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8x³ + 27y³ − 216z³ + 108xyz 

= (2x)³ + (3y)³ +(−6y)³ −3(2x)(3y)(−6z) 

= (2x+3y+(−6z)){ (2x)²+(3y)²+(−6z)² −2x×3y−3y(−6z)−(−6z)2x} 

= (2x+3y−6z) {4x² +9y² +36z² −6xy + 18yz + 12zx}

Let f (x) = x³ - 6x² - 19x + 84 be the given polynomial.

In order to prove that (x + 4), (x – 3) and (x – 7) are factors of f (x), it is sufficient to prove that f (-4) = 0, f (3) = 0 and f (7) = 0 respectively.

Now,

f (x) = x³ - 6x² - 19x + 84

f (-4) = (-4)³ – 6 (-4)² – 19 (-4) + 84

= -64 – 96 + 76 + 84

= 0

f (3) = (3)³ – 6 (3)² – 19 (3) + 84

= 27 – 54 – 57 + 84

= 0

f (7) = (7)³ – 6 (7)² – 19 (7) + 84

= 343 – 294 – 133 + 84

= 0

Hence, 

(x – 4), (x – 3) and (x -7) are the factors of the given polynomial x³ - 6x² - 19x + 84.

1. Eye lens changes its focal length by the ciliary muscle attached to it.

2. By relaxing ciliary muslces, the focal length of the eye lens is reached its maximum value. 

3. By straining ciliary muscles, the focal length of the eye lens is reached its minimum value.

1) Continuously changing atmosphere refracts light from the stars by different amounts from one moment to the other, when atmosphere refracts more starlight towards us and the stars appear to be bright and when the atmosphere refracts less star-light then the stars appear to be dim.

2) However the planets are nearer to us than the stars, they appear to be comparatively bigger to us so they cannot be considered as a point source, hence no twinkling is seen.

Let, f (x) = x³ - 3x² + ax - 10 be the given polynomial. 

By factor theorem, 

If (x – 5) is a factor of f (x) then f (5) = 0 

Now, 

f (x) = x³ - 3x² + ax - 10 

f (5) = (5)³ – 3 (5)² + a (5) – 10 

0 = 125 – 75 + 5a – 10 

0 = 5a + 40 

a = -8 

Hence, 

(x – 5) is a factor of f (x), if a = - 8.

Let, f (x) = 4x⁴ + 2x³ - 3x² + 8x + 5a

f (-2) = 0

4 (-2)⁴ + 2 (-2)³ – 3 (-2)² + 8 (-2) + 5a = 0

64 – 16 – 12 – 16 + 5a = 0

5a = - 20

a = -4

Hence, 

(x + 2) is a factor f (x) when a = -4.

Let f (x) = x³ + ax² - bx + 10 and g (x) = x² - 3x + 2 be the given polynomials.

We have g (x) = x²- 3x+2 = (x – 2) (x – 1)

Clearly, (x -1) and (x – 2) are factors of g (x)

Given that f (x) is divisible by g (x)

g (x) is a factor of f (x)

(x – 2) and (x – 1) are factors of f (x)

From factor theorem

f (x – 1) and (x – 2) are factors of f (x) then f (1) = 0 and f (2) = 0 respectively.

f (1) = 0

(1)³ + a (1)² – b (1) + 10 = 0

1 + a – b + 10 = 0

a – b + 11 = 0 (i)

f (2) = 0

(2)³ + a (2)² - b (2) + 10 = 0

8 + 4a – 2b + 10 = 0

4a – 2b + 18 = 0

2 (2a – b + 9) = 0

2a – b + 9 = 0 (ii)

Subtract (i) from (ii), we get

2a – b + 9 – (a – b + 11) = 0

2a – b + 9 – a + b – 11 = 0

a – 2 = 0

a = 2

Putting value of a in (i), we get

2 – b + 11 = 0

b = 13

Hence, 

a = 2 and b = 13

Let, f (x) = k²x³ - kx² + 3kx - k

By factor theorem, 

If (x – 3) is a factor of f (x) then f (3) = 0 

k² (3)³ – k (3)² + 3 k (3) – k = 0 

27k² – 9k + 9k – k = 0 

k (27k – 1) = 0 

k = 0 or (27k – 1) = 0 

k = 0 or k = \(\frac{1}{27}\)

Hence,

(x – 3) is a factor of f (x) when k = 0 or k = \(\frac{1}{27}.\)

Let f(x) = 5x³ - 7x² - ax - 28 be the given polynomial. 

From factor theorem, 

If (x -4) is a factor of f (x) then f (4) = 0 

f (4) = 0 

0 = 5 (4)³ – 7 (4)² – a (4) – 28 

0 = 320 – 112 – 4a – 28 

0 = 180 – 4a 

4a = 180 

a = 45 

Hence, 

(x – 4) is a factor of f (x) when a = 45.

Let, f (x) = x⁴ + px³ + 2x² - 3x + q be the given polynomial.

And, let g (x) = (x² – 1) = (x – 1) (x + 1)

Clearly,

(x – 1) and (x + 1) are factors of g (x)

Given, 

g (x) is a factor of f (x)

(x – 1) and (x + 1) are factors of f (x)

From factor theorem

If (x – 1) and (x + 1) are factors of f (x) then f (1) = 0 and f (-1) = 0 respectively.

f (1) = 0

(1)⁴ + p (1)³ + 2 (1)² – 3 (1) + q = 0

1 + p + 2 – 3 + q = 0

p + q = 0 (i)

Similarly,

f (-1) = 0

(-1)⁴ + p (-1)³ + 2 (-1)² - 3 (-1) + q = 0

1 – p + 2 + 3 + q = 0

q – p + 6 = 0 (ii)

Adding (i) and (ii), we get

p + q + q – p + 6 = 0

2q + 6 = 0

2q = - 6

q = -3

Putting value of q in (i), we get

p – 3 = 0

p = 3

Hence, 

x² – 1 is divisible by f (x) when p = 3 and q = - 3.

Let, f (x) = ax⁴ + 2x³ - 3x² + bx - 4 and g (x) = x² – 4

We have,

g (x) = x² – 4

= (x – 2) (x + 2)

Given,

g (x) is a factor of f (x)

(x – 2) and (x + 2) are factors of f (x)

From factor theorem if (x – 2) and (x + 2) are factors of f (x) then f (2) = 0 and f (-2) = 0 respectively.

f (2) = 0

a x (-2)⁴ + 2 (2)³ – 3 (2)² + b (2) – 4 = 0

16a – 16 – 12 + 2b – 4 = 0

16a + 2b = 0

2 (8a + b) = 0

8a + b = 0 (i)

Similarly,

f (-2) = 0

a x (-2)⁴ + 2 (-2)³ – 3 (-2)² + b (-2) – 4 = 0

16a – 16 – 12 - 2b – 4 = 0

16a - 2b – 32 = 0

16a – 2b – 32 = 0

2 (8a - b) = 32

8a – b = 16 (ii)

Adding (i) and (ii), we get

8a + b + 8a – b = 16

16a = 16

a = 1

Put a = 1 in (i), we get

8 x 1 + b = 0

b = -8

Hence, 

a = 1 and b = -8.

Let, f (X) = ax³ + x² - 2x + b be the given polynomial.

Given (x + 1) and (x – 1) are factors of f (x).

From factor theorem,

If (x + 1) and (x – 1) are factors of f (x) then f (-1) = 0 and f (1) = 0 respectively.

f (-1) = 0

a (-1)³ + (-1)² – 2 (-1) + b = 0

- a + 1 + 2 + b = 0

- a + 3 + b = 0

b – a + 3 = 0 (i)

f (1) = 0

a (1)³ + (1)² – 2 (1) + b = 0

a + 1 – 2 + b = 0

a + b – 1 = 0

b + a – 1 = 0 (ii)

Adding (i) and (ii), we get

b – a + 3 + b + a – 1 = 0

2b + 2 = 0

2b = - 2

b = - 1

Putting value of b in (i), we get

- 1 - a + 3 = 0

- a + 2 = 0

a = 2

Hence, 

the value of a = 2 and b = - 1.

Let, f (x) = x³ + 3x² - 2αx + β be the given polynomial,

From factor theorem,

If (x + 1) and (x + 2) are factors of f (x) then f (-1) = 0 and f (-2) = 0

f (-1) = 0

(-1)³ + 3 (-1)² – 2 α (-1) + β = 0

-1 + 3 + 2 α + β = 0

2 α + β + 2 = 0 (i)

Similarly,

f (-2) = 0

(-2)³ + 3 (-2)² – 2 α (-2) + β = 0

-8 + 12 + 4 α + β = 0

4 α + β + 4 = 0 (ii)

Subtract (i) from (ii), we get

4 α + β + 4 – (2 α + β + 2) = 0 – 0

4 α + β + 4 - 2 α - β - 2 = 0

2 α + 2 = 0

α = -1

Put α = -1 in (i), we get

2 (-1) + β + 2 = 0

β = 0

Hence, 

α = -1 and β = 0.

If the wavelength of light incident on a convex lens is increased, its focal length will also increase.

(a) Chlorine

(c) Assertion is true and the reason is false

(a) Y > Z > X > A

Correct option is B) Concave

We can write it as

|x + 4| ≥ 0

By taking – ve sign

– |x + 4| ≤ 0

By adding 6 on both sides

– (x + 4) + 6 ≤ 0 + 6

Where f (x) ≤ 6

Here the maximum value of f (x) = 6 which occurs at x = – 4 which is the point of absolute maxima and minimum value of f (x) does not exists.

It is given that

f(x) = x²

By differentiating w.r.t. x

f’(x) = 2x and f”(x) = 2

We know that

f’(x) = 0

By substituting the values

2x = 0 where x = 0

f”(0) = 2 > 0

Hence, x = 0 is point of local minimum and f (0) = 0 is the local minimum value.

For all x

– 1 ≤ sin 2x ≤ 1

By adding 5

5 – 1 ≤ sin 2x + 5 ≤ 5 + 1

On further calculation

4 ≤ f (x) ≤ 6

Here the minimum value of f (x) = 4 and minimum value of f (x) = 6.

It is given that

f (x) = (x – 3)⁴

By differentiating w.r.t. x

f’(x) = 4 (x – 3)³

We know that

f’(x) = 0

By substituting the values

4 (x – 3)³ = 0

So we get

x – 3 = 0 where x = 3

x = 3 is point of local minima and local minimum value is f (3) = 0

For all x

– 1 ≤ sin 4x ≤ 1

By adding 3

3 – 1 ≤ sin 4x + 3 ≤ 1 + 3

On further calculation

2 ≤ sin 4x + 3 ≤ 4

We know that

|2| ≤ |sin 4x + 3| ≤ |4|

2 ≤ f (x) ≤ 4

Here the minimum value of f (x) = 2 and maximum value of f (x) = 4.

max. value = 9

Since the quantity (x -3)² has a –ve sign, the max.

Value it can have is 9.

Also hence it has no minimum value.

It is given that f (x) = (5x – 1)² + 4 with minimum value 4 where 5x – 1 = 0

Here the maximum value cannot be found i.e. when x increases even f (x) increases

Therefore, f (x) does not have a maximum value.

Answer is (C) 16

Answer is (d) √422

Answer is (c) 61

Answer is (a) 18

Let vector a = i - j + k then 

vector |a| = √(1₂ + (-1)² + 1²) = √3

The unit vector in the direction of vector a is given by 

a = vector(a/|a|) = (1 - j + k)/√3 = (1/√3)i - (1/√3)j + (1/√3)k

∴ l = cos α = 1/√3, m = cos β = -1/√3, n = cos γ = 1/√3

α = cos^-1(1/√3), β = cos^-1(-1/√3), γ = cos^-1(1/√3)

∫ sec x dx

∫sec x(sec x + tan x)/(sec x + tan x) dx

∫(sec² x + sec x tan x)/(sec x + tan x) dx

Let, sec x + tan x = t

(sec x + tan x + sec² x) dx = dt

∫(dt/dt) = log_et = log_e(sec x + tan x) + c

Let A (1, 2, 3) and B (3, 2, -1) be the given points. 

Let P(x, y, z) be any point equidistant from A and B, then

PA = PB (i.e., PA² = PB²)

⇒ √((x - 1)² + (y - 2)² + (z - 3)²)

= √((x - 3)² + (y - 2)² + (z + 1)²)

⇒ x² - 2x + 1+ y² – 4y + 4 + z²  + 2z + 1 

⇒ x²- 6x + 9 + y²- 4y + 4 + z²  + 2z + 1 

⇒ 4x – 8z = 0 

⇒ x – 2z = 0 is the required equation.

Answer is (b) log|sec x + tan x| + c

(c) 24

Numbers greater than 60000 will have either 6 or 8 in the TTh place and will consist of 5-digits. 

If the digit 6 occupies the TTh place, the remaining 4 places can be occupied in \(\frac{4!}{2!}\) ways.       (∵ There are two 2’s) 

Number of numbers beginning with 6 = \(\frac{4!}{2!}\) = 12 

Similarly, number of numbers beginning with 8 = \(\frac{4!}{2!}\) = 12 

∴ Number of different numbers greater than 60000 formed with digits 0, 2, 2, 6, 8 

= 12 + 12 = 24.

The curve cuts y-axis at x = 0

When x = 0, y = ae⁰ = a

Now, dy/dx = ae^-xb.(-1/b)

∴ (dy/dx)_(0,a) = ae⁰(-1/b) = -a/b

Hence, required slope of tangent of curve is (-a/b)

We are required to find the number of permutations which begin with the letter A. 

There are 11 letters in the given word of which A – 2, 1-2,N-2, 

When A is fixed in the first place, remaining 10!/2!2!  letters can be arranged in ways.

∴ Required permulations

= (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3)/(2 x 1) = 907200

In the word PERMUTATIONS, there are 2 Ts and all the other letters appear only once.

(i) If P and S are fixed at the extreme ends (P at the left end and S at the right end), then 10 letters are left.

Hence, in this case, required number of arrangements =10!/2!=1814400

(ii) There are 5 vowels in the given word, each appearing only once.

Since they have to always occur together, they are treated as a single object for the time being. This single object together with the remaining 7 objects will account for 8 objects.

These 8 objects in which there are 2 Ts can be arranged in 8!/2! ways.

Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5! ways.

Therefore, by multiplication principle, required number of arrangements in this case

8!/2! x 5!=2419200

(iii) The letters have to be arranged in such a way that there are always 4 letters between P and S.

Therefore, in a way, the places of P and S are fixed. The remaining 10 letters in which there are 2 Ts can be arranged in 10!/2! ways.

Also, the letters P and S can be placed such that there are 4 letters between them in 2 × 7 = 14 ways.

Therefore, by multiplication principle, required number of arrangements in this case

=10!/2! x 14=25401600

Given curve is y = √(4x - 3) - 1

Differentiating, we have

dy/dx = 1/2√(4x - 3) x 4 = 2/√(4x - 3)

Also, slope of tangent at any point on the curve is 2/3

∴ dy/dx = 2/3

2/√(4x - 3) = 2/3

√(4x - 3) = 3

4x - 3 = 9

4x = 12

∴ x = 3

So, y = √(4 x 3 - 3) - 1 = √9 - 1 = 2

So, the require point is (3, 2)

(a) 450

Out of the 10 digits 0, 1, 2, 3, ..., 9, five digits, i.e., 0, 2, 4, 6, 8 are even and five digits, i.e., 1, 3, 5, 7 and 9 are odd. 

The sum of the three digits D₁, D₂, D₃ of the number D₁D₂D₃ will be even if : 

(i) All the three digits are even: 

∴ Number of such numbers = 4 × 5 × 5 = 100 

(Here the position D₁ cannot be occupied by 0) 

(ii) One of the digits is even and the rest two are odd : 

(a) D₁ is even, D₂ is odd, D₃ is odd 

∴ Number of such numbers = 4 × 5 × 5 = 100 

(Again since D₁ ≠ 0, so only 4 choices for D₁) 

(b) D₁ is odd, D₂ is even, D₃ is odd 

∴ Number of such numbers = 5 × 5 × 5 = 125 

(c) D₁ is odd, D₂ is odd, D₃ is even 

∴ Number of such numbers = 5 × 5 × 5 = 125 

∴ Total number of required numbers 

= 100 + 100 + 125 + 125 = 450.

(c) 73

The words coming before RAHUL in the dictionary will have A, H, L or R as their first letters. 

I. When A is the first letter, the rest of the 4 letters H, L, R, U can fill the next 4 places in 4 ! 

∴ Number of words beginning with A = 24 

II. When H is the first letter, the rest of the 4 letters A, L, R, U can fill the next 4 places in 4! 

Number of words beginning with H = 24 

Similarly, the number of words beginning with L = 24 

Now among the words having R as their first letter, there is only one word which comes before RAHUL and that is RAHLU 

Thus there are (24 × 3) + 1 = 73 words before RAHUL.

Since repetition is allowed, each of the 3 places in a 3-digit number can be filled in 5 ways.

∴ Required number of 3-digit numbers = 5 × 5 × 5 = 5³ = 125

The words in a dictionary are written in an alphabetical order, which, here is ADMNOR. 

(i) Starting with A, the remaining letters D, M, N, O, R can be arranged in 5! = 120 ways. So, there are 120 words starting with A. 

(ii) Similarly, number of words starting with M = 120, starting with N = 120, and starting with O =120.

(iii) Now, the number of words starting with R is also 120. Out of these words, one word is RANDOM. First, we find the words starting with RAD and RAM. 

Number of starting with RAD = 3! = 6 

Number of starting with RAM = 3! = 6 

(iv) Thus, so far, 120 + 120 + 120 + 120 + 120 + 6 + 6 = 612 words have been constructed. 

(v) Now, the words starting with RAN will appear. Their number is also = 3! = 6. One of these words is the word RANDOM itself. 

The first word beginning with RAN is RANDMO. It is 613th word and so the next word is RANDOM. RANDOM is the 614th word. 

Hence, rank of RANDOM = 614.

(c) 576

There are 7 letters in the word RAINBOW out of which 4 are consonants and 3 are vowels. 

1  2  3  4  5  6  7 

In order that the vowels may occupy odd places, we first of all arrange any 3 consonants in even places in ⁴P₃ ways and then the odd places can be filled by 3 vowels and the remaining 1 consonant in ⁴P₄ ways. So, 

Required number of words = ⁴P₃ × ⁴P₄ = 24 × 24 = 576.

Since ‘SCHOOL’ has 6 letters, out these 6 letters there is 20’s 

Hence number of permutations 

=\(\frac{6!}{2!}\) = 6 × 5 × 4 × 3 × 2 × 1 2 

= 360

The total number of numbers = 5 ! = 120. Suppose we have 9 in the unit’s place. We will have 4 ! = 24 such numbers. The number of numbers in which we have 1, 3, 5 or 7 in the unit’s place is also 4 ! = 24 in each case.

Hence the sum of the digits in the unit’s place in all the 120 numbers 

= 24 (l + 3 + 5 + 7 + 9) = 600. 

The number of numbers when we have any one of the given digits in ten’s place is also 41=24 in each case. Hence the sum of the digits in the ten’s place 

= 24 (1 + 3 + 5 + 7 + 9) tens = 600 tens = 600 × 10. 

Proceeding similarly, the required sum 

= 600 units + 600 tens + 600 hundreds + 600 thousands + 600 ten thousands 

= 600(1 + 10 + 100 + 1000 + 10000) = 600 × 11111 = 6666600.

There are 5 vowels in 26 alphabets. 

Hence, using all 5 vowels at a time, number of different words (with or without meaning) can be made are = 5! 

= 5 × 4 × 3 × 2 × 1 

= 120

B. 108

Explanation:

The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time =(3+4+5+6)3! =108

Hence, Option (B) 108 is the correct answer.

6 vacant chains can be occupied by 4 people in ⁶p₄ ways = 6 × 5 × 4 × 3 = 360 ways.

The number of ways in which a person can go from A to B & B to C is 4 × 3 = 12 ways.

Given, there are 10 buses from A to B. 

So, a person can select any out of 10, therefore, there are 10 ways to travel from A to B. Similarly, he can come back in 10 ways. 

Therefore, total number of ways to travel from A to B round trip = 10 × 10 = 100 ways