Quantitative Aptitude PYQs

Explanation:

Given:

fog(x) =cos 2 √x and go f(x) = |cos x|. 

If we take option (c),

⇒ f(x) = cos² x; then g(x) = √x 

⇒ fog(x) = cos 2 √x and go f(x) = \(\sqrt{cos^2x}\)

= |cos²x|

Then f(x) = cos² x

∴ Option (c) is correct.

Explanation:

Given:

fog(x) =cos 2 √x and go f(x) = |cos x|. 

If we take option (c),

⇒ f(x) = cos² x; then g(x) = √x 

⇒ fog(x) = cos 2 √x and go f(x) = \(√{cos^2x}\)

= |cos²x|

Then f(x) = cos² x

If f(x) = cos² x then g(x) = √ x

∴ Option (a) is correct.

Explanation:

f(x) = [x]² – [x²]

⇒ f(0.999) = [0.999]² – [0.999² ]

= = 0 – [value lies b/w 0 and 1]

= 0 – 0 = 0

⇒ f(1.001) = [1.001]² – [(1.001)² ]

= 1 – 1 = 0

⇒ f(0.999) + f(1.001) = 0 – 0 = 0

∴ Option (b) is correct.

Explanation:

f(x) = [x]² – [x²] 

At x = 0 

LHL = \(\rm \lim_h\rightarrow 0\) [0 – h]² – [(0 – h)²]

=  (–1)² – 0 = 1

RHL = \(\rm \lim_h\rightarrow0 [0 + h]^2 – [(0 + h)^2] \)

=  0 – 0 = 0 

LHL ≠ RHL; f(x) is not continuous at x = 0

At x = 1

LHL = \(\rm \lim_h\rightarrow 0\) [[1 – h] ² – [(1 – h) ²]

 = 1 – 1 = 0

RHL = \(\rm \lim_h\rightarrow 0\) [1 + h] ² – [(1 + h)²] 

= 1 – 1 = 0

⇒ f(x) is continuous at x = 1; LHL = RHL.

∴ Option (b) is correct.

Explanation:

Given:

 f(x) = cos2x + x on [-π/2, π/2]. 

⇒ f'(x) = – 2 sin 2x + 1

For maxima and minima, f'(x) = 0

Sin2x = 1/2

x = π/2 or 5π/12 (not in interval)

So maximum value of f(x) occurs at x = π/12

\(f(\frac{\pi}{12}) = cos\frac{\pi }{6}+ \frac{\pi }{12} \)

= \(\frac{\sqrt3}{2}+\frac{\pi}{12}\)

∴ Option (b) is correct.

Explanation:

f(x) = cos2x + x on [-π/2, π/2]

\(f (\frac{-\pi}{2}) =-1 - \frac{\pi}{2}\)

\(f (\frac{\pi}{2}) =-1 + \frac{\pi}{2}\)

So minimum value of f(x) \(-1 + \frac{\pi}{2}\)

∴ Option (a) is correct

Explanation:

Given:

The area bounded by the parabola y² = kx and the line x = k, where k > 0, is 4/3 square units

Area bounded = 4/3 sq. units

= \(2 \int_0^kydx\)

⇒ 4/3 = \(2 \int_0^k\sqrt k \sqrt xdx\) 

⇒1 = k²

⇒ k = ±1 (since k>0)

⇒ k = 1

∴ Option (b) is correct.

Explanation:

Given:

The area bounded by the parabola y² = kx and the line x = k, where k > 0, is 4/3 square units

Area bounded = 4/3 sq. units

= \(2 \int_0^kydx\)

⇒ 4/3 = \(2 \int_0^k\sqrt k \sqrt xdx\) 

⇒1 = k²

⇒ k = ±1 (since k>0)

⇒ k = 1

So equation of parabola is y² = x

⇒ 4a = 1

⇒ a - 1/4

Area of parabola y² = x and latus rectum at x = 1/4 is

= \(2 \int_0^\frac{1}{4}\sqrt x dx = 2\times \frac{2}{3}\times(\frac{1}{4})^\frac{3}{2}\)

= 1/6 sq. units

∴ Option (a) is correct

Explanation:

Let ydx + (x – y³ ) dy = 0 be a differential equation.

Then, dy/dx = \(\frac{y}{y^3-x}\)

Thus, degree = 1

∴ Option (a) is correct

Explanation:

ydx + (x – y³ )dy = 0

\(y\frac{dy}{dx}+ x = y^3\)

\(\frac{dx}{dy} + x \frac{1}{y} - y^2\)  (linear differential equation)

L.E = \(e^{\int\frac{1}{y}dy}\) = y

x(L.F) = \(\int(LE)y^2dy\)

xy = \(\int y^3dy\)

⇒ 4xy – y⁴ = c

∴ Option (d) is correct

Explanation:

Given:

f(x) =|x² – x – 2|.

= {x² – x – 2; x ∈ (-∞ –1) ∪  (2,∞ ) – (x² – x – 2); x ∈ [–1, 2]

Let I = - \(\int_0^2(x^2 – x – 2)dx\)

= - \([\frac{x^3}{3}-\frac{x^2}{2} -2x]_2^0\)

= \(- [\frac{8}{3}-\frac{4}{2}+4] +0 =\frac{10}{3}\)

∴ Option (d) is correct

Explanation:

Given:

f(x) =|x² – x – 2|

= {x²- x - 2; x ∈ (-∞, -1) ∪ (2, ∞) 

      - (x2 -x -2 ; x ∈[ -1,2]

Let I = \(\int_1^3f(x)dx\)

= \(- \int_1^2(x^2 -x-2)dx+ \int_3^2(x^2-x-2)dx\)

= \(- [\frac{x^3}{3} -\frac{x^2}{2}-2x]^2_1 + [\frac{x^3}{3} -\frac{x^2}{2}-2x]^3_2\)

= \([\frac{8}{3}-\frac{4}{2}-4] + [[\frac{1}{3}-\frac{1}{2}-2] + [\frac{27}{3}-\frac{9}{2}-6] - [\frac{8}{3}-\frac{4}{2}-4] \)

= \(\frac{20}{6} - \frac{13}{6} -\frac{9}{6} +\frac{20}{6}\) = 3

∴ Option (b) is correct.

Explanation:

Let f(t) = \(ln (t+\sqrt{1+t^2}\)

f(-t) = \(ln (\sqrt{1+t^2} -t\)

= \(ln( \frac{1}{t+ \sqrt{1+t^2}})\)

= \(-ln(t+\sqrt{1+t^2})\) = -f(t)

∴  f(t) is an odd function

Now g(t) = tan f(t)

Then, g(–t) = tan f(–t)

= –tan (f(t)) = –g(t)

So g(t) is also an odd function

Hence, both statements I and II are correct.

∴ Option (c) is correct

Explanation:

Let f(t) = \(ln (t+\sqrt{1+t^2}\)

f(-t) = \(ln (\sqrt{1+t^2} -t\)

= \(ln( \frac{1}{t+ \sqrt{1+t^2}})\)

= \(-ln(t+\sqrt{1+t^2})\) = -f(t)

∴  f(t) is an odd function

Now g(t) = tan f(t)

Then, g(–t) = tan f(–t)

= –tan (f(t)) = –g(t)

So g(t) is also an odd function

When  g(t) is an odd function, then

\(\rm \int_{-\pi}^\pi g(t)dt\) = 0

∴ Option (b) is correct

Explanation:

Let f: (–1, 1)→ R be a differentiable function with f(0) = –1 and f'(0) = 1

⇒ h(x) = f(2f(x) + 2)

⇒ h'(x) = f'(2f(x) + 2)2f'(x)

⇒ h'(0) = f'(2f(0) + 2).2f'(0)

= f'(–2 + 2).2(1)

= f'(0).2 = (1).2 = 2

∴ Option (d) is correct.

Explanation:

Given:

f(0) = -1 and f'(0) = 1 Let h(x) = f(2f(x) +2) and g(x) = (h(x))2 

⇒ g(x) = (h(x))²

⇒ g'(x) = 2(h(x)).h'(x)

⇒ g'(0) = 2(h(0)).h'(0)

= 2f(2f (0) + 2).2

= 2f(0).2 = (–2).2 = –4

∴ Option (a) is correct

Concept:

Integration using Substitution:

• This method involves replacing a complex expression with a single variable to simplify integration.
• Useful for trigonometric integrals where identities and substitutions simplify the expression.
• Here, the identity used is:
• \( \sin x + \cos x = \frac{2\tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \)
• \( 1 + \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2} dx = dt \)
• The integral is solved using limits transformation and properties of logarithmic integrals.

 

Calculation:

Given: \( g(x) = \sin x + \cos x + 1 \)

We evaluate \( \int_{0}^{\frac{\pi}{2}} \frac{dx}{g(x)} = \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x + 1} dx \)

⇒ Apply Weierstrass substitution: \( \tan \frac{x}{2} = t \)

⇒ Use identity: \( \sin x + \cos x = \frac{2\tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \)

⇒ Simplify denominator:

\( \sin x + \cos x + 1 = \frac{2\tan \frac{x}{2} + 1 - \tan^2 \frac{x}{2} + 1 + \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} = \frac{2(1 + \tan \frac{x}{2})}{1 + \tan^2 \frac{x}{2}} \)

⇒ \( \int_{0}^{\frac{\pi}{2}} \frac{1}{2(1 + \tan \frac{x}{2})} \cdot \sec^2 \frac{x}{2} dx \)

Let \( 1 + \tan \frac{x}{2} = t \Rightarrow \frac{1}{2} \sec^2 \frac{x}{2} dx = dt \)

⇒ Limits change:

• \( x = 0 \Rightarrow t = 1 \)
• \( x = \frac{\pi}{2} \Rightarrow t = 2 \)

⇒ \( \int_{1}^{2} \frac{1}{t} dt = [\ln t]_{1}^{2} = \ln 2 - \ln 1 = \ln 2 \)

∴ The final answer is \( \ln 2 \)

Concept:

Property of Definite Integrals (Symmetry):

• If \( f(x) \) is continuous on \( [0, a] \), then \( \int_0^a f(x),dx = \int_0^a f(a - x),dx \).
• This property helps in evaluating integrals involving trigonometric identities when limits are symmetric.
• Standard result used: \( \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x + 1} dx = \ln 2 \).
• Combining two symmetrical integrals can simplify expressions and eliminate complex terms.

 

Calculation:

Let \( I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x + 1} dx \)

⇒ Using substitution \( x \rightarrow \frac{\pi}{2} - x \)

⇒ \( I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x + 1} dx \)

⇒ Adding both expressions, we get:

\( 2I = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x + 1} dx \)

⇒ \( 2I = \int_0^{\frac{\pi}{2}} \left(1 - \frac{1}{\sin x + \cos x + 1} \right) dx \)

⇒ \( 2I = \int_0^{\frac{\pi}{2}} 1 dx - \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x + 1} dx \)

⇒ \( 2I = \frac{\pi}{2} - \ln 2 \)

⇒ \( I = \frac{\pi}{4} - \frac{\ln 2}{2} \)

∴ The value of the integral is \( \frac{\pi}{4} - \frac{\ln 2}{2} \).

Explanation:

|U² (x) – V² (x)| = = |x² – (x² + 1)| = |x² – x² – 1|

= |-1| = 1

∴ Option (b) is correct

Explanation:

U(x).V(x) = \(x\sqrt{x^2+1}\)

 = \(\sqrt{x^4+x^2}\)

∴ Option (a) is correct