NDA Mathematics 22 April 2018 Official Paper (Previous Year Paper)

Concepts:

Suppose there is a given statement P (n) involving the natural number n such that

• The statement is true for n = 1, i.e., P (1) is true, and
• If the statement is true for n = k (where k is some positive integer), then the statement is also true for n = k + 1, i.e., truth of P (k) implies the truth of P (k + 1).

Then, P (n) is true for all natural numbers n

Calculation:

Given:

P (n) = 121ⁿ – 25ⁿ + 1900ⁿ – (-4) ⁿ

Now, P (1) = 121¹ – 25¹ + 1900¹ – (-4)¹

⇒ P (1) = 121 – 25 + 1900 + 4

⇒ P (1) = 2000

Therefore we can say that P (n) is divisible by 2000.

Calculation:

Numbers between 100 and 1000 are three digits number, so we have to fill three places.

Total numbers = 5

(They are 5, 6, 7, 8 and 9)

Number of choices for the 1^st digit = 5

Repetition of digits is not allowed therefore available numbers for 2^nd place is four

⇒ Number of choices for the 2^nd digit = 4

Repetition of digits is not allowed therefore available numbers for 1^st place is three

⇒ Number of choices for the 3^rd digit = 3

∴ Total possible combinations = 5 × 4 × 3 = 60

Concepts:

Let z = x + iy be a complex number.

• Modulus of z = \(\left| \text{z} \right|=\text{ }!!!!\text{ }\sqrt{{{\text{x}}^{2}}+{{\text{y}}^{2}}}=\text{ }!!!!\text{ }\sqrt{\text{Re}{{\left( \text{z} \right)}^{2}}+\text{ }!!!!\text{ Im }!!!!\text{ }{{\left( \text{z} \right)}^{2}}}\)
• arg (z) = arg (x + iy) = ${\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$

Calculation:

We have to find value of ′x′ for which should be a non-zero integral.

Given: ∣1 − 2i∣^x = 5^x

Let first find value of ∣1 − 2i∣

$\Rightarrow {\rm{;}}\mid 1{\rm{;}} - {\rm{;}}2{\rm{i}}\mid {\rm{;}} = {\rm{;}}\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 2} \right)}^2}} {\rm{;}} = {\rm{;}}\sqrt 5$​     ...(∵ z = x + iy ⇒ $\left| {\rm{z}} \right| = {\rm{;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}}$)

Now, ∣1 − 2i∣^x = 5^x

⇒ (√5) ^x = 5^x

⇒ 5^x/2 = 5^x

Base is same so power is also equal

Therefore x/2 = x

⇒ x = 2x

∴ x = 0

Hence no non-zero integral solutions.

Concepts:

• Arithmetic mean of two positive numbers a and b is $\frac{{a; + ;b}}{2}$
• Geometric mean of two positive numbers a and b is $\sqrt {ab}$

Calculation:

Given: Ratio of AM to GM of two positive numbers a and b is 5 : 3

$\Rightarrow ;\frac{{\left( {\frac{{a; + ;b}}{2}} \right)}}{{\sqrt {ab} }} = ;\frac{5}{3}$

$\Rightarrow ;\frac{{a; + ;b}}{{2\sqrt {ab} }} = ;\frac{5}{3}$

$\Rightarrow ;\frac{{\frac{{a; + ;b}}{b}}}{{\frac{{\sqrt {ab} }}{b}}} = ;\frac{{10}}{3}$

$\Rightarrow ;\frac{{\frac{a}{b} + 1}}{{\sqrt {\frac{a}{b}} }} = ;\frac{{10}}{3}$

$\Rightarrow ;\frac{a}{b} + 1 = \frac{{10}}{3}\sqrt {\frac{a}{b}}$

Let a/b = x²     ...(Assume that a > b)

⇒ x² + 1 = (10/3) x

⇒ 3x2 +3 = 10x

⇒ 3x2 -10x + 3 = 0

⇒ 3x2 - 9x - x + 3 = 0

⇒ (3x – 1) (x – 3) = 0

⇒ x = 1/3 and 3

⇒ x² = 1/9 and 9

⇒ a/b = 9     ...(∵ a > b)

Therefore, a : b = 9 : 1.

Concepts:

1. Let z = x + iy be a complex number.

• Modulus of z = $\left| {\rm{z}} \right| = {\rm{;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{;}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{;Im;}}{{\left( {\rm{z}} \right)}^2}}$
• arg (z) = arg (x + iy) = ${\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$
• Conjugate of z =

 = x – iy

• $z\bar z = {\left| z \right|^2}$
• $\left| {\rm{z}} \right| = \left| {\bar z} \right|$
• $\overline {\left( {\bar z} \right)} = z$

2. Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂ be any two complex numbers.

• |z₁z₂| = |z₁||z₂|
• $\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}},;\left| {{z_2}} \right| \ne 0$
• $\overline {{z_1} \pm {z_2}} = ;\overline {{z_1}} \pm \overline {{z_2}}$

 

Calculation:

Given: |α | = 1

We have to find the value of $\left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right|$

Multiple with $\bar \alpha$ in both Numerator and denominator

$\Rightarrow ;\left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right| = \left| {\frac{{\left( {\alpha - \beta } \right); \times ;\bar \alpha }}{{\left( {1 - \alpha \bar \beta } \right); \times ;\bar \alpha }}} \right|$

$= \left| {\frac{{\left( {\alpha \bar \alpha - \beta \bar \alpha } \right);}}{{\left( {1 - \alpha \bar \beta } \right) \times ;\bar \alpha ;}}} \right|$

We know that $z\bar z = {\left| z \right|^2}$

So, $\alpha \bar \alpha = {\left| \alpha \right|^2} = 1$     ...(∵ |α| = 1)

$= \left| {\frac{{\left( {1 - \beta \bar \alpha } \right);}}{{\left( {1 - \alpha \bar \beta } \right); \times ;\bar \alpha ;}}} \right|$

We know that $\overline {\left( {\bar z} \right)} = z;and;\overline {{z_1} - {z_2};} = ;\overline {{z_1}} - ;\overline {{z_2}}$

So, $\left( {1 - \beta \bar \alpha } \right) = \overline {\left( {1 - \alpha \bar \beta } \right)}$

$= \left| {\frac{{\overline {\left( {1 - \alpha \bar \beta } \right)} ;}}{{\left( {1 - \alpha \bar \beta } \right); \times ;\bar \alpha ;}}} \right|$

\(=\frac{\left| \overline{\left( 1-\alpha \bar{\beta } \right)}~ \right|}{\left| \left( 1-\alpha \bar{\beta } \right) \right|\left| {\bar{\alpha }} \right|}\text{ }!!!!\text{ }\left( \because \left| \frac{{{z}{1}}}{{{z}{2}}} \right|=\frac{\left| {{z}{1}} \right|}{\left| {{z}{2}} \right|}and\left| z1z2 \right|=~\left| z1 \right|\left| z2 \right| \right)\)

We know that $\left| {\rm{z}} \right| = \left| {\bar z} \right|$

$\therefore \left| {{\rm{\alpha }}} \right| = {\rm{}}\left| {\bar \alpha } \right| = 1;and;\left| {\left( {1 - \alpha \bar \beta } \right)} \right| = \left| {\overline {\left( {1 - \alpha \bar \beta } \right)} ;} \right|$

$\Rightarrow ;\left| {\frac{{\alpha - \beta }}{{1 - \alpha \bar \beta }}} \right| = \frac{{\left| {\overline {\left( {1 - \alpha \bar \beta } \right)} ;} \right|}}{{\left| {\left( {1 - \alpha \bar \beta } \right)} \right|\left| {\bar \alpha } \right|}} = 1$

Concepts:

Convert decimal to binary

Conversion steps:

• Divide the number by 2.
• Get the integer quotient for the next iteration.
• Get the remainder for the binary digit.
• Repeat the steps until the quotient is equal to 0.

 

Calculation:

Convert 31₁₀ to binary

 

So, binary form of 31 is 11111.

Concepts:

1. Let z = x + iy be a complex number.

• Modulus of z = $\left| {\rm{z}} \right| = {\rm{;}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{;}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{;Im;}}{{\left( {\rm{z}} \right)}^2}}$
• arg (z) = arg (x + iy) = ${\tan ^{ - 1}}\left( {\frac{y}{x}} \right)$

2. Modulus- amplitude form a complex number ⇔ z = r (cos θ + i sin θ)

Where r = |z| and θ = arg z, (-π  < θ < π)

Calculation:

Let z = √3 + i = r (cos θ + i sin θ)

Comparing real and imaginary part,

⇒ r cos θ = √3     ….1)

⇒ r sin θ = 1     ….2)

Squaring and adding above two equations,

⇒ r² (cos² θ + sin² θ) = 3 + 1 = 4

⇒ r² = 4     ...(∵ cos² θ + sin² θ = 1)

∴ r = 2

Now dividing equation 2 by equation 1

$\frac{{{\rm{r;}}\sin {\rm{\theta }}}}{{{\rm{r;}}\cos {\rm{\theta }}}} = {\rm{}}\frac{1}{{\sqrt 3 }}$

⇒ tan θ = tan (π/6)

∴ θ = π/6

Hence z = r (cos θ + i sin θ) = 2 (cos π/6 + i sin π/6)

Concepts:

(1 + x)ⁿ = 1 + ⁿc₁ x + ⁿc₂ x² + ⁿc₃ x³ + …. + ⁿc_n xⁿ

(1 - x)ⁿ = 1 - ⁿc₁ x + ⁿc₂ x² - ⁿc₃ x³ + …. + (-1)^{n n}c_n xⁿ

Total number of terms in the expansion of (1 ± x)ⁿ is n + 1

(1 + x) ⁿ + (1 - x) ⁿ = 2 [1 + ⁿc₂ x² + …]

∴ Total number of non-zero terms in the expansion of (1 + x)ⁿ + (1 - x)ⁿ is (n + 1)/2

Calculation:

We have to find the number of non-zero terms in the expansion of (1 + 2√3x)¹¹ + (1 – 2√3x)¹¹

We know that total number of non-zero terms in the expansion of (1 + x)ⁿ + (1 - x)ⁿ is (n +1)/2

Here n = 11

∴ Total number of non-zero terms in the expansion = (11 + 1)/2 = 6

Concepts:

1. Let Standard form of quadratic Functions: ⇒ f(x) = ax² + bx + c

Steps to Graphing Quadratic Equations

Step-1: Check the sign of a

• If a > 0 the graph will be a convex parabola.
• If a < 0 the graph will be a concave parabola

Step- 2: Vertex = $\rm \left( {\frac{{ - b}}{{2a}},\frac{{ - D}}{{4a}}} \right)$, Where D is Discriminants (D = b² – 4ac)

Step- 3: Parabola is symmetric about the line that intersects x = $\rm \frac{{ - b}}{{2a}}$

Step- 4: Find the coordinates of the points of intersection with the x axis, we must solve the equation f(x) = 0

 

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

 

Example:

Calculation:

We have to find number of solutions of the equation x² - 4x + [x] = 0 in the interval [0, 2]

x² - 4x + [x] = 0

⇒ [x] = -x² + 4x

Let g(x) = [x] and f(x) = -x² + 4x

Graphing of function f(x):

f(x) = -x² + 4x

Above equation is compare with ax² + bx + c,

a = -1, b = 4 and c = 0

Now, Discriminants = D = b² – 4ac = 4² – 0 = 16

Here a < 0 the graph will be a concave parabola

Vertex = $\rm \left( {\frac{{ - b}}{{2a}},\frac{{ - D}}{{4a}}} \right) = ;\left( {\frac{{ - ;4}}{{2; \times ; - 1}},\frac{{ - 16}}{{4; \times ; - 1}}} \right) = \left( {2,;4} \right)$

Points of intersection with the x axis: f(x) = 0

⇒ -x² + 4x = 0

∴ x = 0 and 4

We can see that x = 0 is only one point is common in both curves in the interval [0, 2]

∴ Number of solutions of the equation is one.

Concepts:

• ${\log _e}y = x \Leftrightarrow y = {e^x}$

 

Calculation:

Given: log₁₀ a is negative.

⇒ log₁₀ a < 0

Let log₁₀ a = - b (Here b is positive)

⇒ a = 10^-b

Given a lies between 0 to 1

⇒ 0 < a < 1

⇒ 0 < 10^-b < 1

Hence it is negative power of 10 is between 0 and 1

∴ Option 2 is correct.

Concepts:

Let us consider sequence a₁, a₂, a₃ …. a_n is an G.P.

• Common ratio = r = $\frac{{{a_2}}}{{{a_1}}} = \frac{{{a_3}}}{{{a_2}}} = \ldots = \frac{{{a_n}}}{{{a_{n - 1}}}}$
• n^th term of the G.P. is an = arⁿ⁻¹
• Sum of n terms = s = $\frac{{a;\left( {{r^n} - 1} \right)}}{{r - 1}}$; where r >1
• Sum of n terms = s = $\frac{{a;\left( {1 - {r^n}} \right)}}{{1 - r}}$; where r <1
• Sum of infinite GP = ${{\rm{s}}_\infty } = {\rm{}}\frac{{\rm{a}}}{{1{\rm{}} - {\rm{r}}}}{\rm{}}$; |r| < 1

Where a is 1^st term and r is common ratio.

Calculation:

Given: The third term of a GP is 3

Let 'a' be the first term and 'r' be the common ratio.

∴ T₃ = ar² = 3

We know that T_n = a r^n-1

So, T₁ = a, T₂ = ar, T₃ = ar², T₄ = ar³, T₅ = ar⁴

Now, Product of the first five terms = a × ar × ar² × ar³ × ar⁴ = a⁵r¹⁰ = (ar²)⁵ = 3⁵ = 243

Concepts:

• If a, b, c are in A.P ⇔ b = $\frac{{a; + ;c}}{2}$
• If a, b, c are in G.P ⇔ b² = ac
• If a, b, c are in H.P ⇔ $b = \frac{{2ac}}{{a; + ;c}}$

 

Calculation:

Given: x, 3/2, z are in AP

∴ 2 × (3/2) = x + z

⇒ x + z = 3     ….1)

Now, x, 3, z are in GP

∴ 3² = xz

⇒ xz = 9     ….2)

Let x, y, z are in HP

∴ $y = \frac{{2xz}}{{x + z}} = \frac{{2 \times 9}}{3} = 6$

Hence x, 6, z are in HP

Concepts:

• Sine law ⇔ $\frac{{\rm{a}}}{{\sin {\rm{A}}}} = {\rm{;}}\frac{{\rm{b}}}{{\sin {\rm{B}}}} = {\rm{;}}\frac{{\rm{c}}}{{\sin {\rm{C}}}}$

Where a, b and c are sides and A, B and C are angles.

Here; Side a faces angle A, side b faces angle B and side c faces angle C

Calculation:

Given: a = 2, b = 3 and sin A = 2/3

Apply sine rule,

$\frac{{\rm{a}}}{{\sin {\rm{A}}}} = \frac{{\rm{b}}}{{\sin {\rm{B}}}}$

$\Rightarrow \sin {\rm{B}} = \frac{{b\sin {\rm{A}}}}{a}$

$\Rightarrow \sin {\rm{B}} = \frac{{3; \times \left( {\frac{2}{3}} \right)}}{2} = 1$

∴ B = π/2

Concepts:

• sin (x + y) = sin x cos y + cos x sin y
• sin (x – y) = sin x cos y - cos x sin y

 

Calculation:

Given:

$\frac{{\sin \left( {x; + ;y} \right)}}{{\sin \left( {x; - ;y} \right)}} = ;\frac{{a; + ;b}}{{a; - ;b}}$

$\Rightarrow ;\frac{{\sin x\cos y; + ;\cos x\sin y}}{{\sin x\cos y; - ;\cos x\sin y}} = ;\frac{{a; + ;b}}{{a; - ;b}}$

Using Componendo and dividendo Rule, we get

$\Rightarrow \frac{{(\sin x\cos y + \cos x\sin y) + (\sin x\cos y - \cos x\sin y)}}{{(\sin x\cos y + \cos x\sin y) - (\sin x\cos y - \cos x\sin y)}} = \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{{\left( {a + b} \right) - \left( {a - b} \right)}}$

$\Rightarrow ;\frac{{2{\rm{;sin}}x\cos y}}{{2\cos x\sin y}} = \frac{{2a}}{{2b}}$

$\therefore \frac{{\tan x}}{{\tan y}} = \frac{a}{b}$

Concepts:

• 1 + cos A = 2 cos² (A/2)
• 1 – cos A = 2 sin² (A/2)

Calculation:

We know that 2 cos² (A/2) = 1 + cos A

$\Rightarrow \cos \left( {\frac{A}{2}} \right) = \pm \sqrt {\frac{{1 + \cos A}}{2}}$

Here only one value of cos A possible then cos A must be -1,

Therefore ‘A’ must be odd multiple 180°

Verification:

Let A = 180°

L.H.S = cos (A/2) = cos 90° = 0

R.H.S = $\sqrt {\frac{{1 + \cos A}}{2}} = \sqrt {\frac{{1 + \cos 180}}{2}} = \sqrt {\frac{{1 - 1}}{2}} = 0$

So, L.H.S = R.H.S

Concept:

Ellipse: The locus of a point which moves such that its distance from a fixed point and a fixed straight line is always less than one (Eccentricity = e < 1)

 

Calculation:

Given:

Vertices = (± 5, 0) and foci = (± 4, 0)

⇒ (± a, 0) = (± 5, 0) and (± ae, 0) = (± 4, 0)

∴ a = 5 and ae = 4

⇒ ae = 4

Squaring both sides,

⇒ a²e² = 4² (∵ a²e² = a² – b²)

⇒ a² – b² = 16

⇒ 5² – b² = 16

⇒ 25 – 16 = b²

⇒ b² = 9

Now, Equation of ellipse $\frac{{{\rm{;}}{x^2}}}{{{a^2}}} + \frac{{{{\bf{y}}^2}}}{{{b^2}}} = 1$ (a > b)

⇒ $\frac{{{\rm{;}}{x^2}}}{{25}} + \frac{{{{\bf{y}}^2}}}{9} = 1$

∴ Option 1 is correct.

Concept:

Vector equation of a plane passing through three points with position vectors $\vec a,{\rm{;}}\vec b,{\rm{;}}\vec c$ is

$\left( {\vec r - \vec a} \right) \cdot [\left( {\vec b - \vec a} \right) \times \left( {\vec c - \vec a} \right) = 0$

Cartesian equation of a plane passing through three points:

Let us now consider the coordinates of the three noncollinear points as A(x₁, y₁​, z₁​), B(x₂​, y₂​, z₂​), and C(x₃​, y₃​, z₃​).

Equation of plane is:

\left| {\begin{array}{*{20}{c}} {x - {\rm{;}}{x_1}}&{y - {\rm{;}}{y_1}}&{z - {\rm{;}}{z_1}}\ {{x_2} - {\rm{;}}{x_1}}&{{y_2} - {\rm{;}}{y_1}}&{{z_2} - {\rm{;}}{z_1}}\ {{x_3} - {\rm{;}}{x_1}}&{{y_3} - {\rm{;}}{y_1}}&{{z_3} - {\rm{;}}{z_1}} \end{array}} \right| = 0

If A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right] then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}

Calculation:

We have to find the equation of the plane which is passing through the point (-2, 6, -6), (-3, 10, -9), and (-5, 0, -6)

So, (x₁, y₁, z₁) = (-2, 6, -6), (x₂, y₂, z₂) = (-3, 10, -9) and (x₃, y₃, z₃) = (-5, 0, -6)

We know that the Equation of plane is:

 \left| {\begin{array}{*{20}{c}} {x - {\rm{;}}{x_1}}&{y - {\rm{;}}{y_1}}&{z - {\rm{;}}{z_1}}\ {{x_2} - {\rm{;}}{x_1}}&{{y_2} - {\rm{;}}{y_1}}&{{z_2} - {\rm{;}}{z_1}}\ {{x_3} - {\rm{;}}{x_1}}&{{y_3} - {\rm{;}}{y_1}}&{{z_3} - {\rm{;}}{z_1}} \end{array}} \right| = 0

So, the plane passing through points (-2, 6, -6), (-3, 10, -9) and (-5, 0, -6) is

\left| {\begin{array}{*{20}{c}} {x - \left( { - 2} \right)}&{y - {\rm{;}}6}&{z - \left( { - 6} \right)}\ { - 3 - {\rm{;}}\left( { - 2} \right)}&{10 - {\rm{;}}6}&{ - 9 - {\rm{;}}\left( { - 6} \right)}\ { - 5 - {\rm{;}}\left( { - 2} \right)}&{0 - {\rm{;}}6}&{ - 6 - {\rm{;}}\left( { - 6} \right)} \end{array}} \right| = 0

\left| {\begin{array}{*{20}{c}} {x + 2}&{y - 6}&{z + 6}\ { - 1}&4&{ - 3}\ { - 3}&{ - 6}&0 \end{array}} \right| = 0

Now, Expanding along R₁

(x + 2) (0 – 18) – (y – 6) (0 - 9) + (z + 6) (6 + 12) = 0

⇒ -18 (x + 2) + 9 (y – 6) + 18 (z + 6) = 0

⇒ -9 [2 (x + 2) – (y – 6) – 2 (z + 6)] = 0

⇒ 2x + 4 – y + 6 - 2z – 12 = 0

⇒ 2x – y – 2z – 2 = 0

⇒ 2x – y – 2z = 2

Shortcut Method: Substitute the point (-2, 6, -6), (-3, 10, -9) and (-5, 0, -6) in equation of plane. If points satisfy the equation of the plane then it will be the correct option.

Concept:

1. The intercept form of the line ⇔ $\frac{x}{a} + \frac{y}{b} = 1$

Where a is the x- intercept and b is the y- intercept

1. Section Formula: Section formula is used to determine the coordinate of a point that divides a line into two parts such that ratio of their length is m : n
2. Let P and Q be the given two points (x₁, y₁) and (x₂, y₂) respectively and M(x, y) be the point dividing the line- segment PQ internally in the ratio m: n
3. Internal Section Formula: When the line segment is divided internally in the ration m: n, we use this formula. ⇔ $\left( {x,{\rm{;}}y} \right) = \left( {\frac{{m{x_2}{\rm{;}} + {\rm{;}}n{x_1}}}{{m; + {\rm{;}}n}},\frac{{m{y_2}{\rm{;}} + {\rm{;}}n{y_1}}}{{m; + {\rm{;}}n}}} \right)$
4. External Section Formula: When the point M lies on the external part of the line segment. ⇔ $\left( {x,{\rm{;}}y} \right) = \left( {\frac{{m{x_2} - {\rm{;}}n{x_1}}}{{m; - {\rm{;}}n}},\frac{{m{y_2} - {\rm{;}}n{y_1}}}{{m; - {\rm{;}}n}}} \right)$

 

Calculation:

Intercept form of the line is $\frac{x}{a} + \frac{y}{b} = 1$

At x axis point is A (a, 0)

At y axis point is B (0, b)

Case 1: Given point P (2, 3) divides A and B in ratio of m : n = 3 : 2

Applying Internal Section Formula,

We know that $\left( {x,{\rm{;}}y} \right) = \left( {\frac{{m{x_2}{\rm{;}} + {\rm{;}}n{x_1}}}{{m; + {\rm{;}}n}},\frac{{m{y_2}{\rm{;}} + {\rm{;}}n{y_1}}}{{m; + {\rm{;}}n}}} \right)$

$\Rightarrow ;\left( {2,;3} \right) = ;\left( {\frac{{3\left( 0 \right) + ;2\left( a \right)}}{{3 + 2}},\frac{{3\left( b \right) + 2\left( 0 \right)}}{{3 + 2}}} \right)$

$\Rightarrow ;\left( {2,;3} \right) = ;\left( {\frac{{2a}}{5},\frac{{3b}}{5}} \right)$

Now,

$\frac{{2a}}{5} = 2;and;\frac{{3b}}{5} = 3$

⇒ a = 5 and b = 5

∴ Equation of line is $\frac{x}{5} + \frac{y}{5} = 1$ ⇒ x + y = 5

Case 2: Given point P (2, 3) divides A and B in ratio of m : n = 2 : 3

$\Rightarrow ;\left( {2,;3} \right) = ;\left( {\frac{{2\left( 0 \right) + ;3\left( a \right)}}{{3 + 2}},\frac{{2\left( b \right) + 3\left( 0 \right)}}{{3 + 2}}} \right)$

$\Rightarrow ;\left( {2,;3} \right) = ;\left( {\frac{{3a}}{5},\frac{{2b}}{5}} \right)$

Now,

$\frac{{3a}}{5} = 2;and;\frac{{2b}}{5} = 3$

⇒ a = 10/3 and b = 15/2

∴ Equation of line is \(\frac{x}{\left( {}^{10}!!\diagup!!{}{3}; \right)}+\frac{y}{\left( {}^{15}!!\diagup!!{}{2}; \right)}=1\)

$\Rightarrow {\rm{;}}\frac{{3x}}{{10}} + \frac{2}{{15}} = 1$

⇒ 9x + 4y = 30

Shortcut Method:

• Check through the options: Substitute the point (2, 3) in all options

⇒ Only option 4 satisfied the point (2, 3), so option 4 is correct.

Concept:

• General equation of a sphere cantered at the point (x₀, y₀, z₀) with radius R is given by
• (x – x₀)² + (y – y₀)² + (z – z₀)² = R²

 

Calculation:

Given: Centre is at (-2, 3, 4) and radius is 6

⇒ (x₀, y₀, z₀) = (-2, 3, 4) and R = 6

We know that general equation of sphere is (x – x₀)² + (y – y₀)² + (z – z₀)² = R²

⇒ (x – (-2)) ² + (y – 3)² + (z – 4)² = 6²

⇒ (x + 2) ² + (y – 3)² + (z – 4)² = 6²

⇒ x² + 4x + 4 + y² – 6y + 9 + z² – 8z + 16 = 36

⇒ x² + y² + z² + 4x – 6y – 8z + 29 = 36

⇒ x² + y² + z² + 4x – 6y – 8z = 7

∴ Option 1 is correct.

Concept:

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

 

Calculation:

Given: $\vec a + 2\vec b + 3\vec c = \vec 0$

$\Rightarrow \vec a = ; - \left( {2\vec b + 3\vec c} \right)$

Now, $\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a$

$= - \left( {2\vec b + 3\vec c} \right) \times \vec b + \vec b \times \vec c - \vec c \times \left( {2\vec b + 3\vec c} \right)$

$= - 2\vec b \times \vec b - 3\vec c \times \vec b + \vec b \times \vec c - \vec c \times 2\vec b -\vec c \times 3\vec c$

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

So, $- k\vec c \times \vec b = ;k\vec b \times \vec c$

$= ;0 + ;3\left( {\vec b \times \vec c} \right) + \left( {\vec b \times \vec c} \right) + 2\left( {\vec b \times \vec c} \right) + 0{\rm{;}}\left( {\vec b \times \vec b = 0{\rm{;and;}}\vec c \times \vec c = 0} \right)$

$= 6\left( {\vec b \times \vec c} \right)$

∴ λ = 6

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 ⇔ $\mathop {\lim }\limits_{x \to {{\rm{a}}^ + }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {{\rm{a}}^ - }} f\left( x \right){\rm{}} = {\rm{}}\mathop {\lim }\limits_{x \to {\rm{a}}} f\left( x \right)$

 

Calculation:

Given: f(x) = |x| + |x – 1|

The possible points of discontinuity are x = 0, 1 since the sign of the modulus changes at these points.

For continuity at x = 0

${\rm{LHL}} = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = ;\mathop {\lim }\limits_{x \to {0^ - }} \left| {\rm{x}} \right|{\rm{;}} + {\rm{;}}\left| {{\rm{x;}}-{\rm{;}}1} \right|$

$= ;\mathop {\lim }\limits_{h \to 0} \left| {0 - {\rm{h}}\left| {{\rm{;}} + {\rm{;}}} \right|0 - {\rm{h;}}-{\rm{;}}1} \right|$     ...(∵ x → 0^- ⇒ x = 0 – h)

$= ;\mathop {\lim }\limits_{h \to 0} \left| {\rm{h}} \right|{\rm{;}} + {\rm{;}}\left| {{\rm{h}} + {\rm{;}}1} \right|$     ...(∵ |-x| = x)

= 0 + 1 = 1

${\rm{RHL}} = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = ;\mathop {\lim }\limits_{x \to {0^ + }} \left| {\rm{x}} \right|{\rm{;}} + {\rm{;}}\left| {{\rm{x;}}-{\rm{;}}1} \right|$

$= ;\mathop {\lim }\limits_{h \to 0} \left| {0 + {\rm{;h}}\left| {{\rm{;}} + {\rm{;}}} \right|0 + {\rm{;h;}}-{\rm{;}}1} \right|$    ...(∵ x → 0⁺ ⇒ x = 0 + h)

$= ;\mathop {\lim }\limits_{h \to 0} \left| {\rm{h}} \right|{\rm{;}} + {\rm{;}}\left| {{\rm{h}} - {\rm{;}}1} \right|$

= 0 + 1 = 1     ...(∵ |-x| = x)

Value of f(x) at x = 0

⇒ f(0) = |0| + |0 – 1| = 1

Since LHL = RHL = f(0)

So, Function is continuous at x = 0

For continuity at x = 1

${\rm{LHL}} = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = ;\mathop {\lim }\limits_{x \to {1^ - }} \left| {\rm{x}} \right|{\rm{;}} + {\rm{;}}\left| {{\rm{x;}}-{\rm{;}}1} \right|$

$= \mathop {\lim }\limits_{h \to 0} \left| {1 - {\rm{h}}\left| {{\rm{;}} + {\rm{;}}} \right|1 - {\rm{h;}}-{\rm{;}}1} \right|$     ...(∵ x → 1^- ⇒ x = 1 – h)

$= ;\mathop {\lim }\limits_{h \to 0} \left| {1 - {\rm{;h}}\left| {{\rm{; + }}} \right| - {\rm{h}}} \right|$

$= ;\mathop {\lim }\limits_{h \to 0} \left| {1 - {\rm{;h}}} \right|{\rm{;}} + {\rm{;}}\left| {\rm{h}} \right|$     ...(∵ |-x| = x)

= 1 + 0 = 1

${\rm{RHL}} = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = ;\mathop {\lim }\limits_{x \to {1^ + }} \left| {\rm{x}} \right| + \left| {{\rm{x}}-1} \right|$

$= ;\mathop {\lim }\limits_{h \to 0} \left| {1 + {\rm{;h}}\left| {{\rm{;}} + {\rm{;}}} \right|1 + {\rm{;h;}}-{\rm{;}}1} \right|$     ...(∵ x → 1⁺ ⇒ x = 1 + h)

$= ;\mathop {\lim }\limits_{h \to 0} \left| {1 + {\rm{;h}}} \right|{\rm{;}} + {\rm{;}}\left| {\rm{h}} \right|$

= 1 + 0 = 1

Value of f(x) at x = 1

⇒ f(1) = |1| + |1 – 1| = 1

Since LHL = RHL = f(1)

So, Function is continuous at x = 1

∴ f(x) is continuous at x = 0 and x = 1

Concept:

1. Integration by parts: Integration by parts is a method to find integrals of products.

• The formula for integrating by parts is given by;

$\Rightarrow ;\smallint u;vdx = u;\smallint vdx - ;\smallint \left( {\frac{{du}}{{dx}};\smallint vdx} \right)dx$

Where u is the function u(x) and v is the function v(x)

2. ILATE Rule: Usually, the preference order of this rule is based on some functions such as Inverse, Logarithm, Algebraic, Trigonometric and Exponent.

3. $\smallint {x^n}dx = ;\frac{{{x^{n; + ;1}}}}{{n + 1}} + c$
4. \mathop \smallint \nolimits_a^b f'\left( x \right)dx = ;\left[ {f\left( x \right)} \right]_a^b = \left[ {f\left( b \right) - f\left( a \right)} \right]

Calculation:

We have to find the value of \mathop \smallint \nolimits_1^e x\ln x;dx

\Rightarrow \mathop \smallint \nolimits_1^e (\ln x);x;dx = \left[ {\ln x;\left( {\frac{{{x^2}}}{2}} \right)} \right]_1^e - ;\mathop \smallint \nolimits_1^e \frac{1}{x}\left( {\frac{{{x^2}}}{2}} \right)dx

\Rightarrow \left[ {\ln x;\left( {\frac{{{x^2}}}{2}} \right)} \right]_1^e - \frac{1}{2}\mathop \smallint \nolimits_1^e xdx

$\Rightarrow \left[ {\ln x;\left( {\frac{{{x^2}}}{2}} \right)} \right]_1^e - \frac{1}{2}\left[ {\frac{{{x^2}}}{2}} \right]_1^e$

$\Rightarrow \left[ {\ln e;\frac{{{e^2}}}{2} - ;\ln 1;\frac{{{1^2}}}{2}} \right] - \frac{1}{4}\left[ {{e^2} - 1} \right]$           (∵ loge = 1)

$\Rightarrow \left[ {\frac{{{e^2}}}{2}; - ;0} \right] - \frac{1}{4}\left[ {{e^2} - 1} \right]$

$\Rightarrow \frac{{{e^2} + 1}}{4}$

∴ Option 2 is correct.

Concept:

• \mathop \smallint \nolimits_a^b f\left( x \right)dx = ;\mathop \smallint \nolimits_a^c f\left( x \right)dx + ;\mathop \smallint \nolimits_c^b f\left( x \right)dx
• \mathop \smallint \nolimits_a^b f'\left( x \right)dx = ;\left[ {f\left( x \right)} \right]_a^b = \left[ {f\left( b \right) - f\left( a \right)} \right]
• $\smallint {x^n}dx = ;\frac{{{x^{n + 1}}}}{{n + 1}} + c$

Calculation:

We have to find the value of \mathop \smallint \nolimits_0^{\sqrt 2 } \left[ {{x^2}} \right];dx

Now, f(x) = \left[ {{x^2}} \right]; \Rightarrow ;f\left( x \right) = \left{ {\begin{array}{*{20}{c}} {0,;;0 \le x < 1}\ {1,;;1 \le x < \sqrt 2 } \end{array}} \right.

\mathop \smallint \nolimits_0^{\sqrt 2 } \left[ {{x^2}} \right];dx = ;\mathop \smallint \nolimits_0^1 0;dx + \mathop \smallint \nolimits_1^{\sqrt 2 } 1;dx

$= 0 + \left[ x \right]_1^{\sqrt 2 }$

= √2 – 1

Concept:

Following steps to finding maxima and minima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have find second derivative.

1. f``(x) is less than 0 then the given function is said to be maxima
2. If f``(x) Is greater than 0 then the function is said to be minima

Calculation:

We have to calculate the maximum value of 16 sin θ – 12 sin² θ

Let f(θ) = 16 sin θ – 12 sin² θ

Differentiating, both sides, we get

⇒ f’(θ) = 16 cos θ – 12 × 2 × sin θ × cos θ = 16 cos θ – 24 × sin θ × cos θ

For maxima and minima, f’(θ) = 0

⇒ 16 cos θ – 24 × sin θ × cos θ = 0

⇒ 8 cos θ (2 – 3 sin θ) = 0

⇒ 8 cos θ = 0 and (2 – 3 sin θ) = 0

∴ θ = 90° and sin θ = 2/3 or θ = sin^-1 (2/3)

Now we have to find second derivative,

⇒ f”(θ) = -16 sin θ – 24 [cos² θ – sin² θ]

⇒ f”(θ) = -16 sin θ – 24 [1 - sin² θ - sin² θ] = -16 sin θ – 24 [1 – 2 sin² θ]

When θ = 90° ⇒ ${\left. {{\rm{f}}\left( {\rm{\theta }} \right)} \right|_{{\rm{\theta ;}} = {\rm{;}}90^\circ }}$ = -16 sin 90° – 24 [1 – 2 sin² 90°] = -16 + 24 = 8 > 0

So, function gives minimum value at θ = 90°

When θ = sin^-1 (2/3) ⇒ ${\left. {{\rm{f''}}\left( {\rm{\theta }} \right)} \right|_{{\rm{\theta ;}} = {\rm{;}}{{\sin }^{ - 1}}\left( {\frac{2}{3}} \right)}}$ = -16 sin sin^-1 (2/3) – 24 [1 – 2 sin² sin^-1 (2/3)]

⇒ f”(θ) = -16 (2/3) – 24 [1 – 2 (2/3)²] = -(32/3) – (24/9) < 0

So, function gives maximum value at θ = sin^-1 (2/3) or sin θ = 2/3

⇒ f(θ)_max = 16(2/3) – 12 (2/3)² = (32/3) – (16/3) = 16/3

Concept:

Period of a Function:

• If a function repeats over at a constant period we say that is a periodic function.
• It is represented like f(x) = f(x + T), T is the real number and this is the period of the function.

Calculation:

We have to find the period of the function f(x) = sin x

Now,

f(x + 2π) = sin (x + 2π) = sin x

⇒ f(x + 2π) = f(x)

∴ Period of sin x is 2π

Concept:

• $\frac{{d\left( {{a^x}} \right)}}{{dx}} = ;{a^x};\ln a$
• $\smallint \frac{1}{{\rm{x}}}{\rm{;dx}} = {\rm{}}\ln {\rm{x;}} + {\rm{c}}$

 

Calculation:

We have to find the value of $\smallint \frac{{dx}}{{{2^x} - 1}}$

Let I = $\smallint \frac{{dx}}{{{2^x} - 1;}} = ;\smallint \frac{{dx}}{{{2^x};\left( {1; - ;\frac{1}{{{2^x}}}} \right);}}$

${\rm{I}} = \smallint \frac{{{2^{ - {\rm{x}}}}{\rm{;dx}}}}{{{\rm{;}}\left( {1{\rm{;}} - {\rm{;}}{2^{ - {\rm{x}}}}} \right){\rm{;}}}}$

Let (1 – 2^-x) = t

Now, Differentiating, Both sides, we get

$\Rightarrow {\rm{;}}\left( {0{\rm{;}} + {\rm{;}}{2^{ - {\rm{x}}}}\ln 2} \right){\rm{}} = {\rm{}}\frac{{{\rm{dt}}}}{{{\rm{dx}}}}$

$\Rightarrow ;{2^{ - x}};dx = \frac{{dt}}{{\ln 2}}$

Now, ${\rm{I}} = \frac{1}{{\ln 2}}\smallint \frac{{{\rm{dt}}}}{{{\rm{;}}t{\rm{;}}}}$

$\Rightarrow I = \frac{{\ln t}}{{\ln 2}} + c = \frac{{\ln \left( {1 - ;{2^{ - x}}} \right)}}{{\ln 2}} + c$

∴ Option 2 is correct.

Concept:

• Order: The order of a differential equation is the order of the highest derivative appearing in it.
• Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

y² = 4a (x – a)     ...(1)

Differentiating, both sides, we get

$\Rightarrow 2y;\frac{{dy}}{{dx}} = 4a$

$\Rightarrow \left( {\frac{y}{2}} \right)\frac{{dy}}{{dx}} = a$

Put the value of a in equation 1^st

$\Rightarrow {y^2} = 4\left( {\frac{y}{2}} \right)\frac{{dy}}{{dx}};\left( {(x - \frac{y}{2};\frac{{dy}}{{dx}}} \right)$

$\Rightarrow {y^2} = 2y;\frac{{dy}}{{dx}};\left( {(x - \frac{y}{2};\frac{{dy}}{{dx}}} \right)$

$\Rightarrow {y^2} = 2xy;\frac{{dy}}{{dx}} - {y^2}{\left( {\frac{{dy}}{{dx}}} \right)^2}$

Hence order is 1 and degree is 2.

Concept:

1. Let f(x) and g(x) be two functions such that their limits $\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right)$ and $\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)$ exist.

• Limit of the sum of two functions is the sum of the limits of the functions.

$\Rightarrow \mathop {\lim }\limits_{x \to a} \left[ {{\rm{f}}\left( {\rm{x}} \right) + {\rm{;g}}\left( {\rm{x}} \right)} \right] = {\rm{;}}\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right){\rm{;}} + {\rm{;}}\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)$

• Limit of the difference of two functions is the difference of the limits of the functions.

$\Rightarrow \mathop {\lim }\limits_{x \to a} \left[ {{\rm{f}}\left( {\rm{x}} \right) - {\rm{;g}}\left( {\rm{x}} \right)} \right] = {\rm{;}}\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right){\rm{}} - {\rm{}}\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)$

• Limit of product of two functions is the product of the limits of the functions.

$\Rightarrow \mathop {\lim }\limits_{x \to a} \left[ {{\rm{f}}\left( {\rm{x}} \right){\rm{;}} \times {\rm{;g}}\left( {\rm{x}} \right)} \right] = {\rm{;}}\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right){\rm{;}} \times {\rm{;}}\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)$

• Limit of quotient of two functions is the quotient of the limits of the functions.

$\Rightarrow \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right)}}{{\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)}}$

2. $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$ 3. $\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = 1$

Calculation:

We have to find the value of $\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{\sin 2x}}$

$\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{\sin 2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{\tan x}}{x}} \right); \times x}}{{\left( {\frac{{\sin 2x}}{{2x}}} \right); \times 2x}} = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{\tan x}}{x}} \right)}}{{\left( {\frac{{\sin 2x}}{{2x}}} \right)}}$

We know that $\Rightarrow \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} {\rm{f}}\left( {\rm{x}} \right)}}{{\mathop {\lim }\limits_{x \to a} {\rm{g}}\left( {\rm{x}} \right)}}$

So, $\frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{\left( {\frac{{\tan x}}{x}} \right)}}{{\left( {\frac{{\sin 2x}}{{2x}}} \right)}} = \frac{1}{2}\frac{{\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x}}}{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x}}{{2x}}}} = \frac{1}{2} \times 1 = \frac{1}{2}$

∴ $\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{{\sin 2x}} = \frac{1}{2}$