Quantitative Aptitude PYQs

Concept:

The sides of an equilateral triangle are equal and each angle is equal to 60°

The equation of the line passing through (a,b) and subtending angle θ in positive direction of x axis is (y - b) = tan θ (x - a) where tan θ is the slope of the line. 

Length of latus rectum for parabola x² = 4ay is 4a

Calculation:

Given an equilateral triangle is inscribed in a parabola x2 = \(\sqrt{3}\)y where one vertex of the triangle is at the vertex of the parabola.

For parabola x2 = \(\sqrt{3}\)y, vertex is origin O(0,0)

Let the triangle inscribed be AOB.

The line OB of the triangle passes through origin and subtends an angle of (90 - 30) = 60° with positive x axis.

⇒ Equation of OB is (y - 0) = tan 60° (x - 0)

⇒ y = \(\sqrt{3}\) x,  __(i)

For finding point B (the intersection parabola and line OB), put (i) in parabola,

⇒ x² = \(\sqrt{3}\).\(\sqrt{3}\)x

⇒ x² - 3x = 0 (x ≠ 0)

⇒ x = 3 and y = 3\(\sqrt{3}\)

So coordinates of B = (3,3\(\sqrt{3}\))

The length of the side of the triangle = p = |OB|

⇒ p = \(\sqrt{(3-0)^2 +(3\sqrt 3-0)^2} = \sqrt{9+27 } \)

⇒ p = 6  __(ii)

Now length of latus rectum for parabola  x2 = \(\sqrt{3}\)y (a = \(\sqrt3 \over 4\)) is 

q = 4.\(\sqrt3 \over 4\)

⇒ q = \(\sqrt{3}\) __(iii)

From (ii) and (iii),

p = 2\(\sqrt{3}\)q

∴ The correct option is (3).

Calculation:

Statement 1: The points are the vertices of a rectangle ABCD.

The points A(2, 4, 6), B(−2, −4, −2), C(4, 6, 4), and D(8, 14, 12).

\(AB=\sqrt{(-2-2)^{2}+(-4-4)^{2}+(-2-6)^{2}}\)

\(AB=\sqrt{16+64+64}\)

\(AB=\sqrt{144}\)

AB = 12

\(BC=\sqrt{(4+2)^{2}+(6+4)^{2}+(4+2)^{2}}\)

\(BC=\sqrt{36+100+36}\)

\(BC=\sqrt{172}\)

\(CD=\sqrt{(8-4)^{2}+(14-6)^{2}+(12-4)^{2}}\)

\(CD=\sqrt{16+64+64}\)

\(CD=\sqrt{144}\)

CD = 12

\(DA=\sqrt{(2-8)^{2}(4-14)^{2}+(6-12)^{2}}\)

\(DA=\sqrt{36+100+36}\)

\(DA=\sqrt{172}\)

Hence, AB = CD and BC = DA

So, it could be a parallelogram or a rectangle. To confirm this, we will find the dot product of the vector \(\vec {AB} \ and \ \vec {AD}\) .

Refering to the diagram,

\(\rm \vec{AB}=(-2\hat i-4\hat j-2\hat k)-(2\hat i+4\hat j+6\hat k)\)

\(\rm=(-4\hat i-8\hat j-8\hat k)\)

\(\rm \vec{AD}=(8\hat i+14\hat j+12\hat k)-(2\hat i+4\hat j+6\hat k)\)

\(\Rightarrow\rm \vec{AD}=6\hat i+10\hat j+6\hat k\)

Now,

\(\rm \vec{AB}.\vec{AD}=24-80-48=-152\ne0\)

So, AB & CD are not perpendicular

So, ABCD is a parallelogram.

Therefore, The points are not the vertices of a rectangle ABCD. It is forming the parallelogram.

Hence, statement 1 is incorrect.

Statement 2: The mid-point of A C is the same as that of BD.

The mid-point of AC = \(\left ( \frac{2+4}{2},\frac{4+6}{2},\frac{6+4}{2}\right )\)

= (3, 5, 5)

The mid-point of BD = \(\left ( \frac{-2+8}{2},\frac{-4+14}{2},\frac{-2+12}{2}\right )\)

= (3, 5, 5)

Hence, The mid-point of A C is the same as that of BD.

So, the statement 2 is correct.

∴ The correct option is (2)

Concept:

Equation of Tangent plane to the sphere at a point P(\(\alpha,\beta,\gamma\) ) is given by:

x\(\alpha\) + y\(\beta\) + z\(\gamma\) + u(x + \(\alpha\) ) + v( y + \(\beta\) ) + w(z + \(\gamma\) ) + d = 0.............(equation 1)

When the equation of sphere is given by:

x² + y² + z² + 2ux + 2vy + 2wz + d = 0...............(equation 2)

Calculation:

Given equation of sphere is  x2 + y2 + z2 − 4x − 6y − 8z − 16 = 0

(1).Lets find out the tangent of the given sphere.

For that, Comparing equation 2 i.e.

x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0

with the given equation  i.e. x2 + y2 + z2 − 4x − 6y − 8z − 16 = 0.

We get, 2ux = -4x \(\Rightarrow\) u = -2

2vy = -6y ⇒ v = -3

2wz = -8z ⇒ w = -4

and d = -16

Now, putting the values of u, v, w and d in equation 1,we get,

x\(\alpha\) + y\(\beta\) + z\(\gamma\) - 2(x + \(\alpha\) ) - 3 ( y + \(\beta\) ) - 4(z + \(\gamma\) ) -16= 0,which is clearly not an equation of z-axis.

(2).To find the equation of the plane passing through the centre of the sphere, we firstly need to find the centre of the sphere.

Given equation of the sphere is:

 x2 + y2 + z2 − 4x − 6y − 8z − 16 = 0.

⇒x²- 4x + y² - 6y + z² - 8z = 16

⇒(x² - 2(2)x + 2² )+ (y2 - 2(3)y + 3² )+  (z2 - 2(4)z + 4² ) = 16 + 2² + 3² + 4²

(Squaring 2, 3 and 4 and adding both the sides)

⇒(x - 2)² + (y - 3)² + (z - 4)² = 45 

⇒The centre of the sphere is at (2, 3, 4)

i.e. x = 2, y = 3 and z = 4

Since, the question says the plane to be x + y + z − 9 = 0.

Substituting the values of x, y and z, we get

2 + 3 + 4 - 9 = 0

⇒ 0 = 0,

which is true for the given values of centre(2, 3, 4) i.e. The centre of the sphere lies on the plane x + y + z − 9 = 0.

Concept:

If <a,b,c> are the direction ratios of a line, then direction cosines are given by \(<{a \over \sqrt{a^2 +b^2 +c^2}},{b \over \sqrt{a^2 +b^2 +c^2}},{c \over \sqrt{a^2 +b^2 +c^2}}>\)

If ax + by + cz + d = 0 is the equation of a plane then <a, b, c> are the direction rotios of the normal.

Intercept form:

If a plane cuts intercept a, b, c on the coordinate axis then the equation of the plane is \({x \over a} +{y \over b} + {z \over c} =1\)

Calculation:

Given, a plane cuts intercepts 2, 2, 1 on the coordinate axes.

⇒ The equation of plane is \({x \over 2} +{y \over 2} + {z \over 1} =1\)

⇒ The equation of plane is \({x + y + 2z \over 2} =1\)

⇒ The equation of plane is x + y + 2z = 2

⇒ The direction ratios of the normal to the plane = <1, 1, 2>

⇒ The direction cosines of the normal to the plane = \(<{1 \over \sqrt{1^2 +1^2 +2^2}},{1 \over \sqrt{1^2 +1^2 +2^2}},{2 \over \sqrt{1^2 +1^2 +2^2}}>\)

⇒ The direction cosines of the normal to the plane = \(\left\langle\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right\rangle\)

∴ The correct option is (3).

Concept:

• The direction cosines of x-axis, y-axis and z-axis are <1,0,0>, <0,1,0> and <0,0,1> respectively.
• Direction ratios of a line = k(Direction cosines of a line), for any k ∈ R
• If two perpendicular lines have direction ratios <a,b,c> and <p,q,r> then ap + bq + cr = 0

Calculation:

1. The direction ratios of y-axis can be <0, 4, 0>

The direction cosines of y- axis = <0,1,0>

⇒ The direction ratios of y-axis are of the form k<0,1,0> for any k ∈ R

If k = 4, the direction ratios of y-axis are 4<0,1,0> = <0,4,0>

So,the direction ratios of y-axis can be <0, 4, 0>

Statement 1 is true.

2. The direction ratios of a line perpendicular to z-axis can be <5, 6, 0>

The direction cosines of z-axis = <0,0,1>

⇒ The direction ratios of z-axis = k<0,0,1> = <0,0,k>

A line with direction ratios <5, 6, 0> is perpendicular to z-axis only if <0,0,k>.<5,6,0> = 0

 ⇒ 0(5) + 0(6) + k(0) = 0 

The direction ratios of a line perpendicular to z-axis can be <5, 6, 0>

Statement 2 is true.

∴ The correct option is (3).

Concept:

The diagonals of the parallelogram bisect each other.

In a triangle PQR,  ​\(\overrightarrow{\text{PQ}} + \overrightarrow{\text{QR}}+ \overrightarrow{\text{RP}}=0\) 

For any vector \(\overrightarrow{\text{AB}}, \ \ \overrightarrow{\text{AB}}= -\overrightarrow{\text{BA}}\)

Calculation:

Given:  \(\overrightarrow{\text{PR}}=\vec{\text{a}}\) and \(\overrightarrow{\text{QS}}=\vec{\text{b}}\), 

Let the diagonals of the parallelogram intersect each other at O.

⇒ \(\overrightarrow{\text{PO}} = ​​\overrightarrow{\text{OR}}\ \text {and} \ \overrightarrow{\text{QO}}= \overrightarrow{\text{OS}}\)  __(1)

(∵ The diagonals of the parallelogram bisect each other.)

As,  \(\overrightarrow{\text{PR}}=\vec{\text{a}}\) and \(\overrightarrow{\text{QS}}=\vec{\text{b}}\), 

⇒  \(\overrightarrow{\text{PO}}+ \overrightarrow{\text{OR}}=\vec{\text{a}}\) and \(\overrightarrow{\text{QO}}+ \overrightarrow{\text{OS}}=\vec{\text{b}}\), __(2)

From (1) and (2),

⇒ \(\overrightarrow{\text{PO}} = ​​\overrightarrow{\text{OR}}\ = { \overrightarrow{\text{a}} \over 2}\ \text{and} \ \overrightarrow{\text{QO}}= \overrightarrow{\text{OS}} ={ \overrightarrow{\text{b}} \over 2}\)  __(3)

Now in triangle PQO,

\(\overrightarrow{\text{PQ}}\) + \(\overrightarrow{\text{QO}}\) + \(\overrightarrow{\text{OP}}\) = 0

⇒ \(\overrightarrow{\text{PQ}} = -\overrightarrow{\text{QO}}- \overrightarrow{\text{OP}}\)

⇒ \(\overrightarrow{\text{PQ}} = -\overrightarrow{\text{QO}}+ \overrightarrow{\text{PO}}\)

Putting the values from equation (3),

⇒ \(\overrightarrow{\text{PQ}} = -{ \overrightarrow{\text{b}} \over 2}+ { \overrightarrow{\text{a}} \over 2}\)

⇒ \(\overrightarrow{\text{PQ}}\) = \(\frac{\vec{\text{a}}−\vec{\text{b}}}{2}\)

∴ The correct option. is (4) .

Concept:

• The angle between two vectors  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) is given by, \(\cos θ = {\vec{\text{a}}.\vec{\text{b}} \over |\vec{\text{a}}||\vec{\text{b}}|}\)
• If \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are perpendicular vectors, then \(\vec{\text{a}}\).\(\vec{\text{b}}\) = 0
• For any two vectors \(\vec{\text{a}}\) and \(\vec{\text{b}}\), \(\vec{\text{a}}.\vec{\text{b}} = \vec{\text{b}}.\vec{\text{a}}\) __(i)

Calculation:

Given:  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are two unit vectors such that \(\vec{\text{a}}+2 \vec{\text{b}}\) and \(5\vec{\text{a}}−4\vec{\text{b}}\) are perpendicular.

As  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are two unit vectors

⇒  \(|\vec{\text{a}}| = |\vec{\text{b}}| = 1\)___(ii)

And  \(\vec{\text{a}}+2 \vec{\text{b}}\) and \(5\vec{\text{a}}−4\vec{\text{b}}\) are perpendicular

⇒ (\(\vec{\text{a}}+2 \vec{\text{b}}\)).(\(5\vec{\text{a}}−4\vec{\text{b}}\)) = 0

⇒ \(5\vec{\text{a}}.\vec{\text{a}}-4 \vec{\text{a}}.\vec{\text{b}} + 10\vec{\text{b}}.\vec{\text{a}}-8\vec{\text{b}}.\vec{\text{b}}\) = 0

⇒ \(5|\vec{\text{a}}|^2-4 \vec{\text{a}}.\vec{\text{b}} + 10\vec{\text{a}}.\vec{\text{b}}-8|\vec{\text{b}}|^2\) = 0 {from (i)}

⇒ \(5(1) + 6\vec{\text{a}}.\vec{\text{b}}-8(1)\) = 0 {from (ii)}

⇒ \(\vec{\text{a}}.\vec{\text{b}}= {1 \over 2}\) __(iii)

Now angle between  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) is given by, 

\(\cos θ = {\vec{\text{a}}.\vec{\text{b}} \over |\vec{\text{a}}||\vec{\text{b}}|}\)

⇒ \(\cos θ = {{1 \over 2} \over 1.1} = {1 \over 2}\) (from (ii) and (iii))

⇒ θ = \(\frac{\pi}{3}\)

∴ The correct option is (3).

Concept:

If \(\vec{\text{a}}\) is unit vector, then |\(\vec{\text{a}}\)| = 1

For any vector \(\vec{\text{a}}\), \(\vec{\text{a}}\).\(\vec{\text{a}}\) = |\(\vec{\text{a}}\)|²

For any vector \(\vec{\text{a}}\), \(\vec{\text{a}}\) × \(\vec{\text{a}}\) = 0

For any two vectors \(\vec{\text{a}}\) and \(\vec{\text{b}}\), \(\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{a}}\)

For any two vectors \(\vec{\text{a}}\) and \(\vec{\text{b}}\), \(\vec{\text{a}}×\vec{\text{b}}=-(\vec{\text{b}}×\vec{\text{a}})\)

If  \(\vec{\text{a}}\), \(\vec{\text{b}}\) and \(\vec{\text{c}}\) are three coplanar vectors, then their scalar triple product is 0

\(\vec{\text{a}}.(\vec{\text{b}} \times \vec{\text{c}})= \vec{\text{b}}.(\vec{\text{c}} \times \vec{\text{a}})= \vec{\text{c}}.(\vec{\text{a}} \times \vec{\text{b}}) = 0\)

If any two vectors in scalar triple product are equal, then scalar triple product is 0.

Calculation:

Given,  \(\vec{\text{a}}\), \(\vec{\text{b}}\) and \(\vec{\text{c}}\) be unit vectors lying on the same plane. 

⇒  \(\vec{\text{a}}\), \(\vec{\text{b}}\) and \(\vec{\text{c}}\) are coplanar vectors. 

that is \(\vec{\text{a}}.(\vec{\text{b}} \times \vec{\text{c}})= \vec{\text{b}}.(\vec{\text{c}} \times \vec{\text{a}})= \vec{\text{c}}.(\vec{\text{a}} \times \vec{\text{b}}) = 0\)

and  |\(\vec{\text{a}}\)| = 1,  |\(\vec{\text{b}}\)| = 1 and  |\(\vec{\text{c}}\)| = 1

Now, \({(3\vec{\text{a}} + 2\vec{\text{b}}) × (5\vec{\text{a}} − 4\vec{\text{c}})}⋅(\vec{\text{b}} + 2\vec{\text{c}})\)

Using distributive property,

\({(3\vec{\text{a}}× 5\vec{\text{a}} + 2\vec{\text{b}} × 5\vec{\text{a}} − 3\vec{\text{a}}×4\vec{\text{c}}- 2\vec{\text{b}}×4\vec{\text{c}})}⋅(\vec{\text{b}} + 2\vec{\text{c}})\)

⇒ \({(15\vec{\text{a}}× \vec{\text{a}} + 10\vec{\text{b}} × \vec{\text{a}} − 12\vec{\text{a}}×\vec{\text{c}}- 8\vec{\text{b}}×\vec{\text{c}})}⋅(\vec{\text{b}} + 2\vec{\text{c}})\)

⇒ \({(15(0) + 10(\vec{\text{b}} × \vec{\text{a}}) − 12(\vec{\text{a}}×\vec{\text{c}})- 8(\vec{\text{b}}×\vec{\text{c}})}⋅(\vec{\text{b}} + 2\vec{\text{c}})\)     (∵  \(\vec{\text{a}}\) × \(\vec{\text{a}}\) = 0)

⇒ \({ 10(\vec{\text{b}} × \vec{\text{a}}) − 12(\vec{\text{a}}×\vec{\text{c}})- 8(\vec{\text{b}}×\vec{\text{c}})}⋅(\vec{\text{b}} + 2\vec{\text{c}})\)  

Again using distributive property,

⇒ \({ 10(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{b}} − 12(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{b}}- 8(\vec{\text{b}}×\vec{\text{c}})⋅\vec{\text{b}}} + { 20(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{c}} − 24(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{c}}- 16(\vec{\text{b}}×\vec{\text{c}})⋅\vec{\text{c}}}\) 

As, if any two vectors in scalar triple product are equal, then scalar triple product is 0.

⇒ \({ 10(0) − 12(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{b}}- 8(0)} + { 20(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{c}} − 24(0)- 16(0)}\) 

⇒ \(20(\vec{\text{b}} × \vec{\text{a}})⋅\vec{\text{c}}− 12(\vec{\text{a}}×\vec{\text{c}})⋅\vec{\text{b}}\) 

As,  \(\vec{\text{a}}\), \(\vec{\text{b}}\) and \(\vec{\text{c}}\) are coplanar vectors. 

⇒ 20(0) - 12(0) = 0

∴ The correct option is (4).

Concept:

• The angle between two vectors  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) is given by, \(\cos θ = {\vec{\text{a}}.\vec{\text{b}} \over |\vec{\text{a}}||\vec{\text{b}}|}\)​
• If \(\vec{\text{a}}\) = a₁\(\hat{\text{i}}\) + a₂\(\hat{\text{j}}\) + a₃\(\hat{\text{k}}\) and  \(\vec{\text{b}}\) = b1\(\hat{\text{i}}\) + b2\(\hat{\text{j}}\) + b3\(\hat{\text{k}}\), then  \(\vec{\text{a}}.\vec{\text{b}} = a_1b_1+a_2b_2+a_3b_3\)

Calculation:

Given: The angle between the vectors 2x2\(\hat{\text{i}}\) + 3x\(\hat{\text{j}}\) + \(\hat{\text{k}}\) and \(\hat{\text{i}}\) −2\(\hat{\text{j}}\) + x2\(\hat{\text{k}}\) is obtuse

The angle between the vectors 2x2\(\hat{\text{i}}\) + 3x\(\hat{\text{j}}\) + \(\hat{\text{k}}\) and \(\hat{\text{i}}\) −2\(\hat{\text{j}}\) + x2\(\hat{\text{k}}\) is given by

\(\cos θ = {(2x^2 \hat{\text{i}}+3x \hat{\text{j}}+\hat{\text{k}}).(\hat{\text{i}}-2 \hat{\text{j}}+x^2\hat{\text{k}}) \over |2x^2 \hat{\text{i}}+3x \hat{\text{j}}+\hat{\text{k}}||\hat{\text{i}}-2 \hat{\text{j}}+x^2\hat{\text{k}}|}\)

⇒ \(\cos θ = {(2x^2 (1) +3x(-2)+1(x^2)) \over \sqrt{(2x^2)^2+(3x)^2+(1)^2}\sqrt{(1)^2+(-2 )^2+(x^2)^2}}\)

⇒ \(\cos θ = {3x^2 -6x \over \sqrt{4x^4+9x^2+1}\sqrt{5+x^4}}\)

Since θ is obtuse, 

⇒ cos θ < 0

⇒ 3x² - 6x < 0

⇒ x(x - 2) < 0

⇒ 0 < x < 2

∴ The correct option is (1).

Concept:

• The angle between two vectors  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) is given by, \(\cos θ = {\vec{\text{a}}.\vec{\text{b}} \over |\vec{\text{a}}||\vec{\text{b}}|}\)​
• If \(\vec{\text{a}}\) = a1\(\hat{\text{i}}\) + a2\(\hat{\text{j}}\) + a3\(\hat{\text{k}}\) and  \(\vec{\text{b}}\) = b1\(\hat{\text{i}}\) + b2\(\hat{\text{j}}\) + b3\(\hat{\text{k}}\), then  \(\vec{\text{a}}.\vec{\text{b}} = a_1b_1+a_2b_2+a_3b_3\)
• If position vectors of A and B are a1\(\hat{\text{i}}\) + a2\(\hat{\text{j}}\) + a3\(\hat{\text{k}}\) and b1\(\hat{\text{i}}\) + b2\(\hat{\text{j}}\) + b3\(\hat{\text{k}}\) respectively, then the vector AB = (b1 - a1)\(\hat{\text{i}}\) + (b2 - a2)\(\hat{\text{j}}\) + (b3 - a3)\(\hat{\text{k}}\)

Calculation:

Given: The position vectors of vertices A, B and C of triangle ABC are respectively \(\hat{\text{j}}+\hat{\text{k}}, 3\hat{\text{i}}+\hat{\text{j}+5\hat{\text{k}}}\) and \(3\hat{\text{j}}+3\hat{\text{k}}\).

⇒ \(\vec{\text {AB}}\) = \((3-0)\hat{\text{i}}+(1-1)\hat{\text{j}+(5 -1)\hat{\text{k}}} = 3\hat{\text{i}}+{4\hat{\text{k}}}\)

 \(\vec{\text {BC}}\) = \((0-3)\hat{\text{i}}+(3-1)\hat{\text{j}+(3-5 )\hat{\text{k}}} = -3\hat{\text{i}}+2\hat{\text{j}}-{2\hat{\text{k}}}\)

 \(\vec{\text {CA}}\) = \((0-0)\hat{\text{i}}+(3-1)\hat{\text{j}+(3 -1)\hat{\text{k}}} = 2\hat{\text{j}}+{2\hat{\text{k}}}\)

The angle between the vectors \(\vec{\text {BC}}\) and \(\vec{\text {CA}}\) is angle C

\(\cos C ={ \vec{\text{BC}}.\vec{\text{CA}} \over |\vec{\text{BC}}||\vec{\text{CA}}|}\)

⇒ \(\cos C ={(-3\hat{\text{i}}+2\hat{\text{j}}-{2\hat{\text{k}}}).(2\hat{\text{j}}+{2\hat{\text{k}}}) \over |-3\hat{\text{i}}+2\hat{\text{j}}-{2\hat{\text{k}}}||2\hat{\text{j}}+{2\hat{\text{k}}}|}\)

⇒ \(\cos C= {-3(0)+2(2)-2(2) \over \sqrt{3^2 +2^2+(-2)^2}\sqrt{2^2 +2^2}}\) = 0

⇒ cos C = 0

⇒ C = \(\frac{\pi}{2}\)

∴ The correct option is (4).

Concept:

• The greatest integer function of a number rounds off the number to the integer less than the number
• Every integer x can be witten as x = [x] + {x}, where [x] is the integer part of x and {x} is the fractional part of x
• 0 ≤ {x} < 1
• If x is an integer, then {x} = 0

Property of greatest integer function:

• [-x] = -[x] , if x ∈ Z
• [-x] = -[x] - 1, if x ∉ Z

Calculation:

Given, z = [y] and y = [x] − x, where [.] is the greatest integer function

 If x is not an integer but positive, 

⇒ x > 0 and x ∉ Z

⇒ y = [x] − x =  [x] − ([x] + {x})

⇒ y = -{x} 

Now given z = [y] 

Putting the value of y,

⇒ z =[-{x}]

⇒ z = -[{x}] - 1   ({x} ∉ Z)

⇒ z = -0 - 1  (∵ 0 ≤ {x} < 1))

⇒ z = -1 

∴ The correct option is (1).

Explanation:

Given: f(x) = 4x + 1 and g(x) = kx + 2 such that fog(x) = gof(x)

fog(x) = f(g(x)) = 4g(x) + 1

⇒ fog(x) = 4(kx + 2) + 1 = 4kx + 9  __-(i)

And gof(x) = g(f(x)) = kf(x) + 2

⇒ gof(x) = k(4x + 1) + 2 = 4kx + k + 2 __(ii)

Since  fog(x) = gof(x), so equating (i) and (ii),

⇒ 4kx + 9 = 4kx + k + 2

⇒ 9 = k + 2 

⇒ k = 7

∴ The correct option is (1).

Concept:

The function f(x) has minimum value at x = a, if f '(a) = 0 and f "(a) > 0

Formula used:

• \(\log_b a = {\ln a \over \ln b}\)
• \({d \over dx} ({ p \over q})= {p'q -pq' \over q^2}\)

Calculation:

Given, f(x) = log10(x2 + 2x + 11) = \({1 \over \ln 10}\) ln (x2 + 2x + 11) 

⇒ f '(x) = \({1 \over \ln 10}\)({1 \over x^2 + 2x + 11}\)(x2 + 2x + 11)'

⇒ f '(x) = \({1 \over \ln 10}\)({2x +2 \over x^2 + 2x + 11}\)__(i)

⇒ f '(x) = 0 when 2x + 2 = 0 → x = -1

So, f '(-1) = 0

Now again differentiating (i),

⇒ f "(x) = \({1 \over \ln 10}\)(\Big({2 (x^2 + 2x + 11) - (2x +2)(2x+2)\over (x^2 + 2x + 11)^2}\Big)\) 

⇒ f "(x) = \({1 \over \ln 10}\)(\Big({2 x^2 + 4x + 22 - 4x^2 -8x -4\over (x^2 + 2x + 11)^2}\Big)\)  

⇒ f "(x) = \({1 \over \ln 10}\)(\Big({-2 x^2 -4 x + 18 \over (x^2 + 2x + 11)^2}\Big)\) 

⇒ f "(-1) = \({1 \over \ln 10}\)(\Big({-2 -4(-1) + 18 \over (1 + 2(-1) + 11)^2}\Big)\) 

⇒ f "(-1) = \({1 \over \ln 10}\)(\times {20 \over 100}\) 

⇒  f "(-1) = \({1 \over 5\ln 10}\)> 0

So, f '(-1) = 0 and f "(-1) > 0

⇒ f(x) has minimum value at x = -1

The minimum value of f(x) = f(-1) = log10((-1)2 + 2(-1) + 11)

⇒ The minimum value of f(x) = log10(10) = 1

∴ The correct option is (2).

Calculation:

Given I = ∫(xx)2 (1 + ln x)dx

⇒ I = ∫x2x (1 + ln x)dx  __(1)

Consider x2x = u 

Taking log both sides, 

⇒ 2x ln x = ln u

Differentiating both sides,

⇒ (2ln x + 2x.\(1 \over x\))dx = \(1 \over u\)du

⇒ (1 + ln x)dx  = \(1 \over 2u\)du  __(2)

Putting the value from (2) to (1)

⇒ I = ∫u.\(1 \over 2u\)du = \(1 \over 2\)∫du

⇒ I = \(1 \over 2\)u + c

⇒ I = \(1 \over 2\)x2x + c

∴ The correct option is (2).

Concept:

Integral of the product of two functions

\(∫ (f.g) \ dx = f(x) ∫ gdx - ∫ (f' ∫ gdx ) \ dx\)

Calculation:

Given, I = ∫ex{1 + ln x + x ln x}dx

⇒ I = ∫ex dx + ∫ e^x ln x dx + ∫ e^x x ln x dx

⇒ I = ex + ∫ ex ln x dx + ∫ ex x ln x dx

Computing the last integral  ∫ ex x ln x dx by taking f(x) = x ln x and g(x) = e^x

⇒ I = ex + ∫ ex ln x dx + [x ln x ∫ ex dx - ∫(x ln x)' ∫ ex dx dx]

⇒ I = ex + ∫ ex ln x dx + x ln x ex - ∫(1 + ln x)ex dx

⇒ I = ex + ∫ ex ln x dx + x ln x ex - ∫ex dx - ∫ex ln x dx

⇒ I = ex + x ln x ex - ex 

⇒ I = x ln x ex + c

∴ The correct option is (1).

Given,

I = \(\int\frac{(\cos x)^{1.5}−(\sin x)^{1.5}}{\sqrt{\sin x⋅\cos x}}dx\)

⇒ I = \(\int\frac{\cos x}{\sqrt{\sin x}}- \frac{\sin x}{\sqrt{\cos x}}dx\)

⇒ I = \(\int\frac{\cos x}{\sqrt{\sin x}}dx- \int \frac{\sin x}{\sqrt{\cos x}}dx\) = I₁ - I₂ (let)

Computing I₁,

I₁ = \(\int\frac{\cos x}{\sqrt{\sin x}}dx\)

Put sin x = t → cos x dx = dt

⇒ I₁ = \(\int\frac{dt}{\sqrt{t}} = {t^{-{1 \over 2}+1} \over{-{1 \over 2}+1}} + c_1\)

⇒ I1 = \(2\sqrt t + c_1\)

⇒ I1 = \(2\sqrt {\sin x} + c_1\)

Similarly, 

I2 = \(\int\frac{\sin x}{\sqrt{\cos x}}dx\)

Put cos x = t → - sin x dx = dt

⇒ I2 = \(\int\frac{dt}{\sqrt{t}} = -{t^{-{1 \over 2}+1} \over{-{1 \over 2}+1}} + c_2\) 

⇒ I2 = \(-2\sqrt t + c_2\) 

⇒ I2 = \(-2\sqrt {\cos x} + c_2\) 

Putting these value in I,

⇒ I = I1 - I2 = \(2\sqrt {\sin x} + c_1\) - \((-2\sqrt {\cos x}) + c_2\)

⇒ I = \(2\sqrt{\sin \text{x}}+2 \sqrt{\cos \text{x}}\) + c

∴ The correct answer is option (3).

Concept:

• \({d \over dx} (p(x) q(x)) = p'(x) q(x) + p(x)q'(x)\)
• \({d \over dx} (f(g(x)) = f'(g(x))g'(x)\)

Calculation:

If y = \(\frac{x \sqrt{x^2−16}}{2} − 8 \ln\left|x + \sqrt{x^2−16}\right|\), 

Differentiating both sides with respect to x,

⇒ \(\frac{\text{dy}}{\text{dx}}\) = \({1 \over2}[{(x)' \sqrt{x^2−16}}+ x(\sqrt{x^2−16})'] − 8 {1 \over \left(x + \sqrt{x^2−16}\right)}.\left(x + \sqrt{x^2−16}\right)'\)

⇒ \(\frac{\text{dy}}{\text{dx}}\) = \({1 \over2}[{ \sqrt{x^2−16}}+{x\over 2(\sqrt{x^2−16})}(x^2−16)'] − 8 {1 \over \left(x + \sqrt{x^2−16}\right)}.(1 + {1\over 2(\sqrt{x^2−16})}(x^2−16)')\)

⇒ \(\frac{\text{dy}}{\text{dx}}\) = \({1 \over2}[{ \sqrt{x^2−16}}+{2x.x\over 2(\sqrt{x^2−16})}] − 8 {1 \over \left(x + \sqrt{x^2−16}\right)}.(1 + {2x\over 2(\sqrt{x^2−16})})\)

⇒ ​ \(\frac{\text{dy}}{\text{dx}}\) = \({1 \over2}[{ \sqrt{x^2−16}}+{x^2\over \sqrt{x^2−16}}] − {8 \over \left(x + \sqrt{x^2−16}\right)}.(1 + {x\over \sqrt{x^2−16}})\)​

⇒ ​ \(\frac{\text{dy}}{\text{dx}}\) = \({1 \over2}[{ {{x^2−16}}+x^2\over \sqrt{x^2−16}}] − {8 \over \left(x + \sqrt{x^2−16}\right)}.{\left(x + \sqrt{x^2−16}\right)\over \sqrt{x^2−16}}\)

⇒ ​ \(\frac{\text{dy}}{\text{dx}}\) = \([{ {{2x^2−16}}\over 2\sqrt{x^2−16}}] − {8 \over \sqrt{x^2−16}}\)

⇒ ​ \(\frac{\text{dy}}{\text{dx}}\) = \([{ {{x^2−8 - 8}}\over \sqrt{x^2−16}}] \)

⇒ ​ \(\frac{\text{dy}}{\text{dx}}\) = \( { \sqrt{x^2−16}}\)

∴ The correct answer is option (3).

Concept:

• \({d \over dx} (p(x) q(x)) = p'(x) q(x) + p(x)q'(x)\)
• \({d \over dx} (f(g(x)) = f'(g(x))g'(x)\)

Calculation:

If y = (xx)x, 

Taking log both sides,

⇒ log y = log (xx)x, 

⇒ log y = x log (xx), 

⇒ log y = x.x log x,

⇒ log y = x² log x,

Now differentiating both sides with respect to x,

⇒ \({1 \over y}{dy \over dx} = 2x \log x + x^2 .{1 \over x}\)

⇒ \({dy \over dx} = 2xy \log x + xy\)

⇒ ​\({dy \over dx} - xy (1 + 2\ln x) = 0\)​

∴ The correct answer is option (2).

Concept:

• sin3x = 3sinx - 4sin³x
• cos3x = 4cos³x - 3cosx

Calculation:

f(x) = 3(sin x − cos x) + 4(cos3 x − sin3 x)

f(x) = 3sinx - 3cosx + 4cos³x - 4sin³x

f(x) = (3sinx - 4sin³x) + (4cos³x - 3cosx)

f(x) = sin3x + cos3x

f'(x) = 3cos3x - 3sin3x

Now, f'(x) = 0

⇒  3cos3x - 3sin3x = 0

⇒ cos3x - sin3x = 0

⇒ sin3x = cos3x

⇒ tan3x = 1

⇒ \(3x = \frac{\pi }{4}\)

Hence, the maximum value of given function is = \(sin\frac{\pi }{4}+cos\frac{\pi }{4}\)

= \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)

∴ The correct option is (2)

Calculation:

Curve y = \(\sqrt{1−\text{x}^2}\), y = x and y = 0

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

= \(\int_{0}^{\frac{1}{\sqrt{2}}}xdx+\int_{\frac{1}{\sqrt{2}}}^{1}\sqrt{1-x^{2}}dx\)

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

= \(\frac{1}{4}+\left [ 0+\frac{1}{2} sin^{-1}(1)-\frac{1}{2\sqrt{2}}.\frac{1}{\sqrt{2}}-\frac{1}{2}sin^{-1}\left ( \frac{1}{\sqrt{2}} \right )\right ]\)

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

= \(\frac{\pi }{4}-\frac{\pi }{8}\)

= \(\frac{\pi }{8}\)

∴ The correct option is (3)