x² – 7x + 12 = (x – 4)(x – 3) 

p(x) = 0 (x – 4) ( x – 3) = 0 

x = 4, x = 3

P(1)=3 x 13 – 2 x 12 – 3 x 1 + 2 = 0

Therefore x – 1 is a factor.

a. p(1/2).

b. p(1).

p(x) = x³ – 6x² + 11x – 1

p(1) = 1 – 6 + 11 – 1 = 5 

p(2) = 8 – 24 + 22 – 1 = 5 

p(3) = 27 – 54 + 33 – 1 = 5 

p(x) – p(1) = x³ – 6x² + 11x – 6 

p(x) – p(2) = x³ – 6x² + 11x – 6

p(x) – p(3) = x³ – 6x² + 11x – 6 

p(x) – p(1) = x³ – 6x² + 11x – 6 = 0

If x = 1, 2, 3 then p(x) – p(1) = 0.

Factors of the equations are (x – 1), (x – 2), (x – 3).

Solutions of the equations are 1, 2, 3.

p (x) = x² + x – 1, remainder p(2) (1) 

p (2) = (2)² + 2 – 1 = 5 (1) 

For x – 2 to become a factor of p (2) must be equal to zero. 

For p (2) = 0 here we have to substract 5 from p(x). (1) 

That is, add –5 to p(x) for (x – 2) become a factor. (1)

p(x) = x¹³  – 1 

p(1)= 1¹³  – 1 = 1 – 1=0 

p(1) = 0

(x – 1) is a factor of p(x)

p(x) = x² + ax + b = (x – 3)(x – 2) 

= x²  – 5x + 6

a = –5, b = 6 (3)

The remainder obtained when p(x) is

divided by (ax + b) = \(p =\left(\cfrac{-b}{a}\right)\)

When \(p =\left(\cfrac{-b}{a}\right)\) = 0, then (ax + b) will be a factor of p(x).

When a polynomial p(x) is divided by (x – a), if p(a) = 0 then (x – a) is a factor of p(x). 

When a polynomial p(x) is divided by (x + a),

if p(–a) = 0 then (x + a) is a factor of p(x).

p(x) = x² + 7x + 10 p(–2) = 4 – 14 + 10 = 0 (1)

\(\therefore\) x + 2 is a factor (1)

Remainder p(5) = (5)² + 7(5) +10 (1) 

= 25 + 35 + 10 ≠ 0 (1) 

\(\therefore\) x – 5 is not a factor

p (x) = x² + kx + 6

If (x – 1) is a factor of p(x)

then p(1) = 0

p(1) = 12 + k x 1 + 6 = 7 + k

7 + k = 0

k= –7

\(\therefore\) p(x) = x² – 7x + 6

a + b = 7

ab = 6

a= 1, b = 6

factors are (x – 1 )(x – 6)

Second factor is (x – 6)

a. p(x) = x³ – 4x² – 12x – 45 

remainder = p(2) = 2³ – 4(2)² + 12 x 2 – 45 

= 32 – 61 = –29

b. p(x) = 2x³ + 4x² – 10x + k; p(l) = 0

\(\Rightarrow\) 2 x 13 + 4 x 12 – 10 x 1 + k = 0

2 + 4 – 10 + k = 0 – 4 + k = 0; k = 4

Let (x – 1) be a factor, then p(1)=0

p(1) = 12 + a x 1 + b = 0

i.e., a + b = –1

p(x) = 5x³ +3x² + ax + b

\(\therefore\) x²-1 is a factor then p(1), p(–1) will be zero. 

p(1) = 5 x 1 + 3 x 1 + a x 1 + b = 0

= 5x – 1 + 3 x 1 + ax – 1 + b = 0 a + b = –8 (1)

= 5x – 1 + 3 x 1 + ax – 1 + b = 0 –a + b = 2 (2)

Find the solutions of the equation 

b = –3, a = –5 added polynomial = –5x – 3

i. p(x) = x² – 3x + 5

If x – 4 is a factor, p(4) = 0 

p(4) = (4)² – 3 x 4 + 5 = 9

For x – 4 to become a factor of p (4) must be equal to zero.

For p (4) = 0 here we have to subtract 9 from p(x).

That is, add -9 to p(x) for (x – 4) become a factor.

\(\therefore\) p(x) = x² – 3x + 5 – 9 = x² – 3x – 4

x² – 3x – 4 = (x – a) (x – b)

= x² – (a + b) x + ab

a + b = 3

ab = –4

(a – b)² = (a + b)² – 4 ab

= (3)² – 4x – 4 = 25

a – b = 5

a + b = 3

a = 4; b = –1

x² – 3x – 4 = (x – 4)(x + 1)

Second factor is (x +1)

ii. p(x) = x² – 3x + 5 

If x + 4 is a factor, p(–4) = 0

p(–4) = (–4)² – 3x – 4 + 5 = 33

For x + 4 to become a factor of p (–4) must be equal to zero.

For p (–4) = 0 here we have to subtract 33 from p(x). 

That is, add –33 to p(x) for (x + 4) become a factor.

\(\therefore\) p(x) = x² – 3x + 5 – 33 = x² – 3x – 28

x² – 3x – 28 = (x – a) (x – b)

= x² – (a + b) x + ab

a + b = 3

ab = -28

(a – b)² = (a + b)² – 4ab

= (3)³ – 4x – 28 = 121

a – b= 11

a + b = 3

a = 7;

b = -4 x² – 3x – 28 = (x – 7)(x + 4)

Second factor is (x – 7)

iii. p(x) = x² + 5x – 7

If x – 1 is a factor, p(1) = 0

p(1) = (1)² +5 x 1 – 7 = –1

For x – 1 to become a factor of p (1) must be equal to zero.

For p (1) = 0 here we have to subtract –1 from p(x).

That is, add 1 to p(x) for (x – 1) become a factor.

p(x) = x² + 5x – 7 + 1 = x² + 5x – 6

x² + 5x – 6 = (x – a) (x – b)

a + b = 5

ab = –6

a = –6; b= 1

x² + 5x – 6 = (x + 6)(x – 1)

Second factor is (x + 6)

iv. p(x) = x² – 4x – 3

If x – 1 is a factor, p(1) = 0

p(1) = (1)² – 4 x 1 – 3 = –6

For x – 1 to become a factor of p (1) must be equal to zero.

For p (1) = 0 here we have to subtract -6 from p(x).

That is, add 6 to p(x) for (x – 1) become a factor.

\(\therefore\) p(x) = x² – 4 x – 3 + 6 = x² – 4x +3

x² – 4x + 3 = (x – a) (x – b)

a + b = –4 ab = 3

a =1; b = 3

x² – 4x +3 = (x – 1)(x – 3)

Second factor is (x – 3)

b² – 4ac = 2² – 4 x 1 x 2 = –4 < 0

\(\therefore\) x² + 2x + 1 cannot be written as the product of first degree polynomials.

The correct answer is Increases by 4 times.

• The kinetic energy of an object is defined as the energy that it possesses due to its motion.

• An object that has motion possesses kinetic energy.
• The amount of kinetic energy that an object has depends upon two variables-
  • the mass (m) of the object 
  • the speed (v) of the object
• If the velocity of the moving body is doubled, then the kinetic energy becomes 4 times because the kinetic energy of a body is directly proportional to the square of its velocity when the mass of the body is kept constant.

i. p (x) = x² – 7x + 12 

a + b = –7, ab = 12 

a = –3, b = –4 

x² – 7x + 12 = (x – 3) (x – 4) 

x² – 7x + 12 = 0 

(x – 3) (x – 4) = 0 

x – 3 = 0, x – 4 = 0 

x = 3, x = 4

ii. p(x) = x² + 7x + 12 

a + b = 7, ab= 12 a = 3, b = 4 

x² + 7x + 12 = (x + 3) (x + 4) 

x² + 7x + 12 = 0 

(x + 3)(x + 4) = 0 

x = 3, x = 4

iii. p(x) = x² - 8x + 12 

a + b = 8, ab= 12 

a = 6, b = –2 

x² – 8x + 12 = (x – 6) (x – 2) 

(x – 6) (x – 2) = 0 

x = 6, x = 2 

iv. p(x) =x² + 13x + 12 

a + b = 13, ab = 12

a =12, b=1 

(x +13x + 12) = (x + 12) (x + 1) 

x² +13x + 12 = 0 

(x+ 12) (x+ 1) = 0 

x+ 12 = 0, x+ 1 = 0 

x = -12 ,x = –1

v. p(x) = x² + 12x – 13 

a = –13, b = 1 

x² + 12x – 13 = (x + 13)(x – 1) 

x + 13 = 0, x – 1 =0 

x = –13, x= 1 .

vi. p(x) = x² – 12x – 13

x² – 12x – 13 = (x – a) (x – b)

= x² – (a + b) x + ab

a + b= 12 ab = –13

(a – b)² =(a + b)² – 4ab

= (12)² – 4x – 13 = 196

a – b = 14

a + b= 12

a= 13; b = –1

x² – 12x – 13 = (x – 13)(x + 1)

x – 13 = 0, x + 1 =0

x= 13 ,x = –1

The correct answer is Capillarity.

• Capillarity- Capillary action is the ability of a liquid to flow against gravity in a narrow space such as a thin tube. 
  • Surface tension is an important factor in the phenomenon of capillarity.
  • If a capillary tube is dipped in a liquid, liquid ascends or descends in the capillary tube because of capillarity.
  • Thus, the melted wax in the candle rises up to the wick due to capillary action.
  • This spontaneous rising of a liquid is because of two opposing forces.

• Cohesion –These are the attractive forces between similar molecules or atoms
• Adhesion –These are the attractive forces between dissimilar molecules or atoms

x² + 4x + 1 = (x – a) (x – b) = x² – (a + b) x + ab(1)

a + b = –4, ab = 1, a – b = 2√3 

a = –2 + √3, b = –2 – √3 (1)

x² + 4x + 1 =(x + 2+ √3 ) (x + 2 – √3 ) (1) 

x² + 4x + 1 = 0 =>(x + 2 + √3 )(x + 2 –√3 ) = 0 (1)

x = –2 – √3 ,or x = –2 + √3

Concept:

• Sum of the first n Natural Numbers \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)
• Sum of the Square of the first n Natural Numbers \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
• Sum of the Cubes of the first n Natural Numbers \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

Calculation:

Here, we have to find the sum of the series 12 ⋅ 2 + 22⋅3 + 32⋅4 + 42⋅5 +.....+up to n terms

nth term of the given series is T_n = n² ⋅ (n + 1) = n3 + n² 

Sum of n terms of the series,  \(\rm S_n = \sum T_n= \sum (n^3+n^2)=\sum n^3+\sum n^2\)

As we know that, 

\(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )^2+\frac{n(n+1)(2n+1)}{6}\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{n(n+1)}{2}+\frac{2n +1}{3} \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{3n^2+7n+2}{6} \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{(3n+1)(n+2)}{6} \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)(3n+1)(n+2)}{12}} \right )\)

Hence, option 2 is the correct answer.

Answer is (b) 5/8

Calculation:

3/5 = 0.6

7/9 = 0.77

11/13 = 0.846

1/2 = 0.5

∴ 1/2 is the smallest fraction

Answer is (d) Statement-I is false, but Statement-II is true.

Answer is (b) Both the statements are true, but Statement-II is not the correct explanation of Statement-l

Answer is (a) P(A).P(B)

Answer is (a) P(A) + P(B) + P(C)

Answer is (a) Both the statements are true and Statement-II is the correct explanation of Statement-I.

Answer is (a) n(E)/n(S)

Answer is (c) P(A/B) = 0

Correct option is C) filled

Answer is (a) 13/24

Answer is (a) 3/5

(a) Area of parallelogram = Base x Height

= (12 x 7) cm²

= 84 cm² 

(b) r = Radius of semi-circle = 3.5 cm

∴ Area of semi-circle = \(\frac12 \pi r^2\)

\(= \frac 12 \times \frac{22}7 \times 3.5\times 3.5\)

\(= 19.25\, cm^2\)

(c) Area of shape/figure = Area of semi-circle + Area of triangle

\(= \frac12\pi r^2 + \frac12\times \) Base \(\times\) Height

\(= \frac 12 \times \frac{22}7 \times 3.5\times3.5+\frac 12 \times 3.5\times 3\)

\(= 19.25 + 5.25\)

\(= 24.50 \,cm^2\)

We have the equation

(b-c)x²+(c-a)x+(a-b)=0

Comparing with quadratic equation

Ax²+Bx+C=0

A=(b-c),
B=c-a,
C=a-b

Discriminate when roots are equal

D=B²-4AC=0

D=(c-a)²−4(b-c)(a-b)=0

D=(c²+a²−2ac)-4(ba-ac-b²+bc)=0

D=c²+a²−2ac-4ab+4ac+4b²-4bc=0

c²+a²+2ac-4b(a+c)+4b²=0

(a+c)²-4b(a+c)+4b²=0

[(a+c)-2b]²=0

a+c=2b

Let numbers are 2x & 3x.

\((2x + 15):(3x + 15) = 7:8\)

⇒ \(\frac{2x + 15}{3x + 15} = \frac 78\)

⇒ \(16 x + 120 = 21x + 105\)

⇒ \(5x = 120-105\)

⇒ \(5x = 15\)

⇒ \(x = 3\)

∴ Numbers are 6, 9

∴ Sum of numbers = 6 + 9 = 15

Given -
D = 7
Dx = 49
Dy = -63

Value of x can be calculated by -
x = Dx / D
x = 49 / 7
x = 7

Value of y can be calculated by -
y = Dy / D
y = -63 / 7
y = -9

Given - D = 7 Dx = 49 Dy = -63 Value of x can be calculated by - x = Dx / D x = 49 / 7 x = 7 Value of y can be calculated by - y = Dy / D y = -63 / 7 y = -9

Answer (b) Retina