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Correct answer is (B) equal 

Correct answer is (D) 29/99

We have A = IA

\(\begin{bmatrix}2&1&1\\3&2&1\\2&1&2\end{bmatrix}=\) \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)A

Applying R₃ → R₃ — R₁

R₂ → 2R₂ — 3R₁

\(\begin{bmatrix}2&1&1\\0&1&-1\\0&0&1\end{bmatrix}=\)\(\begin{bmatrix}1&0&0\\-3&2&0\\-1&0&1\end{bmatrix}\)A

Applying R₂ → R₂ + R₃

\(\begin{bmatrix}2&1&1\\0&1&0\\0&0&1\end{bmatrix}=\)\(\begin{bmatrix}1&0&0\\-4&2&1\\-1&0&1\end{bmatrix}\)A

Applying R₁ → R₁ — R₂ — R₃

\(\begin{bmatrix}2&0&0\\0&1&0\\0&0&1\end{bmatrix}=\) \(\begin{bmatrix}6&-2&-2\\-4&2&1\\-1&0&1\end{bmatrix}\)A

Applying R₁ → \(\frac{R_1}2\)

\(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\)\(\begin{bmatrix}3&-1&-1\\-4&2&1\\-1&0&1\end{bmatrix}\)A

Hence A^-1 = \(\begin{bmatrix}3&-1&-1\\-4&2&1\\-1&0&1\end{bmatrix}\)

\(3\sqrt{[-2(x +3)] - 1} = |x + 3| + a\)

For existence

\([-2(x + 3)] - 1 \ge 0\)

⇒ \([-2(x +3)] \ge 1\)

⇒ \(-2(x + 3) \ge 1\)

⇒ \(x + 3 \le \frac{-1}2\)

⇒ \(x \le \frac {-1}2 -3\)

⇒ \(x \le \frac{-5}2\)

Also \(|x + 3| \ge \frac 12\)

\(|x + 3| + a \ge \frac 12 + a\)

√3 sinθ = cosθ 

\(\frac{sin\theta}{cos\theta}\) = \(\frac{1}{\sqrt{3}}\) 

tanθ = \(\frac{1}{\sqrt{3}}\) 

θ = 30˚

Concept used :-

Trick to multiply two numbers ending in 5 with difference of 10 between them.

Method :-

• Write 75 at the end.
• Add 1 to ten’s place of the bigger number.
• Now multiply the increased digit with the ten’s digit of the smaller number.
• Write the product so obtained on the left side of 75

Calculation :-

Let the answer be A75.

A = (4 + 1) × 3 = 15 (using step 2 and 3)

We need to find the Greatest Common Factor (HCF).

Thus, (3663 – 2295) = 1368

(6399- 2295) = 4104

(6399 – 3663) = 2736

So, factors of 1368 = 8 × 9 × 19 = 8 × 171

Factors of 4104 = 8 × 19 × 27 = 24 × 171

Factors of 2736 = 16 × 9 × 19 = 16 × 171

Therefore,

The three digit number is 171.

So, Numbers are –

2295 = (171 × 13) + 72

3663 = (171 × 21) + 72

6399 = (171 × 37) + 72

∴ The required three digit number is 171 and the remainder is 72.

(b) Sir Thomas Roe

Given that cos\(\theta\) = 0.6. 

Therefore sin\(\theta\) = \(\sqrt{1 − cos^2\theta}\) = \(\sqrt{1 − (0.6)^2}\) = \(\sqrt{1 − 0.36}\) = \(\sqrt{0.64 }\) = 0.8. 

And tan\(\theta\) = \(\frac{sin \theta}{cos \theta}\) = \(\frac{0.8}{0.6}\) =\(\frac{8}{6}\) = \(\frac{4}{3}\) . 

Now, 5 sin\(\theta\) – 3tan\(\theta\) = 5 × 0.8 − 3 × \(\frac{4}{3}\) = 4 – 4 = 0.

Hence Proved

sin²15° + sin²75°

⇒ [sin (90 – 75)]² + sin²75°

⇒ cos²75° + sin²75

⇒ 1

Correct option is B) am/have taken

\(S_n = 1 + 4 + 1 +...+ 3n - 2\)    \(\begin{pmatrix}\because \text{nth term of sequence is}\\a_n = a+ (n - 1)d\\= 1+(n-1)3\\= 3n- 2\end{pmatrix}\) 

\(= \sum (3n - 2)\)

\(= 3\sum n - 2\sum 1\)

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

\(= \frac{3n^2+ 3n - 4n}{2}\)

\(= \frac{3n^2 - n}{2}\)

\(= \frac{n(3n -1)}{2}\)

The main business of an investment company is to hold and manage securities for investment purposes, but they typically offer investors a variety of funds and investment services, which include portfolio management, recordkeeping, custodial, legal, accounting and tax management services.

(a) If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so during new Moon we can observe solar eclipse.

But Moon’s orbit is tilted 5° with respect to Earth’s orbit. Due to this 5° tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment.

(b) The common misconception is that ‘Earth revolves around the Sun, so when the Earth is very far away, it is winter and when the Earth is nearer, it is summer’.

Actually, the seasons in the Earth arise due to the rotation of Earth around the Sun with 23.5° tilt. Due to this 23.5° tilt, when the northern part of Earth is farther to the Sun, the southern part is nearer to the Sun. So when it is summer in the northern hemisphere, the southern hemisphere experience winter.

(I) The vector product of any two vectors is always another vector whose direction is perpendicular to the plane containing these two vectors, i.e., orthogonal to both the vectors \(\vec A\) and \(\vec B\) , even though the vectors A and B may or may not be mutually orthogonal.

(II) The vector product of two vectors is not commutative, i.e., \(\vec A\)x \(\vec B\) ≠ \(\vec B\) x \(\vec A\). But \(\vec A\) x \(\vec B\) = -\(\vec B\) x \(\vec A\).

Here it is worthwhile to note that |\(\vec A\)×\(\vec B\)| = |\(\vec B\) ×\(\vec A\) | = AB sin θ i.e., in the case of the product vectors \(\vec A\)x \(\vec B\) and \(\vec B\) x \(\vec A\), the magnitudes are equal but directions are opposite to each other.

(III) The vector product of two vectors will have maximum magnitude when sin 0 = 1, i.e., θ = 90° i.e., when the vectors \(\vec A\) and \(\vec B\) are orthogonal to each other.

 (\(\vec A\)×\(\vec B\))_max = AB\(\hat n\)

(IV) The vector product of two non-zero vectors will be minimum when sin 0 = 0, i.e., 0 = 0° or 180°  (\(\vec A\)×\(\vec B\))_max = 0

i.e., the vector product of two non-zero vectors vanishes, if the vectors are either parallel or antiparallel.

(V) The self-cross product, i.e., product of a vector with itself is the null vector 

\(\vec A\)×\(\vec A\) = AA sin 0° \(\hat n\) = \(\vec 0\) 

In physics the null vector \(\vec 0\) is simply denoted as zero.

After taking approximation equation will be 60 ÷ 12 x 6 = 30

\(59.88\div 12.21 \times 6.35\)

\(= \frac{59.88}{12.21}\times 6.35\)  (If brackets are not given then we first solve division, then multiplication)

\(\simeq4.9 \times 6.35\)

= 31.115

3x - 7y + 10 = 0  ------(1)

y - 2x -3 = 0  -----(2)

from (2) , we get 

y = 2x + 3

put y = 2x + 3 in equation (1) , we get

3x - 14x - 21 + 10 = 0

⇒ -11x - 11 = 0

⇒ x = 11/-11 = -1

∴ y = -2 + 3 = 1

    2x+3y-6=0

       a=2 ,    b=3 ,      c=-6

And 

   14x+21y-42=0

      a=14 ,     b=21 ,   c=-42

a1/a2=b1/b2=c1/c2

2/14=3/21=-6/-42

1/7=1/7=1/7

  Hence ,the pair of linear equation is parallel and they will not intersect at any point.

∵ x + y = 12

घन करने पर x³ + y³ + 3xy (x + y) = 1728

x³ + y³ + 3 × 27 (12) = 1728

x³ + y³ + 972 = 1728

x³ + y³ = 1728 – 972 = 756

One term – (v) 5x²y, viii) 3x/7y

Two terms –  (i) x + y , (iv) x²z + 3, (vii) x – 11/3, (ix) 2z – y (x) 3x + 5

Three terms – (ii) 11x – 3y – 5, (iii) 6x² + 5x – 4, (vi) x + 3 + y

0.4x + 0.3y = 1.7 ----(1)

0.7x + 0.2y = 0.8 -----(2)

from (2) we get y = 0.8 - 0.7x / 0.2

substitute y = 0.8 - 0.7x/0.2 in equation (1), we get 

0.4x + 0.24 - 0.21x / 0.2 = 1.7

⇒ 0.08x + 0.24 - 0.21x = 0.34

⇒ -0.13x = 0.34 - 0.24 = 0.10 

⇒ x = 0.10 / - 0.13 = -10 / 13

∴ y = \(\frac{0.8-0.7\times\frac{-10}{13}}{0.2}\) 

= \(\frac{8+\frac{70}{13}}{2}=\frac{174}{26}=\frac{87}{13}\) 

\(\frac{x}{x-y}+\frac{y}{x+y}+\frac{2xy}{x^2+y^2}\) 

 = \(\frac{x(x+y)+y(x-y)}{(x-y)(x+y)}+\frac{2xy}{x^2+y^2}\) 

 = \(\frac{x^2+xy+xy-y^2}{x^2-y^2}+\frac{2xy}{x^2+y^2}\) 

 = \(\frac{x^2+2xy-y^2+(x^2+y^2)+2xy(x^2-y^2)}{(x^2-y^2)(x^2+y^2)}\)

 = \(\frac{x^4+2x^3y-x^2y^2+x^2y^2+2xy^3-y^4+2x^3y+2xy^3}{x^4-y^4}\) 

 = \(\frac{x^4+4x^3y-y^4}{x^4-y^4}\)

(a + b)² – 2ab = a² + b²

= (a + b)² – 2ab

= a² + 2ab + b² – 2ab

= a² + b²

(i) (x + 3)(x + 7) 

= x² + x(3 + 7) + 21 

= x² + 10x + 21

(ii) (6a + 9)(6a - 5) 

= (6a)² + 6a(9 - 5) +9x - 5 

= 36a² + 24a - 45

(iii) (4x + 3y) (4x + 5y) 

= (4x)² + 4x(3y + 5y) + 3y x 5y 

= 16x² + 32xy + 15y²

(iv) (8 + pq)(pq + 7)

= (pq + 8)(pq + 7) 

= (pq)² + pq(8 + 7) + 8 x 7

= p²q² + 15pq + 56

(n + 1)² - n²

(n + 1)(n + 1) - n²

n(n + 1)+ 1(n + 1) - n²

n² + n + 1(n + 1) - 1n²

n² + n + n + 1 - 1n²

n² + 2n + 1 - 1n²

2n + 1

a² + b - ab - a

= a² - ab + b - a

= a(a - b) - (a - b)

= (a - b) (a - 1)

1. Astronomy 

2. Geology