Quantitative Aptitude PYQs

Explanation:

Types of matrices with 4 entries are 2 × 2, 4 × 1 and 1 × 4.

Number of 2 × 2 matrices using 1, 2, 3, 4 is 4⁴ = 256

Similarly, number of 4 × 1 matrices is 256 and number of 1 × 4 is 256.

So, total number of different matrices is 256 + 256 + 256 = 768.

Hence option (4) is true.

Concept:

One-one function: A function is said to be one-one function if it maps each element of the domain to a unique element of the range. 

On-to function: A function is said to be on-to function if every element in range has a pre-image in the domain.

Bijective function: A function is said to be bijective function if it is both one-one and on-to,

Explanation:

A = {x ∈ R: -1 < x < 1}.

1. f(x) = x|x|

 

From the graph we can see that each element from (-1,1) maps to a unique element. So it is one-one function.

Also, each element of range has a pre-image so it is on-to.

Hence f(x) = x|x| is bijective function.

1. is true.

2. g(x) = cos(πx)

\(-\frac12, \frac12\) ∈ A are two different element 

But g(-\(\frac12\)) = cos(-π/2) = 0 and g(\(\frac12\)) = cos(π/2) = 0

So, they maps to same element of the range.

Therefore, g(x) is not one-one and so not bijective.

2. is false.

Hence option (1) is correct.

Concept:

A relation R in a set A is said to be 

(i) reflexive if (x, x) ∈ R for every x ∈ A

(ii) Symmetric if (x, y) ∈ R implies (y, x) ∈ R for x, y ∈ A.

(iii) transitive if (x, y) ∈ R, (y, z) ∈ R implies (x, z) ∈ R for x, y, z ∈ A.

A relation R is said to be equivalence relation if it is reflexive, symmetric and transitive.

Explanation:

R is a relation on the open interval (-1, 1) and is given by

R = {(x, y) : |x + y| < 2}.

(i) - 1 < x < 1

⇒ -1 - 1 < x + x < 1 + 1

⇒ -2 < x + x < 2

⇒ |x + x| < 2

So, (x, x) ∈ R.

R is reflexive.

(ii) Let (x, y) ∈ R then |x + y| < 2 ⇒ |y + x| < 2

So, (y, x) ∈ R

R is symmetric.

(iii) Let (x, y) ∈ R, (y, z) ∈ R

then |x + y| < 2 and |y + z| < 2 for x, y, z ∈ (-1, 1)

⇒ |x + z| < 2

So, (x, z) ∈ R.

R is transitive.

Therefore R is equivalence relation.

Hence option (4) is true.

Explanation:

Venn diagram of {(A - B) ∪ (B - A) ∪ (A ∩ B)} is

Hence we can see that {(A - B) ∪ (B - A) ∪ (A ∩ B)} = A ∪ B

So, (A ∪ B) - {(A - B) ∪ (B - A) ∪ (A ∩ B)} = (A ∪ B) - (A ∪ B) = ϕ 

Hence (A ∪ B) - {(A - B) ∪ (B - A) ∪ (A ∩ B)} is equal to null set.

Option (1) is true.

Concept:

Cosine rule: Let ABC be a triangle with angles A, B, C are corresponding sides a, b, c then 

Cos A = \({b^2+c^2-a^2 \over 2bc}\)

Explanation:

\(\left|\begin{array}{ccc} a^2 & b \sin A & c \sin A \ b \sin A & 1 & \cos A \ c \sin A & \cos A & 1 \end{array}\right|\)

= a²(1 - cos²A) + b sin A(c sin A cos A - b sin A) + c sin A(b sin A cos A - c sin A)

= a2(1 - cos2A) + bc sin² A cos A - b² sin² A + bc sin² A cos A - c² sin² A

= a2sin2A + 2bc sin2 A cos A - (c² + b2)sin2 A

= - sin²A(b² + c² - a² - 2bc cos A)

= - sin2A(2bc cos A - 2bc cos A)

= 0 

Hence option (1) is true.

Concept:

Arithmetic progression (AP): Three number a, b, c are called in AP if 2b = a + c.

Geometric progression (GP): Three numbers a, b, c are called in GP if b² = ac.

Harmonic progression (HP): Three numbers a, b, c are called in HP if \({2\over b}=\frac1a+\frac1c\).

Explanation:

If we take a = 2, b = 4, c = 6, d = 9, e = 18 then 

a, b, c are in AP

b, c, d are in GP and

c, d, e are in HP

Now, c² = 6² = 36 = 2 × 18 = ae

So, a, c, e are in GP

1. is true.

\({1\over c^2}=\frac1{36}\)

and \(\frac1a.\frac1e=\frac{1}{36}\) 

So, \(\frac{1}{a}, \frac{1}{c}, \frac{1}{e}\) are in GP

2. is true.

Hence option (3) is the correct answer.

Concept:

(i) log_ax exits only if x > 0

(ii) \(\log_{b^n}a=\frac1n\log_ba\)

(iii) log_axⁿ = n log_ax 

Explanation:

log4(x - 1) = log2(x - 3)

⇒ log2²(x - 1) = log2(x - 3)

⇒ \(\frac12\)log₂(x - 1) = log₂(x - 3)

⇒ log2(x - 1)^1/2 = log2(x - 3)

⇒ (x - 1)1/2 = (x - 3)

Squaring both sides

⇒ x - 1 = x² - 6x + 9

⇒ x2 - 7x + 10 = 0

⇒ (x - 5)(x - 2) = 0

⇒ x = 2, 5

If we take x = 2 then the right hand side will be log₂(-1) which is not possible.

So x can't be 2

Hence x = 5 is the only solution i.e., number of solution is one.

Hence option (2) is true.

Concept:

(i) logax exits only if x > 0

(ii) \(\log_{b^n}a=\frac1n\log_ba\)

(iii) logaxn = n logax 

(iv) log_a x + log_a y = log_a xy

(v) loga x - loga y = loga (x/y)

(vi) log_ax = \(\log x\over \log a\)

Explanation:

\(\log _x\left(\frac{x}{y}\right)+\log _y\left(\frac{y}{x}\right)\) = k

⇒ \({\log\left(\frac{x}{y}\right)\over \log x}+{\log\left(\frac{y}{x}\right)\over \log y}\) = k

⇒ \({\log x-\log y\over \log x}+{\log y-\log x\over \log y}\) = k

⇒ \({\log x.\log y-(\log y)^2+\log x\log y-(\log x)^2\over \log x\log y}\) = k

⇒ \({-{(\log x)^2-2\log x\log y+(\log y)^2}\over \log x\log y}\) = k

⇒ \({-(\log x-\log y)^2\over \log x\log y}\) = k

Now, since \(x ≥slant y>1\) so

log x > 0, log y > 0

also, (log x - log y)² ≥ 0 for any x, y

Hence k can be zero or negative i.e., k can never be positive

Hence option (4) is true.

Concept:

(i) A square matrix A is called symmetric matrix if A = AT and is called skew symmetric if A = - A^T.

(ii) A square matrix A is called orthogonal if AAT = I.

Explanation:

A = \(\left[\begin{array}{ccc}\sin 2 θ & 2 \sin ^2 θ-1 & 0 \ \cos 2 θ & 2 \sin θ \cos θ & 0 \ 0 & 0 & 1\end{array}\right]\)

   = \(\left[\begin{array}{ccc}\sin 2 θ & -\cos 2θ & 0 \ \cos 2 θ & \sin 2θ & 0 \ 0 & 0 & 1\end{array}\right]\) 

(as 2sin x cos x = sin2x and 2sin²x = 1- cos 2x)

A^T = \(\left[\begin{array}{ccc}\sin 2 θ & \cos 2θ & 0 \ -\cos 2 θ & \sin 2θ & 0 \ 0 & 0 & 1\end{array}\right]\)

i.e., A ≠  A^T

So, A is not symmetric matrix. 

2. is false.

Now, |A| = \(\begin{vmatrix}\sin 2 θ & -\cos 2θ & 0 \ \cos 2 θ & \sin 2θ & 0 \ 0 & 0 & 1\end{vmatrix}\) = 1(sin²θ + cos²θ) = 1

and adj(A) = \(\left[\begin{array}{ccc}\sin 2 θ & \cos 2θ & 0 \ -\cos 2 θ & \sin 2θ & 0 \ 0 & 0 & 1\end{array}\right]\)

So, A^-1 = \({1\over |A|}adj(A)\) = \(\left[\begin{array}{ccc}\sin 2 θ & \cos 2θ & 0 \ -\cos 2 θ & \sin 2θ & 0 \ 0 & 0 & 1\end{array}\right]\)

Hence, A-1 = AT and A-1 = adj(A) 

1. is true and 3. is true.

Hence option (3) is correct.

Explanation:

\(\left(2-x^2-\frac{1}{x^2}\right)^{-5}\)

= - \(\left(x^2+\frac{1}{x^2}-2.x.\frac1x\right)^{-5}\)

= - \(\left[(x-\frac1x)^2\right]^{-5}\)

= - \(\left(x-\frac1x\right)^{-10}\)

= - \(\left(\frac{x^2-1}x\right)^{-10}\)

= - \(\left(\frac{x}{x^2-1}\right)^{10}\) = - \(\frac{x^{10}}{(x^2-1)^{10}}\)

So, \(\left(1-x^2\right)^{20}\left(2-x^2-\frac{1}{x^2}\right)^{-5}\)

= - (x² - 1)²⁰\(\frac{x^{10}}{(x^2-1)^{10}}\)

= - x¹⁰(x2 - 1)10

Hence the coefficient of x10 is

= - (constant term of the expression (x2 - 1)10)

= - 1

Option (1) is true.

Concept:

(i) The general term of the expression (a + b)ⁿ is

T_r+1 = \(^nC_ra^{n-r}b^{r}\)

(ii) \(^nC_r={n!\over r!(n-r)!}\)

Explanation:

4^th term in the expansion of \(\left(m x+\frac{1}{x}\right)^n \) is 

T3+1 = \(^nC_3(mx)^{n-3}.(\frac1x)^{3}\) = \(^nC_3m^{n-3}.x^{n-6}\)

By given condition, 4^th term is \(\frac{5}{2}\) which is independent of x. 

So, n - 6 = 0 i.e., n = 6

Therefore

\(^6C_3m^{3}\) = \(\frac{5}{2}\)

⇒ 20m³ = 5/2

⇒ m3 = 1/8

⇒ m = 1/2

Therefore mn = 6 × \(\frac12\) = 3

Hence option (2) is true.

Explanation:

a, b, c are in GP so b² = ac...(i)

Also, a > 0 and c > 0 so, ac > 0

Now, roots of  ax2 + bx + c = 0 are

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) = \({-b \pm \sqrt{ac-4ac} \over 2a}\) = \({-b \pm \sqrt{-3ac} \over 2a}\)

Since ac > 0 so -3ac < 0

So roots are imaginary.

1. is true.

ratio of the roots

= \({-b + \sqrt{-3ac} \over 2a}\over {-b - \sqrt{-3ac} \over 2a}\) 

= \({-\sqrt{ac} + \sqrt{-3ac} \over -\sqrt{ac}-\sqrt{-3ac}}\) (using (i))

= \({-1 + \sqrt{3}i \over -1-\sqrt{3}i}\) 

= \(2ω\over 2ω^2\) = \(1\over \omega\)

(as ω is a cube root of unity i.e., ω = \(-1+\sqrt3i\over 2\) and ω² = \(-1-\sqrt3i\over 2\))

2. is true.

The product of the roots

= \({-b + \sqrt{-3ac} \over 2a}\).\({-b- \sqrt{-3ac} \over 2a}\) 

= \({b^2 -(-3ac) \over 4a^2}\) 

= \({b^2 -(-3ac) \over 4a^2}\) 

= \({b^2 +3b^2 \over 4a^2}\) (using (i))

= \({4b^2 \over 4a^2}\) = \({b^2 \over a^2}\)

3. is true.

Hence option (4) is correct option

Explanation:

x2 + mx + n

= x2 + 2.\(\frac m2\).x + \(m^2\over4\)+ n - \(m^2\over4\)

= (x + \(\frac m2\))² + n - \(m^2\over4\)

So, x2 + mx + n will be an integer if

x + \(\frac m2\) and n - \(m^2\over4\) will be an integer for all integral value of x

Now, x + \(\frac m2\) is integer implies \(\frac m2\) is integer (say k)

Them m = 2k i.e., m is an even integer.

1. is true 

n - \(m^2\over4\) is integer implies n - \(4k^2\over 4\) is integer i.e., n - k² is integer so n is integer.

2. is true.

Hence option (3) is correct.

Concept: 

The rth term of the expansion (a + b)ⁿ is ⁿC_ra^n-rb^r.

Explanation:

 (x + y)2n + 1(x - y)2n + 1

=  {(x + y)(x - y)}2n + 1

= (x² - y²)2n + 1

if we take n = 1 then expression is (x2 - y2)3

Number of terms = 4 which is even

So, the expansion has 2 middle terms.

2^nd term and 3^rd term

2^nd term is - 3C1(x²)2y2 = - 3x⁴y²

3^rd term is 3C2x2y4 = 3x2y4

By give condition,

• 3x4y2 + 3x2y4 = 0

⇒ x4y2 = x2y4

⇒ x² = y² ⇒ \(\left(\frac{x^2}{y^2}\right)\) = 1

Option (1) is true.

Concept:

Number of onto function from a set A (containing m elements) to a set B (containing n elements) (m ≥ n) is given by

n^m - ⁿC₁(n - 1)^m + nC2(n - 2)m - ...

Explanation:

A = {1, 2, 3, 4, 5}, B = {6, 7}

n(A) = m = 5 and n(B) = n = 2

So number of onto functions from A to B is

2⁵ - 2C1(2 - 1)5 

= 32 - 2 = 30

Option (3) is true.

Concept:

Formula: sin a cos b - cos a sin b = sin(a - b)

sin2x = 2 sin x cos x

Explanation:

\(​\frac{\sqrt{3} \cos 10^{\circ}-\sin 10^{\circ}}{\sin 25^{\circ} \cos 25^{\circ}}\)

= \(​\frac{{\sqrt{3}\over2} \cos 10^{\circ}-\frac12\sin 10^{\circ}}{\frac12\sin 25^{\circ} \cos 25^{\circ}}\)

= \(​\frac{\sin60^{\circ} \cos 10^{\circ}-\cos60^{\circ}\sin 10^{\circ}}{\frac12\sin 25^{\circ} \cos 25^{\circ}}\)

= \(​\frac{2\sin(60^{\circ}-10^{\circ})}{\sin 25^{\circ} \cos 25^{\circ}}\) (using the formula)

= \(​\frac{2\sin50^{\circ}}{\sin 25^{\circ} \cos 25^{\circ}}\)

= \(​\frac{2\sin(2\times25^{\circ})}{\sin 25^{\circ} \cos 25^{\circ}}\)

= \(​\frac{2\times 2\sin 25^{\circ} \cos 25^{\circ}}{\sin 25^{\circ} \cos 25^{\circ}}\)

= 4 

Option (4) is true.

Concept:

cos 36° = \(\sqrt5+1\over 4\)

Explanation:

(sin 9° - cos 9°)

= \(\sqrt2\)(\(\frac1{\sqrt2}\)sin 9° - \(\frac1{\sqrt2}\)cos 9°)

= \(\sqrt2\)(sin 9° cos 45° - cos 9° sin 45°)

= \(\sqrt2\)sin(-36°) (sin a cos b - cos a sin b = sin(a - b))

= - \(\sqrt2\) sin(36°)

= - \(\sqrt2\) \(\sqrt{1-\cos^236^{\circ}}\)

= - \(\sqrt2\)(\sqrt{1-({\sqrt5+1\over 4})^2}\)

= - \(\sqrt2\)(\sqrt{{16-5-2\sqrt5+1\over 16}}\)

= - \(\sqrt2\)(\sqrt{{10-2\sqrt5\over 16}}\)

= - \(\sqrt{{20-4\sqrt5\over 16}}\)

= \(-\frac{\sqrt{5-\sqrt{5}}}{2}\)

Hence option (1) is true.

Concept:

(i) sine rule: Let ABC be a triangle with angles A, B, C and corresponding sides a, b, c then 

\({\sin A\over a}={\sin B\over b}={\sin C\over c}\)

(ii) x³ + y³ + z³ = 3xyz then x + y + z = 0

Explanation:

ABC is a triangle with sides a, b, c and angles A, B, C then 

\({\sin A\over a}={\sin B\over b}={\sin C\over c}\) = k (say)

⇒ sin A = ka, sin B = kb, sin C = kc...(i)

Now, sin3 A + sin3 B + sin3 C = 3 sin A sin B sin C 

⇒ sin A + sin B + sin C = 0

⇒ ka + kb + kc = 0 (using (i))

⇒ a + b + c = 0....(ii)

\(\left|\begin{array}{lll}a & b & c \ b & c & a \ c & a & b\end{array}\right|\)

= \(\left|\begin{array}{lll}a+b+c & b & c \ a+b+c & c & a \ a+b+c & a & b\end{array}\right|\) (C₁ = C₁ + C₂ + C₃)

= \(\left|\begin{array}{lll}0 & b & c \ 0 & c & a \ 0 & a & b\end{array}\right|\) (using (ii))

= 0

Option (4) is true.

Explanation:

cos-1 x = sin-1 x

⇒ sin-1 \(\sqrt{1-x^2}\) = sin-1 x

⇒ \(\sqrt{1-x^2}\) = x

⇒ 1 - x² = x² (squaring both sides)

⇒ 2x² = 1

⇒ x = \(\frac12\)

⇒ x = \(\frac{1}{\sqrt{2}}\)

Option (3) is true.

Concept:

(i) sin2θ + cos2θ = 1

(ii) 2sin θ cos θ = sin 2θ

Explanation:

(sin θ - cos θ)2 = 2 where -π < θ < π

⇒ sin²θ + cos²θ - 2sin θ cos θ = 2

⇒ 1 - 2sin θ cos θ = 2

⇒ 2sin θ cos θ = -1

⇒ sin 2θ = -1

⇒ θ = nπ + \(3\pi\over 4\), n ∈ \(\mathbb Z\)

Now, -π < θ < π

So, solution θ = - \(\pi\over 4\), \(3\pi\over 4\)

Therefore number of solution is two.

Option (2) is true.