \pi+2}{4}\)

∴ The value of A1 is \(\frac{3 \pi+2}{4}\).

The correct answer is Option 4.

Calculation:

Given, x2 + y2 - 2x = 0

⇒ x2 + y2 - 2x + 1 = 1

⇒ (x - 1)2 + y2 = 1, which is a circle with centre at (- 1, 0) and radius = 1

A1 = \(\frac{3 π+2}{4}\)

A₂ = π - \(\frac{3\pi+2}{4}\)

= \(\frac{π-2}{4}\)

∴ \(\frac{2\left(A_1+A_2\right)}{A_1-3 A_2}\)

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

= \(\frac{2\left(\frac{4\pi}{4}\right)}{\left(\frac{2}{4}+\frac{6}{4}\right)}\)

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

= π 

∴ The value of \(\frac{2\left(A_1+A_2\right)}{A_1-3 A_2}\) is π.

The correct answer is Option 1.

Explanation -

Given the equation:

\(3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1\) .....

(i) Substitute x with 1/x :

\(3f\left(\frac{1}{x}\right) + f(x) = x + 1\)......

(ii) We now have two equations:

\(3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1\)

\(3f\left(\frac{1}{x}\right) + f(x) = x + 1\)

Let: f(x) = a and f(1/x) = b

The system of equations becomes:

3a + b = 1/x + 1

3b + a = x + 1

Multiply the first equation by 3:

9a + 3b = 3/x + 3

Subtract the second equation from the modified first equation:

9a + 3b - (3b + a) = 3/x + 3 - (x + 1)

⇒ 8a = 3/x + 3 - x - 1

⇒ 8a = 3/x - x + 2

⇒ a =\( \frac{1}{8} \left(\frac{3}{x} - x + 2\right)\)

Therefore: f(x) = \( \frac{1}{8} \left(\frac{3}{x} - x + 2\right)\)  

Thus, the function f(x) = \(\frac{3}{8x} - \frac{x}{8} + \frac{1}{4}\)

Explanation -

Given the equation:

\(3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1\) .....

(i) Substitute x with 1/x :

\(3f\left(\frac{1}{x}\right) + f(x) = x + 1\)......

(ii) We now have two equations:

\(3f(x) + f\left(\frac{1}{x}\right) = \frac{1}{x} + 1\)

\(3f\left(\frac{1}{x}\right) + f(x) = x + 1\)

Let: f(x) = a and f(1/x) = b

The system of equations becomes:

3a + b = 1/x + 1

3b + a = x + 1

Multiply the first equation by 3:

9a + 3b = 3/x + 3

Subtract the second equation from the modified first equation:

9a + 3b - (3b + a) = 3/x + 3 - (x + 1)

⇒ 8a = 3/x + 3 - x - 1

⇒ 8a = 3/x - x + 2

⇒ a =\( \frac{1}{8} \left(\frac{3}{x} - x + 2\right)\)

Therefore: f(x) = \( \frac{1}{8} \left(\frac{3}{x} - x + 2\right)\)  

Thus, the function f(x) = \(\frac{3}{8x} - \frac{x}{8} + \frac{1}{4}\)

Now  \(8 \int_1^2 f(x) d x\)

= \(8 \int_1^2 [ \frac{3}{8x} - \frac{x}{8} + \frac{1}{4} ]d x\)

= \(8 [ \frac{3}{8}ln (x) - \frac{x^2}{16} + \frac{x}{4} ]_1^2 \)

= \( [ 3ln (x) - \frac{x^2}{2} + 2x ]_1^2 \)

= \( [ 3ln (2) - \frac{4}{2} + 4 ] - [ - \frac{1}{2} + 2]\)

= \( [ 3ln (2) + 2 ] - \frac{3}{2}\)

= \( 3ln (2) + \frac{1}{2}\)

= \( 3ln (2) + \frac{1}{2}ln e\)

= ln 8 + ln√e

= ln 8√e

Hence Option (1) is Correct.