Quantitative Aptitude PYQs

Calculation:

Let x, y and z represent the number on die-I, die-II and die-III respectively.

Given: x > y > z

It means the smaller number is shown on Die-III (z) than Die-II (y)

 

 

Total number of ways = 4 + 3 + 2 + 1 + 3 + 2 + 1 + 2 + 1 + 1 = 20

∴ Option 4 is correct.

Concept:

• Median can be obtained from ogive.

Hence Option 2 is correct.

Concept:

• \({\rm{M}}.{\rm{D}} = {\rm{;}}\frac{4}{5}{\rm{;S}}.{\rm{D}}\)

Where M.D is mean deviation and S.D is standard deviation

Calculation:

Given: x = M.D., y = S. D

We know that,

\({\rm{M}}.{\rm{D}} = {\rm{;}}\frac{4}{5}{\rm{;S}}.{\rm{D}}\)

⇒ \({\rm{x}} = {\rm{}}\frac{4}{5}{\rm{;y}}\)

∴ x < y

Hence Option 4^th is correct.

Concept:

• \({\left| {\vec a - \vec b} \right|^2} + ;{\left| {\vec a + \vec b} \right|^2} = ;2; \times ;\left( {{{\left| {\vec a} \right|}^2} + ;{{\left| {\vec b} \right|}^2}} \right)\)

 

Calculation:

Given:  \(|\vec a|\) = 3, \(\left| {\vec b} \right| = 4\) and \(\left| {\vec a - \vec b} \right| = 5,\)

We know that,

\({\left| {\vec a - \vec b} \right|^2} + ;{\left| {\vec a + \vec b} \right|^2} = ;2; \times ;\left( {{{\left| {\vec a} \right|}^2} + ;{{\left| {\vec b} \right|}^2}} \right)\)

\(\Rightarrow {5^2} + ;{\left| {\vec a + \vec b} \right|^2} = ;2; \times ;\left( {{3^2} + ;{4^2}} \right)\)

\(\Rightarrow {\left| {\vec a + \vec b} \right|^2} = ;50 - 25 = 25\)

∴ \(\left| {\vec a + \vec b} \right| = 5\)

Concept:

Let \({\rm{\vec a}} = {\rm{x;\vec i}} + {\rm{y;\vec j}} + {\rm{z;\vec k}}\)

• Magnitude of vector of a = \(\left| {{\rm{\vec a}}} \right| = {\rm{;}}\sqrt {{{\rm{x}}^2} + {\rm{;}}{{\rm{y}}^2} + {{\rm{z}}^2}}\)
• Unit vector = \(\hat a = {\rm{;}}\frac{{\vec a}}{{\left| {\vec a} \right|}}\)

Calculation:

Given: \({\rm{\vec a}},{\rm{\vec b;and;\vec c}}\) be three mutually perpendicular vectors each of unit magnitude

Therefore \({\rm{\vec a}} \cdot {\rm{\vec b}} = ;{\rm{\vec b}} \cdot {\rm{\vec c}} = ;{\rm{\vec c}} \cdot {\rm{\vec a}} = 0\) and \(\left| {{\rm{\vec a}}} \right| = \left| {{\rm{\vec b}}} \right| = {\rm{;}}\left| {{\rm{\vec c}}} \right| = 1\)  

\({\rm{\vec A}} = {\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}\)

\(\Rightarrow {\left| {\vec A} \right|^2} = ;{\rm{\vec A}}; \cdot ;{\rm{\vec A}} = ;\left( {{\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}} \right); \cdot \left( {{\rm{\vec a}} + {\rm{\vec b}} + {\rm{\vec c}}} \right)\)

\(\Rightarrow {\left| {\vec A} \right|^2} = ;{\left| {{\rm{\vec a}}} \right|^2} + ;{\left| {{\rm{\vec b}}} \right|^2} + ;{\left| {{\rm{\vec c}}} \right|^2} + 2\left( {{\rm{\vec a}} \cdot {\rm{\vec b}} + ;{\rm{\vec b}} \cdot {\rm{\vec c}} + ;{\rm{\vec c}} \cdot {\rm{\vec a}}} \right)\)

\(\Rightarrow {\left| {\vec A} \right|^2} = 1 + 1 + 1 + 0\)

∴ \(\left| {{\rm{\vec A}}} \right| = \sqrt 3 \)

Now, \({\rm{\vec B}} = {\rm{\vec a}} - {\rm{\vec b}} + {\rm{\vec c}}\)

\(\Rightarrow {\left| {\vec B} \right|^2} = ;{\rm{\vec B}}; \cdot ;{\rm{\vec B}} = ;\left( {{\rm{\vec a}} - {\rm{\vec b}} + {\rm{\vec c}}} \right); \cdot ;\left( {{\rm{\vec a}} - {\rm{\vec b}} + {\rm{\vec c}}} \right)\)

\(\Rightarrow {\left| {\vec B} \right|^2} = ;{\left| {{\rm{\vec a}}} \right|^2} + ;{\left| {{\rm{\vec b}}} \right|^2} + ;{\left| {{\rm{\vec c}}} \right|^2} + 2\left( { - {\rm{\vec a}} \cdot {\rm{\vec b}} - ;{\rm{\vec b}} \cdot {\rm{\vec c}} + ;{\rm{\vec c}} \cdot {\rm{\vec a}}} \right)\)

\(\Rightarrow {\left| {{\rm{\vec B}}} \right|^2} = 1 + 1 + 1 + 0\)

∴ \(\left| {{\rm{\vec B}}} \right| = \sqrt 3\)

Similarly, \({\rm{\vec C}} = {\rm{\vec a}} - {\rm{\vec b}} - {\rm{\vec c}}\)

\(\Rightarrow {\left| {\vec C} \right|^2} = ;{\rm{\vec C}}; \cdot ;{\rm{\vec C}} = ;\left( {{\rm{\vec a}} - {\rm{\vec b}} - {\rm{\vec c}}} \right); \cdot \left( {{\rm{\vec a}} - {\rm{\vec b}} - {\rm{\vec c}}} \right)\)

\(\Rightarrow {\left| {\vec C} \right|^2} = ;{\left| {{\rm{\vec a}}} \right|^2} + ;{\left| {{\rm{\vec b}}} \right|^2} + ;{\left| {{\rm{\vec c}}} \right|^2} + 2\left( { - {\rm{\vec a}} \cdot {\rm{\vec b}} + ;{\rm{\vec b}} \cdot {\rm{\vec c}} - ;{\rm{\vec c}} \cdot {\rm{\vec a}}} \right)\)

\(\Rightarrow {\left| {{\rm{\vec C}}} \right|^2} = 1 + 1 + 1 + 0\)

∴ \(\left| {{\rm{\vec C}}} \right| = \sqrt 3\)

Hence \(\left| {{\rm{\vec A}}} \right| = \left| {{\rm{\vec B}}} \right| = \left| {{\rm{\vec C}}} \right|\)

Concept:

• \(\vec a\) and \(\vec b\) are two vectors parallel to each other ⇔  \(\vec a \times \vec b = 0\)
• Cross product of parallel vectors are zero ⇔  \(\vec a \times \vec a = 0,{\rm{;}}\vec b \times \vec b = 0{\rm{;and;}}\vec c \times \vec c = 0\)
• A cross or vector product is not commutative ⇔ \(\vec a \times \vec b = - {\rm{;}}\vec b \times \vec a\)

Calculation:

We have to find the value of \(\left( {{\rm{\vec a}} - {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} + {\rm{\vec b}}} \right)\)

\(\Rightarrow \left( {{\rm{\vec a}} - {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} + {\rm{\vec b}}} \right) = {\rm{;\vec a;}} \times {\rm{;\vec a}} + {\rm{;\vec a;}} \times {\rm{;\vec b}} - {\rm{;\vec b;}} \times {\rm{;\vec a}} - {\rm{;\vec b;}} \times {\rm{;\vec b}}\)

We know that \({\rm{\vec a;}} \times {\rm{;\vec b}} = - {\rm{;\vec b;}} \times {\rm{;\vec a}}\)

\(\Rightarrow \left( {{\rm{\vec a}} - {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} + {\rm{\vec b}}} \right) = {\rm{;}}0 + {\rm{;\vec a;}} \times {\rm{;\vec b}} + {\rm{;\vec a;}} \times {\rm{;\vec b}} - {\rm{;}}0\)                \(\because \left( {{\rm{\vec a;}} \times {\rm{;\vec a}} = ;{\rm{\vec b;}} \times {\rm{;\vec b}} = 0} \right)\) 

\(\Rightarrow \left( {{\rm{\vec a}} - {\rm{\vec b}}} \right) \times \left( {{\rm{\vec a}} + {\rm{\vec b}}} \right) = {\rm{;}}2;\left( {{\rm{;\vec a;}} \times {\rm{;\vec b}}} \right)\)

∴ Option 3 is correct.

Concept:

Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

From above statement,

\(\overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}} = \overrightarrow {{\rm{AC}}} \)

\( \Rightarrow \overrightarrow {{\rm{AB}}} + \overrightarrow {{\rm{BC}}} = - \overrightarrow {{\rm{CA}}} \)

\(\therefore \overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CA} = \vec 0\)

Only statement

(1) is correct

Concept:

• cos^-1 (cos x) = x
• Slope = tan θ = dy/dx

 

Calculation:

Given: Slope = tan θ

 y = cos^-1 (sin x)

⇒ y = cos^-1 (cos (π/2 -x))

⇒ y = π/2 –x                                        (∵ cos^-1 (cos x) = x)

Differentiating both sides with respect to x, we get

dy/dx = -1

So, Slope = dy/dx

⇒ tan θ  = -1

∴ θ = 3π/4

Calculation:

Given: \({\rm{f}}\left( {\rm{x}} \right) = \frac{{\sqrt {x - 1} }}{{x - 4}}\)

For Domain of function,

Denominator ≠ 0

⇒ x – 4 ≠ 0

∴ x ≠ 4        ….

(1) Now, \(\sqrt {x - 1} \)

⇒ x – 1 ≥ 0

∴ x ≥ 1      …. (2) From equation 1 and 2, we get

⇒ x ∈ [1, 4) ∪ (4, ∞ )

Hence option 4 is correct.

Concept:

• A function f(x) is said to be continuous at a point x = a, in its domain if \(\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right) = f\left( a \right)\) exists or its graph is a single unbroken curve.
• f(x) is Continuous at x = a ⇔ \({\rm{;}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ + }} f\left( x \right){\rm{;}} = {\rm{;}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ - }} f\left( x \right){\rm{;}} = {\rm{;}}\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right)\)

Calculation:

Given:

\(f\left( x \right) = \left{ {\begin{array}{*{20}{c}} {\frac{{sin2x}}{{5x}};;;;;;;;;if;x \ne 0;;;;;;;;;;}\ {\frac{2}{{15}};;;;;if;x = 0} \end{array}} \right.\)

Let’s check continuity at x = 0

\(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = ;\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{5x}}\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{2x}}; \times \frac{2}{5}; = 1; \times ;\frac{2}{5} = ;\frac{2}{5}; \ne ;\frac{2}{{15}}\)

∴ It is not continuous at x = 0

Concept:

• \(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\)
• \(1 - {\rm{;}}\cos 2{\rm{\theta }} = 2{\rm{;}}{\sin ^2}{\rm{\theta }}\)
• \(1 - {\rm{;}}\cos {\rm{\theta }} = 2{\rm{;}}{\sin ^2}\frac{{\rm{\theta }}}{2}\)

Calculation:

We have to find the value of \(\mathop {\lim }\limits_{{\rm{\theta }} \to 0} \frac{{\sqrt {1 - \cos {\rm{\theta }}} }}{{\rm{\theta }}}\) 

\(\Rightarrow \mathop {\lim }\limits_{{\rm{\theta }} \to 0} \frac{{\sqrt {2{\rm{;}}{{\sin }^2}\frac{{\rm{\theta }}}{2}} }}{{\rm{\theta }}} = ;\mathop {\lim }\limits_{{\rm{\theta }} \to 0} \frac{{\sqrt {2{\rm{;}}} {\rm{;}}\sin \frac{{\rm{\theta }}}{2}}}{{\rm{\theta }}}\)

\(\Rightarrow ;\mathop {\lim }\limits_{{\rm{\theta }} \to 0} \frac{{\sqrt {2{\rm{;}}} {\rm{;}}\sin \frac{{\rm{\theta }}}{2}}}{{\frac{{\rm{\theta }}}{2}; \times 2}} = ;\frac{{\sqrt 2 }}{2} = ;\frac{1}{{\sqrt 2 }}\)

Concept:

Greatest Integer Function: Greatest Integer Function [x] indicates an integral part of the real number x which is nearest and smaller integer to x. It is also known as floor of x

• In general, If \(n \le x; \le n + 1\) Then [x] = n     (n ∈ Integer)
• Means if x lies in [n, n+1) then the Greatest Integer Function of x will be n.

Example:

 

• + [-x] = -1  if x ∉ Z
• + [-x] = 0  if x ∈ Z

 

Calculation:

We have to find the value of \(\mathop \smallint \nolimits_n^b \left[ x \right]dx + \mathop \smallint \nolimits_a^b \left[ { - x} \right]dx\)

Let I = \(\mathop \smallint \nolimits_a^b \left[ x \right]dx + \mathop \smallint \nolimits_a^b \left[ { - x} \right]dx = ;\mathop \smallint \nolimits_a^b \left( {\left[ x \right] + ;\left[ { - x} \right]} \right);dx\)

Case I:

If x ∈ Z   then [x] + [-x] = 0 

⇒ I = 0

Case II:

if x ∉ Z   then [x] + [-x] = -1 

⇒ I = \(\mathop \smallint \nolimits_a^b \left( {\left[ x \right] + ;\left[ { - x} \right]} \right);dx = ;\mathop \smallint \nolimits_a^b \left( { - 1} \right);dx = ;\left[ { - x} \right]_a^b\)

⇒ I = a – b

∴ Option 2 and 3 both are correct.

Concept:

• \(\smallint {{\rm{x}}^{\rm{n}}}{\rm{dx}} = {\rm{;}}\frac{{{{\rm{x}}^{\rm{n}+1}}}}{{\rm{n+1}}} + {\rm{c}}\)

 

Calculation:

Let I = \(\smallint {\sin ^3}x\cos x;dx\)

Let sin x = t

Now differentiating both sides, we get

⇒ cos x dx = dt

Now,

\({\rm{I;}} = \smallint {\sin ^3}{\rm{x}}\cos {\rm{x;dx}} = {\rm{;}}\smallint {{\rm{t}}^3}{\rm{dt}} = {\rm{;}}\frac{{{{\rm{t}}^4}}}{4} + {\rm{c}}\)

\(\Rightarrow {\rm{I;}} = {\rm{;}}\frac{{{{\sin }^4}{\rm{x}}}}{4} + {\rm{c}} = {\rm{;}}\frac{{{{\left( {1 - {{\cos }^2}{\rm{x}}} \right)}^2}}}{4} + {\rm{c;}}\)

∴ Option 4 is correct answer

Concept:

• If f′(x) > 0 then the function is said to be increasing.
• If f′(x) < 0 then the function is said to be decreasing.

Calculation:

Given:

f(x) = x² – 5x + 6

Now differentiating both sides, we get

⇒ f’(x) = 2x – 5

For decreasing function, f’(x) < 0

⇒ 2x – 5 < 0

∴ x < 5/2

x ∈ (-∞ ,2.5)

So option 1^st comes under this intervals.

Hence option 1^st is correct.

Concept:

• Order: The order of differential equation is equal to the number of arbitrary constants in the general equation

Calculation:

Given:

y = a cos x + b sin x + ce^-x + d

Here a, b, c and d are arbitrary constants.

Order = 4     (∵ No. of arbitrary constants = 4)

Concept:

• \(\smallint {{\rm{e}}^{{\rm{ax}}}}{\rm{dx}} = {\rm{;}}\frac{{{{\rm{e}}^{{\rm{ax}}}}}}{{\rm{a}}} + {\rm{c}}\)

Calculation:

Given differential equation is

\(\log \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) = {\rm{ax}} + {\rm{by}}\)

\( \Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = ;{e^{\left( {{\rm{ax}} + {\rm{by}}} \right)}}\)

\(\Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = ;{e^{ax}}{e^{by}}\)

\(\Rightarrow \frac{{{\rm{dy}}}}{{{e^{by}}}} = ;{e^{ax}}{\rm{dx}}\)

Integrating both sides, we get

\(\Rightarrow \smallint {e^{ - by}}dy = ;\smallint {e^{ax}}{\rm{dx;}}\)

\( \Rightarrow {\rm{;}}\frac{{{{\rm{e}}^{ - {\rm{by}}}}}}{{ - {\rm{b}}}} = {\rm{;}}\frac{{{{\rm{e}}^{{\rm{ax}}}}}}{{\rm{a}}} + {\rm{c}}\)

\(\Rightarrow \frac{1}{{\rm{a}}}{{\rm{e}}^{{\rm{ax}}}} + \frac{1}{{\rm{b}}}{{\rm{e}}^{ - {\rm{by}}}} = - {\rm{c}}\)

\(\Rightarrow \frac{1}{{\rm{a}}}{{\rm{e}}^{{\rm{ax}}}} + \frac{1}{{\rm{b}}}{{\rm{e}}^{ - {\rm{by}}}} = {\rm{c}}\)       (∵ -c = c)

Hence option 4 is correct.

Concept:

Suppose that we have two functions f(x) and g(x) and they are both differentiable.

• Chain Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right)} \right] = {\rm{;f'}}\left( {{\rm{g}}\left( {\rm{x}} \right)} \right){\rm{g'}}\left( {\rm{x}} \right)\)
• Product Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {{\rm{f}}\left( x \right);{\rm{g}}\left( {\rm{x}} \right)} \right] = {\rm{;}}{{\rm{f}}^{{\rm{'}}\left( {\rm{x}} \right)}}{\rm{g}}\left( {\rm{x}} \right) + {\rm{f}}\left( x \right){\rm{;}}{{\rm{g}}^{{\rm{'}}\left( {\rm{x}} \right)}}\)
• Quotient Rule: \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left[ {\frac{{{\rm{f}}\left( x \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}}} \right] = \frac{{{\rm{g}}\left( {\rm{x}} \right){\rm{;f'}}\left( {\rm{x}} \right) - {\rm{f}}\left( x \right){\rm{;g'}}\left( {\rm{x}} \right)}}{{{{\left[ {{\rm{g}}\left( {\rm{x}} \right)} \right]}^2}}}\)

Calculation:

Given:

y = sin (ln x)    ….

(1) Now differentiating both sides, we get

\(\Rightarrow {\rm{;}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{;}}\cos \left( {\ln {\rm{x}}} \right){\rm{;}} \times {\rm{;}}\frac{1}{{\rm{x}}}\)

\(\Rightarrow {\rm{;x}}\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) = {\rm{;}}\cos \left( {\ln {\rm{x}}} \right){\rm{;}}\)

Again differentiating both sides, we get

\(\Rightarrow \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) + {\rm{x;}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{}} - \sin \left( {\ln {\rm{x}}} \right) \times {\rm{;}}\frac{1}{{\rm{x}}}\)

From equation 1^st,

\(\Rightarrow \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) + {\rm{x;}}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{}} - {\rm{y}} \times {\rm{;}}\frac{1}{{\rm{x}}}\)

\(\Rightarrow {{\rm{x}}^2}\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} + {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}} = 0\)

Hence Option 3 is correct.

Calculation:

Given:

y = p cos (ax) + q sin (ax)       ….

(1) Now differentiating both sides, we get

\(\Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{}} - {\rm{pa;}}\sin \left( {{\rm{ax}}} \right) + {\rm{qa;}}\cos \left( {{\rm{ax}}} \right)\)

Again differentiating both sides, we get

\(\Rightarrow \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{}} - {\rm{p}}{{\rm{a}}^2}\cos \left( {{\rm{ax}}} \right) - {\rm{q}}{{\rm{a}}^2}{\rm{;}}\sin \left( {{\rm{ax}}} \right) = - {{\rm{a}}^2}\left( {{\rm{p;cos;}}\left( {{\rm{ax}}} \right){\rm{}} + {\rm{;q;sin;}}\left( {{\rm{ax}}} \right)} \right)\)

From equation 1^st,

\(\Rightarrow \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = {\rm{}} - {{\rm{a}}^2}{\rm{y;}}\)

\(\therefore \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} + {\rm{;}}{{\rm{a}}^2}{\rm{y}}\)

Calculation:

Given:

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

\(\Rightarrow \smallint {\rm{dy}} = {\rm{}}\smallint \left( { - {{\rm{x}}^2} - \frac{1}{{{{\rm{x}}^3}}}} \right){\rm{dx}}\)

\(\Rightarrow \smallint {\rm{dy}} = {\rm{}}\smallint \left( { - {{\rm{x}}^2} - {{\rm{x}}^{ - 3}}} \right){\rm{dx}}\)

\(\Rightarrow y = {\rm{;}}\frac{{ - {{\rm{x}}^3}}}{3} - ;\frac{{{{\rm{x}}^{ - 2}}}}{{ - 2}} + c\)

\(\Rightarrow y = {\rm{;}}\frac{{ - {{\rm{x}}^3}}}{3} + ;\frac{{{{\rm{x}}^{ - 2}}}}{2} + c\)    ….

(1) Given: curve passing through the point (-1, -2) which satisfies the above equation

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

∴ c = -17/6

Put the value of c in equation 1^st,

\(\Rightarrow y = {\rm{;}}\frac{{ - {{\rm{x}}^3}}}{3} + ;\frac{{{{\rm{x}}^{ - 2}}}}{2} - ;\frac{{17}}{6}\)

⇒ 6x²y + 17x² + 2x⁵ – 3 = 0

Hence Option 2 is correct.

Concept:

• If f′(x) > 0 then the function is said to be increasing.
• If f′(x) < 0 then the function is said to be decreasing.

Calculation:

Given:

f(x) = x sin x + cos x + \(\frac{1}{2}\) cos2 x

Differentiating both sides, we get

⇒ f’(x) = x cos x + sin x – sin x + ½ × (2 cos x ) × - sin x

⇒ f’(x) = x cos x – cos x sin x = cos x (x – sin x)

For critical points:

cos x = 0

∴ x = π /2

And (x – sin x) = 0

∴ x = 0

f’(x) > 0 in x ∈ \(\left( {0,\frac{\pi }{2}} \right)\)

Therefore f(x) is increasing in \(\left( {0,\frac{\pi }{2}} \right)\)