If find x. 1/(9!) + 1/(10!) = x/(11!).

 

Here n! is read as n factorial. It can be written as n(n-1)! or n(n-1)(n-1)!

Expand the following using suitable identity (i) (x+2y+4z)² (ii) (-2x+3y+2z)². [ncert]

(i) (x + 2y + 4z) * iii * (- 2x + 3y + 2z) ^ 2

We know that  (a + b + c)² = a²+ b²+ c²+ 2ab + 2bc + 2ca

(i) (x + 2y + 4z)² = (x)² + (2y)²+ (4z)²+ 2(x)(2y)+2(2y)(4z) + 2(4z)(x)

= x² + 4y²+ 16z² + 4xy + 16yz + 8zx  Ans.

(ⅱ) (- 2x + 3y + 2z)²=(-2x)²+(3y)²+(2z)² +2(-2x)

(3y)+2(3y)(2z) + 2(2z)(- 2x)

 = 4x²+9y²+4z²-12xy+12yz-8zx  Ans.

The sum of three numbers in A.P. is -3. and their product in 8. Find the numbers. [ncert]

The sum of three numbers in A.P. is -3. and their product in 8. Find the numbers.

Let the three numbers are a-d, a and a+d. Here a is the first term and d  the Common difference.

Now given that (a - d) + a + (a + d) = - 3

=> 3a = - 3 => a = - 1

Now (a - d) * a* (a + d) = 8

=> a(a - d)(a + d) = 8 => a(a² - d²) = 8 

 => (-1){(-1)²-d²)}=8

=> (-1)(1-d²)=8 => -1+d²=8

=> d²=9 => d=±3

Now take a = - 1 and d = 3 

=> a-d, a, a+d= -1-3, -1, -1+3 = -4,-1,2 Ans.

Now take a= -1 and d= -3, then a-d, a, a+d becomes, -1-(-3), -3, -1+(-3)

= 2, -3, -4. Ans.

Matrix multiplication.

 

Find the square root of -15-8i. [ncert]

Describe the set of complex number z for which |(x-3)/(x+3)|=2 geometrically. [ncert]

Define matrix

Prove that intg.dx/(x²+a²)=(1/a)tan^-1(x/a)+C.

Prove that Intg.(x²-a²)dx=1/2a(log|(x-a)/(x+a)|. [ncert 12th]

Solve the quadratic equation 2x²+ax-a²=0 for x. [cbse 10th 2014]

Integrating (1/(1-cosx)) -(1/(1+cosx)) w.r.t x.

 

The altitude of a right triangle is 7 cm less than its base. If the Hypotenuse is 13 cm, find the other two sides. [ncert 10th]

Let the base of the given rt|_d triangle=x cm.

Then its altitude = x-7 cm.

[Here Altitude=Height=perpendicular length] 

Hypotenuse = 13 cm.

Now in right-angled triangle, 

(Hypotenuse)²=(Based)²+(Altitude)²

=> (13)²=x²+(x-7)²

=> 169=x²+x2+49-14x

=> 2x²+49-14x-169=0

=> 2x²-14x-120=0

Dividing both sides by 2, we get,

x²-7x-60=0

=> x²-12x+5x-60=0

=> (x²-12x)+(5x-60)=0

=> x(x-12)+5(x-12)=0

=> (x-12)(x+5)=0

=> x=12, -5

But length of a side of a triangle cannot be negative. So x=12

=> Base=12 cm. Ans.

Altitude=x-7=12-7=5cm. Ans.

Rohan's mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan's present age. [ncert 10th]

Let the present age of Rohan= x years

Then present age of Rohan's mother =x+26 years.

After 3 years, age of Rohan = x+3 years.

After 3 years, Rohan's mother age = x+26+3 = x+29 years.

Now according to the given condition, 

(x+3)(x+29)=360

 => x²+32x+87=360

=> x²+32x+87-360=0

=> x²+32x-273=0

=> x²-7x+39x-273=0

=> (x²-7x)+(39x-273)=0

=> x(x-7)+39(x-7)=0

=> (x-7)(x+39)=0

=> x=7 or x= -39

Ignore x= -39 [∵ age cannot be negative]

=> Present age of Rohan = x = 7 years.

Find the two Numbers, whose sum is 27 and product is 182. [ncert 10th]

Find two consecutive positive integers, sum of whose squares is 365. [ncert 10th.]

Let the two consecutive positive integers

are  x and x+1.

Given that x²+(x+1)² = 365

Using (a+b)=a²+b²+2ab, we get,

=> x²+x²+1+2x=365.  

 => 2x²+2x+1-365=0

=> 2x²+2x-364=0

Dividing both sides by 2 (non zero), we get,

=> x²+x-182=0

=> x²+14x-13x-182=0

=> (x²+14x)-(13x+182)=0

=> x(x+14)-13(x+14)=0

=> (x+14)(x-13)=0

=> Either x+14=0 or x-13=0

=> Either x= -14 , 13

But we have to find positive integer, so -14 must be rejected.

So x= 13 is the required positive integer.

Now two consecutive positive integers shall be x, x+1 

=> 13 and 13+1=14

So 13 and. 14 are the required positive consecutive integers. Ans.

If A=[(1,3),(3,4)], B=[(-2,1),(-3,2)] and A²-5B²=5C. Find matrix C where C is a 2 by 2 matrix. [icse2017].

What must be subtracted from 16x ^ 3 - 8x ^ 2 + 4x + 7 that the resulting expression has (2x + 1) as a factor? [icse2017]

If (x+2) and (x+3) are factors of x^3+ax+b, find the values of 'a' and 'b'. [icse 2018]

 

If A=[(3,x),(0,1)] and B=[(9,16), (0,-y)], find x and y, when A²=B. [icse 2015]

Solve the following inequation and write down the solution set. 11x-4 < 15x+4 <= 13x+14 ; xεW. Represent the solution on a number line. [icse 2019]

Question: Solve the following inequation and write down the solution set.

11x - 4 < 15x + 4 ≤ 13x + 14 ; x εW

Represent the solution on the number line.

Given that:

11x - 4 < 15x + 4 ≤13x + 14 - - - - - - - (1)

=> 11x - 4 < 15x + 4 and 15x + 4 ≤13x + 14

Case I.

11x - 4 < 15x + 4

Subtracting 11x on both sides, we get,

-4 < 4x +4

Subtracting '4' on both sides, we get,

-8 < 4x

Dividing both sides by 4, we get,

-2 < x 

or x > -2. - - - - - - (2)

Case II.

15x + 4 ≤ 13x + 14

Subtracting both sides by 13x, we get,

2x + 4 ≤ 14

Subtracting 4 on both sides, we get,

2x  ≤10

Dividing both sides by 2, we get

x ≤ 5 - - - - - - - (3)

From,(1), (2) and (3), the solution set is

- 2 < x ≤ 5 :xεW

=(-2,5], here W is a whole number.

W = 0, 1, 2, 3 ,- - - - - - - - .

So the solution set will be {0,1,2,3,4,5}

Solve the following inequation and represent the solution set on a number line.(-8*1/2)< -1/2 -4x<=7*1/2 ; x€I [icse 2017]

Solve the following inequation and represent the solution set on a number line:

- 8*1/2 < - 1/2 - 4x ≤ 7 *1/2 , , x εI

Solutions:

Given that:- 

-8 *1/2 < - 1/2 - 4x≤7 *1/2

⇒ Convert mixed fractions to improper fractions:

- 17/2 < - 1/2 - 4x ≤15/2

Adding 1/2 to each term, we get:

- 17/2 + 1/2 < - 4x ≤ 15/2 + 1/2

Simplify the fractions:

- 16/2 < - 4x ≤16/2

=> - 8 < - 4x ≤ 8

Dividing each term by -4, we get:

2 > x >= - 2

[Sign reverses when each term is divided by same negative number]

We can write it as

- 2 ≤ x < 2

The solution set is {x:  - 2 ≤ x < 2; x Ɛ I}

= {-2,-1,0,1}

Use the remainder theorem to factorise the following expression: 2x^3+x^2-13x+6. [icse 2010]

 

Rules for solving Inequations

Rules for Solving a Linear Inequality

<, >, ≤ and ≥ are known as signs or an inequality.

Rule 1

If a positive number is added to both sides of an inequality, the sign of inequality remains the same.

(a) x+3<5 => x + 3 + 2 < 5 + 2

(b) x+1>3 => x + 1 + 1 > 3 + 1

(c) x+3≥7 =>  x + 3 + 1≥7 +1

(d) x+5≤9 => x + 5 + 2 <= 9 + 2

Rule 2:-

If a negative number is subtracted on both sides of an inequality, the sign of that  ineq -uality remains the same.

(a) x+1<5 => x+1-2< 5-2

(b) x+3>7 => x+3-2>7-2

(c) x + 2≼ 11 =>x+2-1≼ 11-1

(d) x + 5≥9 => x+5-2 ≥ 9-2

Rule 3:- 

If we multiply the both sides of an inequality with the same positive number, then the sign of the inequality remains the same.

(a) x < 2 => x×5<2×5 

(b) x > 5 => x×2>5×2

(c)x ≤ 3 => x× 2 ≤ 2×3

(d) x ≤7 => x ×3≤ 7×3

Rule 4 :-

If we multiply both sides of an inequality with the same negative number, then the inequality will reverse i.e. the sign of inequality reversed.

(a) x < 5 => (-2)× x > (-2)×5

(b) x > 4 =>  (-3) × x < (-3) × 4 

(c) x ≤ 9 => (-1) × x ≥ (-1) × 9 

(d) x≽3 => (-2)×x ≼ (-2)×3

Rule 5:- 

If both sides of an inequality are divided by the same positive number, then the sign of inequality remains the same.

(a) x<5  => x/2 < 5/2

(b)x>3 => x/3 > 3/3

(c) x ≤2  =>   x/2 ≤  2/2

(d) x≥6 => x/3 ≥6/3

Rule 6 :-

If both the sided of an inequality are divided by same negative number, then the sigh of inequality will reverse. 

(a) x<4 => x/(-2) > 4/(-2)

(b) x>5 => x/(-3) < 5/(-3)

(c) x≤6 => x/(-2) ≥ 6/(-2)

(d) x≥8 => x/(-4)≤ 8/(-4)

Find the value of x, which satisfy the inequation - 2*5/6 less than1/2 -2x/3 less than aur equal to 2 and x€W and represent the solution on the numbe line.[icse 10th 2014]

Find the value of x, which satisfy the inequation and x ∈ W and represent the solution on the number line. 

Given that - 2 *5/6 < 1/2 - (2x)/3 ≤ 2 

=>  - 17/6 < 1/2 - (2x)/3 ≤ 2 

=> - 17/6 < 1/2 - (2x)/3 and 1/2 - (2x)/3≤ 2 

Case I  - 17/6 < 1/2 - (2x)/3

Multiplying both sides by 6, we get, 

- 17 < 3 - 4x

Subtracting both sides by 3, we get,

- 17 - 3 < 3 - 4x - 3

- 20 < - 4x

Dividing both sides by -4, we get,

5>x

or x < 5  -  - - - - - (1) 

[It both sides of  an inequality is divided by the same negative number,then sign of inequality reverses.]

Case 2:- 1/2 -2x/3 ≤ 2 

Subtracting both sides  1/2 we get,

=> 1/2 - (2x)/3 - 1/2 ≤ 2 - 1/2

=>  - (2x)/3 ≤ 3/2

Multiplying both sides by 3, we get

- 2x ≤ 9/2

Dividing both sides by -2, we get,

x ≥ - 9 /4   =>  - 9/4 ≤ x   - - - - - - - - (2)

From (1) and (2), we get,

- 9/4 ≤ x < 5

=> x=[ -9/4, 5)

But x is a whole Numbers, i.e  {W=0, 1, 2, 3, 4, 5, ...}

So the solution set ={x ; x = 0, 1, 2, 3, 4}

Solve the following inequation, write the solution set and represent on the number line. x+1 -3(x - 7) ≥ 15 - 7x > X€R [icse 2016]

Given that-

 3(x - 7) ≥3 15 - 7x > (x + 1)/3

=> - 3(x - 7)  ≥ 15 - 7x and 15 - 7x > (x + 1)/3

Case I

When - 3(x - 7) ≥ 15 - 7x

=>  - 3x + 21 ≥ 15 - 7x

Adding 7x - 21 on both sides, we get,

- 3x + 21 + 7x - 21  ≥  15 - 7x + 7x - 21

=> 4x  ≥ - 6

Dividing both sides by 4, we get,

(4x)/4  ≥  - 6/4

=> x  ≥ - 3/2

=>x  ≥  - 1.5

=> -1.5 ≤ x - - - - - - - (1)

 Case II 

15 - 7x > (x + 1)/3

Multiplying both sides by 3, we get,

3(15 - 7x) > x + 1

45 - 21x > x + 1

Adding both sides by 21x - 1, we get,

45 - 21x + 21x - 1 > x + 1 + 21x - 1

=>  44 > 22x

Dividing both sides by 22, we get

44/22 > 22/22 * x

=> 2 > x

=> x < 2. - - - - - - - - - (2)

combining (1) & (2),the solution set is

{x : - 1.5  ≤   x < 2, x ε{R}}

= [-1.5, 2)

Area of a triangle in terms of determinant.

 

Using Componendo and dividendo, find the value of x: (√(3x+4) +√(3x-5))/(√(3x+4) - √(3x-5)) = 9. [icse 2011]

 

If x/a = y/b = z/c , then prove that (x^3/a^3)+(y^3/b^3)+(z^3/c^3) = (3xyz)/(abc). [icse 2016]

Define Componendo and dividendo in ratio and proportion.

 

Alternendo and Invertendo properties in Ratio and Proportion.

 

If x,y,z are in continued proportion; then prove that (x+y)²/(y+z)² = x/z. [icse 2010]

 

If (7m + 2n)/(7m - 2n) = 5/3 ; use the properties of proportion to find:(i) m:n. (ⅱ) (m ^ 2 + n ^ 2)/(m ^ 2 - n ^ 2). [icse 2017]

(i) m:n 

Given that ,

(7m + 2n)/(7m - 2n) = 5/3

Applying componendo and dividendo, we get, ((7m + 2n) + (7m - 2n))/((7m + 2n) - (7m - 2n)) = (5 + 3)/(5 - 3)

[By using (Num. + Deno.)/(Num. - Deno.) on both sides]

=) (7m + 2n + 7m - 2n)/(7m + 2n - 7m + 2n) = 8/2 

=> 14m/4n = 8/2 => (7m)/(2n) = 4/1

Multiplying both sides by 2/7 , we get,

 2/7 * 7/2 * m/n = 2/7 * 4/1 

=> m/n = 8/7 => m:n = 8:7

(ii) Given that (7m + 2n)/(7m - 2n) = 5/3

Applying componendo and dividendo, we get, 

 ((7m + 2n) + (7m - 2n))/((7m + 2n) - (7m - 2n)) = (5 + 3)/(5 - 3)

=> (7m + 2n + 7m - 2n)/(7m + 2n - 7m + 2n) = 8/2

=> 14m/4n = 4 => 7/2(m/n) = 4/1 Multiplyning both sides by 2/7 , 

we get. m/n = 8/7

Squaring sides, we get (m²)/(n²) = 64/49

Applying componendo and devedando, we get,

m²+n²/m²-n² = 64+49/ 64-49

=> (m²+n²)/(m²-n²) = 113/15 Ans.

If (x-2) is a -factor of the expression 2x ^ 3 + a * x ^ 2 + bx - 14 and when the expression in divided by (x - 3) , it leaves 52, find a the values of a and b. [icse 2013]

What least number must be added to each of the numbers 6, 15, 20 and 43 to make them proportional ? [ICSE 2013]

 

Find the equation of a line passing through the points (0,-3) and (5,0) by the method of determinant.

 

Solve the following equation for x correct to 2 decimal places, 4x²-5x-3=0 [ICSE 2017].

 

Find the values of p and q for which the following system of equations has infinite number of solutions: 2x + 3y = 7 , (p + q) * x + (2p - q) * y = 21 [cbse 10th 2001]

 

Solve x/a + y/b = 2 ax - by = a ^ 2 - b ^ 2 [CBSE.10th 2005 ]

 

Find the value of x if 0.75:x :: 5:8.