Quantitative Aptitude PYQs

Calculation:

Curve x - |y| = 0 and x - 2 = 0

The area of the region bounded by the curve is 

= \(\int_{x=0}^{x=2}\int_{y=-x}^{y=x}dydx\)

= \(\int_{x=0}^{x=2}x-(-x)dx\)

= \(\int_{x=0}^{x=2}2xdx\)

= \(2\int_{x=0}^{x=2}xdx\)

= \(2\times \left [ \frac{x^{2}}{2} \right ]_{x=0}^{x=2}\)

= \(2\times \frac{4}{2}\)

= 4

∴ The correct option is (3)

Concept:

1 + tan²x = sec²x

\(\tan (x+y) = {\tan x + \tan y \over 1 - \tan x \tan y}\)

Calculation:

Given, f(α) = \(\sqrt{\sec^2α−1}\)

⇒ f(α) = \(\sqrt{\tan^2α}\), 

⇒ f(α) = tan α __(i)

⇒ f(β) = tan β  

Put values in \(\frac{f(\alpha)+f(\beta)}{1−f(\alpha) f(\beta)}\),

⇒ \(\frac{f(\alpha)+f(\beta)}{1−f(\alpha) f(\beta)} = {\tan \alpha+\tan\beta \over 1 - \tan \alpha\tan\beta }\)

⇒ \(\frac{f(\alpha)+f(\beta)}{1−f(\alpha) f(\beta)}\) = tan(α + β)

From (i),

⇒ \(\frac{f(\alpha)+f(\beta)}{1−f(\alpha) f(\beta)}\) = f(α + β)

∴ The correct option is (2).

Concept:

• (x + y)(x - y) = x² - y²
• \(\ln { 1\over a} = \ln 1 - \ln a = 0 - \ln a = -\ln a\)

Calculation:

Given: f(x) = ln (x + \(\sqrt{1+\text{x}^2}\)),__(i)

Replace x by (-x) in (i),

⇒  f(-x) = ln (-x + \(\sqrt{1+(\text{-x})^2}\)),

⇒  f(-x) = ln (-x + \(\sqrt{1+\text{x}^2}\)),

Multiply and divide by (x + \(\sqrt{1+\text{x}^2}\)) inside the ln function,

⇒  f(-x) =  \(\ln (-x + \sqrt{1+\text{x}^2}.{ x + \sqrt{1+\text{x}^2}\over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(\ln ({- x^2 + {1+\text{x}^2}\over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(\ln ({1 \over x + \sqrt{1+\text{x}^2}})\),

⇒  f(-x) =  \(-\ln ({ x + \sqrt{1+\text{x}^2}})\),

From (i),

⇒  f(-x) =  - f(x)\

⇒  f(-x) + f(x) = 0

∴ The correct answer is option (1).

Concept:

Limit of Trigonometric Functions:

• To evaluate limits involving trigonometric functions near zero, we use standard limits and trigonometric identities.
• Standard Limit: lim_x→0 (sin x / x) = 1
• Trigonometric Identity: 1 − cos θ = 2 sin²(θ/2)
• These help simplify expressions that are otherwise indeterminate forms like 0/0.
• Use rationalization or substitution to reduce complex expressions into standard limits.

 

Calculation:

Given,

Expression: lim_x→0 x / √(1 − cos 4x)

⇒ 1 − cos 4x = 2 sin²(2x)

⇒ √(1 − cos 4x) = √(2 sin²(2x)) = √2 × sin(2x)

⇒ lim_x→0 x / (√2 × sin(2x))

⇒ 1/√2 × lim_x→0 x / sin(2x)

⇒ 1/√2 × lim_x→0 1 / (2 × (sin(2x)/2x))

⇒ 1/√2 × 1/2 × 1 = 1 / (2√2)

∴ The value of the limit is 1 / (2√2).

Concept:

L's Hospital Rule-

If \(\displaystyle\lim_{x \rightarrow a} f(x) = \displaystyle\lim_{x \rightarrow a} g(x) = 0\)  then,

\(\displaystyle\lim_{x \rightarrow a}{ f(x) \over g(x)} = \displaystyle\lim_{x \rightarrow a}{ f'(x) \over g'(x)}\)

Calculation:

Given \(\displaystyle\lim_{x\rightarrow \frac{\pi}{2}} \frac{4x−2\pi}{\cos x}\)  {0/0 form}

By L's Hospital Rule,

⇒ \(\displaystyle\lim_{x\rightarrow \frac{\pi}{2}} \frac{4x−2\pi}{\cos x} =\displaystyle\lim_{x\rightarrow \frac{\pi}{2}} \frac{(4x−2\pi)'}{(\cos x)'} \)

⇒ \(\displaystyle\lim_{x\rightarrow \frac{\pi}{2}} \frac{4x−2\pi}{\cos x} =\displaystyle\lim_{x\rightarrow \frac{\pi}{2}} \frac{4}{-\sin x} \)

⇒ \(\displaystyle\lim_{x\rightarrow \frac{\pi}{2}} \frac{4x−2\pi}{\cos x} =- \frac{4}{\sin {\pi \over 2}} \) = -4

∴ The correct option is (1).

Concept:

|x| = x when x > 0 and |x| = -x when x < 0.

\(\displaystyle\lim_{x \rightarrow a} f(x) \) exists only if the left-hand limit is equal to the right-hand limit.

that is, \(\displaystyle\lim_{x \rightarrow a^-} f(x) = \displaystyle\lim_{x \rightarrow a^+} f(x)\)

Calculation:

Given, f(x) = \(\frac{x^2+x+|x|}{x}\), 

⇒ f(x) = \(\frac{x^2+x+x}{x}\text { when} \ x > 0\), 

and f(x) = \(\frac{x^2+x-x}{x}\text { when} \ x < 0\), 

⇒ f(x) = x + 2 when x > 0, 

and f(x) = x when x < 0, 

⇒ \(\displaystyle\lim_{x \rightarrow 0^+} f(x) =0 + 2 = 2\) 

and  \(\displaystyle\lim_{x \rightarrow 0^-} f(x) =0 \)

⇒ \(\displaystyle\lim_{x \rightarrow 0^-} f(x) \neq \displaystyle\lim_{x \rightarrow 0^+} f(x)\) 

⇒  \(\displaystyle\lim_{x \rightarrow 0}\) f(x) does not exist

∴ The correct option is (4).

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. \(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\frac{0}{0}\)

II. \(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔ \(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =\lim\limits_{x \to a} \frac{f'(x)}{g'(x)}\)

Note:

We have to differentiate both the numerator and denominator with respect to x unless and until

\(\lim\limits_{x \to a} \frac{f(x)}{g(x)} =l\neq \frac{0}{0}\) where l is a finite value. 

Solution:

\(\displaystyle\lim_{h \rightarrow 0} \frac{\sin^2(x+h)−\sin^2x}{h}\)

Form of limit \(\frac{0}{0}\)

Applying the L-Hospital rule, 

= \(\displaystyle \lim_{h \to 0}\frac{2sin(x+h)cos(x+h)-0}{1}\)

= 2sin(x + 0)cos(x + 0)

= 2sinx.cosx

= sin2x

∴ The correct option is (3)

Explanation:

Given,  f'(x) = g(x) __(i)

and f''(x) = −f(x)  __(ii)

h(x) = {f(x)}2 + {g(x)}2.__(iii)

Differentiating (i) gives, f " (x) = g'(x)

From (ii), ⇒ g'(x) = -f(x) __(iv)

1. h'(3) = 0

Differentiating (iii) with respect to x,

⇒ h'(x) = 2f(x)f '(x) + 2g(x)g'(x)

⇒ h'(x) = 2f(x)g(x) + 2g(x)g'(x) {From (i)}

⇒ h'(x) = 2f(x)g(x) + 2g(x){-f(x)} {From (iv)}

⇒ h'(x) = 0 for every x  __(v)

⇒ h'(3) = 0 

So statement 1 is true.

2. h(1) = h(2)

From (v), h'(x) = 0 for every x

Integrating both sides, 

⇒ h(x) = c {constant function}

⇒ h(1) = h(2)

So statement 2 is true.

∴ The correct option is (3).

Concept:

• Chain rule: \({d \over dx} (f(g(x)) = f'(g(x) )g'(x)\)
• \({ d \over dx} ({p(x) \over q(x)}) = {p'(x)q(x) - p(x) q'(x) \over (q(x))^2}\)

Calculation:

Given y =  ln2\(\left(\frac{x^2−x+1}{x^2+x+1}\right)\),

By Chain Rule,

⇒ \(dy \over dx\) = 2 ln\(\left(\frac{x^2−x+1}{x^2+x+1}\right)\)(d \over dx\)[ln\(\left(\frac{x^2−x+1}{x^2+x+1}\right)\)],

Again applying the chain rule,

⇒ \(dy \over dx\) = 2 ln\(\left(\frac{x^2−x+1}{x^2+x+1}\right)\)( 1\over \left(\frac{x^2−x+1}{x^2+x+1}\right)\)(\left(\frac{x^2−x+1}{x^2+x+1}\right)'\),

⇒ \(dy \over dx\) = 2 ln\(\left(\frac{x^2−x+1}{x^2+x+1}\right)\).\( \left(\frac{x^2+x+1}{x^2-x+1}\right)\)(\left(\frac{(2x - 1)(x^2 + x+ 1) - (x^2 -x + 1) (2x +1)}{(x^2+x+1)^2}\right)\),

Put x = 0,

⇒ \(dy \over dx\) = 2 ln\(\left(\frac{0-0+1}{0+0+1}\right)\).\( \left(\frac{0+0+1}{0-0+1}\right)\)(\left(\frac{(2(0) - 1)(1) - ( 1) (2(0) +1)}{(1)^2}\right)\),

⇒ \(dy \over dx\) = 2 (1)(0) ln 1 = 0

∴ The correct option is (2).

Concept:

• Chain rule: \({d \over dx} (f(g(x)) = f'(g(x) )g'(x)\)
• \({ d \over dx} ({p(x) \over q(x)}) = {p'(x)q(x) - p(x) q'(x) \over (q(x))^2}\)

Calculation:

Given, \(\frac{d}{d x}\left(\frac{1+x^4+x^8}{1−x^2+x^4}\right)\) = ax + bx3,

Solving LHS,

⇒ \(\frac{d}{d x}\left(\frac{1+x^4+x^8}{1−x^2+x^4}\right) = \frac{d}{d x}\left(\frac{(1-x^2 +x^4) + x^2+x^8}{1−x^2+x^4}\right) \)

⇒ \(\frac{d}{d x}\left(\frac{1+x^4+x^8}{1−x^2+x^4}\right) = \frac{d}{d x}\left(1 + \frac{ x^2+x^8}{1−x^2+x^4}\right) \)

⇒ \(\frac{d}{d x}\left(\frac{1+x^4+x^8}{1−x^2+x^4}\right) = \frac{d}{d x}\left(1 + \frac{ x^2((1 - x^2 + x^4) +x^2 - x^4+ x^6)}{1−x^2+x^4}\right) \)

⇒ \(\frac{d}{d x}\left(\frac{1+x^4+x^8}{1−x^2+x^4}\right) = \frac{d}{d x}\left(1 +x^2 + \frac{ x^4 - x^6+ x^8}{1−x^2+x^4}\right) \)

⇒ \(\frac{d}{d x}\left(\frac{1+x^4+x^8}{1−x^2+x^4}\right) = \frac{d}{d x}\left(1 +x^2 + x^4\right) \)

(or it can be solved by performing long division)

⇒ \(\frac{d}{d x}\left(\frac{1+x^4+x^8}{1−x^2+x^4}\right) =\left(0 +2x + 4x^3\right) \)

⇒ \(\frac{d}{d x}\left(\frac{1+x^4+x^8}{1−x^2+x^4}\right) =\left(2x + 4x^3\right) =\left(ax + bx^3\right) \)

⇒ a = 2 and b = 4

⇒ 2a = b

∴ The correct option is (4).

Concept:

• 1 + tan²x = sec²
• 1 + cot²x = cosec²x
• f(x) is minimum at x = a, if f '(a) = 0 and f "(a) > 0
• Chain Rule: (f(g(x))' = f '(g(x)).g'(x)
• (f(x)g(x))' = f '(x).g(x) + f(x).g'(x)

Calculation:

Given: f(x) = (p sec x)2 + (q cosec x)2 

⇒ f(x) = p² sec²x + q² cosec²x

⇒ f(x) = p2 + p2tan2x + q2 + q2cot2x

⇒ f '(x) = 0 + p2(2tan x sec²x) + q2(2cot x (-cosec²x)

⇒ f '(x) = 2p2tan x sec2x - 2q2cot x cosec2​x __(i)

Again differentiating (i),

⇒ f ''(x) = 2p2((tan x)' sec2x + tan x (sec²x)') - 2q2((cot x)' cosec2​x + cot x (cosec²x)')

⇒ f ''(x) = 2p2(sec4x + tan x ( 2sec x tan x sec x)) - 2q2(-cosec4​x + cot x (-2cosec x cosec x cot x))

⇒ f ''(x) = 2p2(sec4x + 2tan²x sec²x) + 2q2(cosec4​x + 2cot²x cosec²x )

⇒ f :"(x) > 0 for every x {∵ every term is of even power}

And from (i),

f ' (x) = 0 when 2p2tan x sec2x - 2q2cot x cosec2​x = 0

⇒ p2tan x sec2x = q2cot x cosec2​x

⇒ \({\tan x \sec^2 x \over \cot x\ \text {cosec}^2 x} = {q^2 \over p^2}\)

⇒ \({\tan^2 x \sin^2 x \over \text {cos}^2 x} = {q^2 \over p^2}\)

⇒ \({\tan^4 x } = {q^2 \over p^2}\)

⇒ tan2 x = \(\frac{q}{p}\)

so when tan2 x = \(\frac{q}{p}\), f "(x) = 0 and f "(x) > 0

⇒ f(x) is minimum when tan2 x = \(\frac{q}{p}\)

∴ The correct option is (1).

Concept:

• 1 + 2 + 3 + ... + n = \(n(n+1) \over 2\)
• 12 + 22 + 32 + ...+ n² = \(n(n+1)(2n+1) \over 6\)
• f(x) is minimum at x = a if f '(a) = 0 and f "(a) > 0

Calculation:

Given, f(x) = \(\displaystyle\sum_{j=1}^7\)(x − j)2 

⇒ f(x) = (x − 1)2 + (x − 2)2 + (x − 3)2 + (x − 4)2 + (x − 5)2 + (x − 6)2 + (x − 7)2 

⇒ f(x) = (x² − 2x + 1²) + (x2 − 4x + 2²) + (x2 − 6x + 3²) + (x2 − 8x + 4²) + (x2 − 10x + 5²) + (x2 − 12x + 6²) +(x2 − 14x + 7²)

⇒ f(x) = 7x² - 2(1 + 2 + 3 + 4 + 5 + 6 + 7)x + (1² + 22 + 32 + 42 + 52 + 62 + 72)

⇒ f(x) = 7x2 - 2(\(7(7+1) \over 2\))x + (\(7(7+1)(2(7)+1) \over 6\))

⇒ f(x) = 7x2 - 2(7(4))x + (\(7(8)(15) \over 6\))

⇒ f(x) = 7x2 - 56x + 140

⇒ f '(x) = 14x - 56 

⇒ f "(x) = 14

So f '(x) = 0 at x = 4 and f "(4) > 0

⇒ f(x) is minimum at x = 4

∴ The correct option is (2).

Explanation:

Plotting the function f(x) , we get,

0 < |x| ≤ 3

Therefore, 

x ∈ [-3 , 0) ∪ (0, 3]

The maximum value of graph is at x = 3 & x = -3 

Hence, statement 1 is wrong.

And by looking at the graph, its minimum value is at x = 0.

As on putting x = 0, f(x) = 1 ⇒the minimum value of function.

Concept:

• \(\ln { a\over b} = \ln a - \ln b\)
• Property of definite integral: \(\displaystyle\int_a^b f(x) dx = \displaystyle\int_a^b f(a+b - x) dx\)

Calculation:

Given, I = \(\displaystyle\int_0^1 \ln \left(\frac{1}{x}−1\right)dx\)

⇒ I = \(\displaystyle\int_0^1 \ln \left(\frac{1-x}{x}\right)dx\) 

⇒ I = \(\displaystyle\int_0^1 [\ln \left(1-x\right)- \ln \left(x\right)]dx\) 

⇒ I = \(\displaystyle\int_0^1 \ln \left(1-x\right)dx- \displaystyle\int_0^1 \ln \left(x\right)dx\) 

In the second term, using the property of definite integral,

⇒ I = \(\displaystyle\int_0^1 \ln \left(1-x\right)dx- \displaystyle\int_0^1 \ln \left(1+0 -x\right)dx\) 

⇒ I = \(\displaystyle\int_0^1 \ln \left(1-x\right)dx- \displaystyle\int_0^1 \ln \left(1-x\right)dx\) 

⇒ I = 0

∴ The correct option is (2).

Concept:

If f(x) is a periodic function with period T, then \(\displaystyle\int_0^{nT}f(x) dx =n \displaystyle\int_0^{T}f(x) dx\)

If f(x) is periodic with period T, then f(ax) is periodic with period \(T \over |a|\)

Calculation:

Let f(x) = (sin4 x + cos4 x)

⇒ f(x) = (sin2x)² + (cos2x)² + 2sin²x cos²x - 2sin²x cos²x

⇒ f(x) = (sin2x + cos2x)2 - 2sin2x cos2x

⇒ f(x) = \(1 - {\sin^2 2x \over 2}\)

⇒ f(x) = \(1 - {1 - \cos 4x \over 4}\)

⇒ f(x) = \({3 \over 4} + { \cos 4x \over 4}\)

Since cos x is a function of period 2π 

⇒ cos 4x is periodic with period \({2\pi \over 4} = {\pi \over 2}\)

⇒ f(x) = (sin4 x + cos4 x) is periodic with period \(\pi \over 2\)

Given,  \(\displaystyle\int_0^{\pi/2}\)(sin4 x + cos4 x)dx = k ___(i)

So, \(\displaystyle\int_0^{20 \pi}\)(sin4 x + cos4 x)dx =  \(\displaystyle\int_0^{40({ \pi\over 2})}\)(sin4 x + cos4 x)dx

As f(x) = (sin4 x + cos4 x) is periodic with period \(\pi \over 2\)

By the property of definite integral,

 \(\displaystyle\int_0^{40({ \pi\over 2})}\)(sin4 x + cos4 x)dx = 40\(\displaystyle\int_0^{\pi/2}\)(sin4 x + cos4 x)dx

⇒  \(\displaystyle\int_0^{20 \pi}\)(sin4 x + cos4 x)dx = 40k { Using (i)}

∴ The correct option is (4).

Concept:

 \(\displaystyle\int_{−a}^{a}\)f(x) dx = 0, if f(x) is an odd function.

Calculation:

Given, I =\(\displaystyle\int_{−\pi/2}^{\pi/2}\)(ecos x sin x + esin x cos x)dx

⇒ I = \(\displaystyle\int_{−\pi/2}^{\pi/2}\)ecos x sin x dx + \(\displaystyle\int_{−\pi/2}^{\pi/2}\)esin x cos x dx

⇒ I = I₁ + I₂

where I₁ =  \(\displaystyle\int_{−\pi/2}^{\pi/2}\)ecos x sin x dx and I₂ =  \(\displaystyle\int_{−\pi/2}^{\pi/2}\)esin x cos x dx

Solving I₁,

Let f(x) = ecos x sin x

⇒ f(-x) = ecos (-x) sin (-x) = - ecos x sin x

⇒ f(-x) = f(x), So f(x) is an odd function.

 \(\displaystyle\int_{−\pi/2}^{\pi/2}\)f(x) dx = 0,

that is, I1 =  \(\displaystyle\int_{−\pi/2}^{\pi/2}\)ecos x sin x dx = 0

So, I =  \(\displaystyle\int_{−\pi/2}^{\pi/2}\)esin x cos x dx

Let sin x = t → cos x dx = dt

when x = -\({\pi \over 2}\), t = -1 and when x = \({\pi \over 2}\), t = 1

⇒  I =  \(\displaystyle\int_{−1}^{1}\)et dt

⇒  I =  et \(|_{-1}^1\) = e - e^-1

⇒  I =  \(\frac{e^2−1}{e}\)

∴ The correct option is (1).

Calculation:

Curves x2 + y2 = π2, y = sin x and x = 0 

The area of the region (in the first quadrant) is

= \(\int_{x=0}^{x=\pi }\int_{y=sinx}^{y=\sqrt{\pi ^{2}-x^{2}}}dydx\)

= \(\int_{x=0}^{x=\pi }\sqrt{\pi ^{2}-x^{2}}-sinxdx\)

= \(\int_{x=0}^{x=\pi }\sqrt{\pi ^{2}-x^{2}}dx-\int_{x=0}^{x=\pi }sinxdx\)

= \(\left [ \frac{x}{2}\sqrt{\pi ^{2}-x^{2}} +\frac{\pi ^{2}}{2}sin^{-1}\frac{x}{\pi }\right ]_{0}^{\pi }-(-cosx)_{0}^{\pi }\)

= \(\left ( 0+\frac{\pi ^{2}}{2}sin^{-1}(1)-0-0\right )+cos(\pi )-cos(0)\)

= \(\frac{\pi ^{2}}{2}.\frac{\pi }{2}-1-1\)

= \(\frac{\pi ^{3}}{4}-2\)

∴ The correct option is (2)

Concept:

The order of the differential equation is the order of the highest derivative in the differential equation.

The degree of the differential equation is the power of the highest order derivative in the differential equation.

The differential equation must be a polynomial equation in derivatives for the degree to be defined.

Calculation:

1. The degree of the differential equation \(\frac{\text{dy}}{\text{dx}} + \cos \left(\frac{\text{dy}}{\text{dx}}\right)\) = 0 is 1.

The given differential equation \(\frac{\text{dy}}{\text{dx}} + \cos \left(\frac{\text{dy}}{\text{dx}}\right)\) = 0 is not a polynomial equation in derivatives.

⇒ The degree of this differential equation is not defined.

So statement 1 is false.

2. The order of the differential equation \(\left(\frac{\text{d}^2\text{y}}{\text{dx}^2}\right)^3 + \cos \left(\frac{\text{dy}}{\text{dx}}\right)\) = 0 is 2.

The highest order of the differential equation \(\left(\frac{\text{d}^2\text{y}}{\text{dx}^2}\right)^3 + \cos \left(\frac{\text{dy}}{\text{dx}}\right)\) = 0 is \(\frac{\text{d}^2\text{y}}{\text{dx}^2}\)

⇒ The order of the differential equation is 2.

So statement 2 is true.

∴ The correct option is (2).

Concept:

The equation of the parabola having a vertex at the (h,k) and axis along the positive y-axis is (x - h)² = 4a(y - k)

Calculation:

The equation of the family of parabolas having vertex at origin and axis along the positive y-axis

⇒ (x - 0)² = 4a(y - 0) 

⇒ x² = 4ay  __(i)

Differentiating both sides with respect to x,

⇒ 2x = 4a\(\frac{\text{dy}}{\text{dx}}\)

⇒ x = 2a \(\frac{\text{dy}}{\text{dx}}\) __(ii)

Putting the value of a from (i) in (ii),

⇒ x = 2\(x^2 \over 4y\) \(\frac{\text{dy}}{\text{dx}}\)

⇒ 2xy = x²\(\frac{\text{dy}}{\text{dx}}\)

⇒ x\(\frac{\text{dy}}{\text{dx}}\) − 2y = 0

 

∴ The correct option is (2).

Concept:

• 1 + cos 2x = 2cos²x
• 1 - cos 2x = 2sin²x
• 1 + tan²x = sec²x
• ∫ sec²ax dx = \(1 \over a\)tan ax + c

Calculation:

Given differential equation (dy − dx) + cos x(dy + dx) = 0

⇒ (1 + cos x) dy = (1 - cos x) dx

⇒ \({ dy \over dx } = { 1- \cos x \over 1 + \cos x}\)

⇒ \({ dy } = { 1- \cos x \over 1 + \cos x}dx\)

⇒ \({ dy } = { 2\sin^2{ x\over 2} \over 2 \cos^2 {x\over 2}}dx\)

⇒ \({ dy } = { \tan^2{ x\over 2}}dx\)

⇒ \({ dy } = { (\sec^2{ x\over 2}-1)}dx\)

Integrating both sides,

⇒ \(\int { dy } = \int{ (\sec^2{ x\over 2}-1)}dx\)

⇒ \(y =2{ \tan { x\over 2}-x + c}\)

∴ The correct option is (3).