Kamakhya's Some Important Question For 10th Class

Polynomials: Polynomials are a particular type of algebraic expression. 

Polynomial with one term is Monomial

Polynomial with two terms is Binomial

Polynomial with three terms is Trinomial 

Degree Of A Polynomial : The highest power of the variable in a polynomial.

Polynomial of degree 1 is called linear polynomial.

Polynomial of degree 2 is called quadratic polynomial.

Polynomial of degree 3 is called cubic polynomial. 

 

Zeroes Of A Polynomial: Zero is a special digit which satisfied equation.

★A zero of a polynomial need not to be 0. 

★0 may be zero of a polynomial.

★Every linear polynomial has one and only one zero.

★A polynomial can have more than one zero. 

Remainder Theorem: If p(x) is any polynomial of degree greater than or equal to 1 and p(x) is divided by the linear polynomial x − a, then the remainder is p(a). 

🌺★★★★★★🌺🌹KSIQ🌹🌺★★★★★🌺

🌺Kamakhya's Some Important Question🌺

🌺★★★★★★🌺🌹KSIQ🌹🌺★★★★★🌺

Kamakhya's Some Important Question For 10th Class

KSIQ 01. In fig, OAPB is a sector of a circle of radius 3.5 cm with the centre at O and ∠AOB = 120°. Find the length of OAPBO.

Sol. 

AC = 28cm 

BC = Side√2

        = 28√2

BC = 28√2cm (by Pythagoras Theorem).

Radius = (BC)/2

            = (28√2)/2

            = 14√2cm

Shaded region 

= Area of semicircle – Area of segment BCD

      = 392cm²

★★Kamakhya's Some Important Question ★★

KSIQ 02. Find the area of the shaded region shown in the fig.

Sol. 

KSIQ 04. Find the ratio in which the line x + 3y–14=0 divides the line segment joinig the points A(-2,4) and B(3,7).

Sol: Let P(x, y) be the point which the line x+3y–14=0 divides the line segment in the ratio k:1. 

Given points are A(-2,4) and B(3,7).

By section formula 

So, x = (3k–2)/(k+1); y = (7k+4)/(k+1) 

Placing the value of x and y into the equation 

x + 3y–14=0

(3k–2)  + 3 (7k+4)  –14 = 0

(k+1).           (k+1)

Multiplying by (k + 1) into the whole question

(3k–2)(k+1) + 3 (7k+4)(k+1) –14(k+1)=0×(k+1)

        (k+1).                  (k+1)

=> (3k–2) +3[(7k+4) +14(k+1) =0

=> 3k–2 + 21k + 12–14k –14 = 0,

=> 10k – 4 = 0

=> 10k =4

=> k =4/10

k =2/5

Therefore, ratio is 2:5

KSIQ 05. Find the centre of the circle passing through (5, -8), (2, -9) and (2, 1).

Sol.

Let the centre of the circle be O( x ,y). 

The points on the circle are 

A(5, - 8), 

B(2, - 9) and 

C(2,1) .

                                            

OA = √(5 – x) ² + (–8 –y)²

                                            

OB = √ (2 – x)² + (–9 –y)²

                                         

OC=  √(2 –x)² + (1 – y)² 

=> OA= OB = OC = radius

If AO = BO

                                                                     

=> √(5–x)²+(–8 −y)² = √(2−x)² + (−9 − y)² 

=> (5 – x)² + (8 + y)² =(2–x)² + (9 + y)²

=>  25 –10x + x² + 64 + 16y + y² = 4 – 4x + x² + 81 + 18y + y²

=>  89 – 10x + 16y = 85 – 4x + 18y

=>  – 6x –2y = –4

=>  3x + y = 2           ..............(i) 

If BO = CO

                                                                     

=> √(2–x)²+(–9–y)² = √(2 − x)² + (1 − y)²

=> (2–x)²+(–9–y)² = (2 − x)² + (1 − y)²

=> 4–4x+ x² + 81+18y+ y² =4 –4x + x²+1–2y+y²

=> 4–4x+ x²+81+18y+ y²–4+4x–x²–1+2y–y²= 0

=> 81 + 18y – 1 + 2y=0

=> 80 + 20y =0

=> 20y = –80

=> y = –80/20

      y=–4          ................(ii)

Substituting or placing y = –4 in (i)

= 3x–4=2

=> 3x = 6

=> x =6/3

=>  x = 2

Therefore the coordinates of centre of the circle O = (2, - 4)

KSIQ 06. What number should be added to the polynomial x ² – 5 x + 4 so that 3 is the zero of resulting polynomial.

Sol. 

      x² – 5 x + 4

=> If one zero is 3. Then 3 satisfied the whole equation.

=> Placing one 3 in equation => x² – 5 x + 4

=> (3)² – 5(3) + 4

=> 9 – 15 + 4

=> –6 + 4

=> –2 

Two satisfied it we should subtract +2 in the equation.

=> –2 –(–2)

=> –2 + 2

=> 0

KSIQ 07. What number should be subtracted subtracted to the polynomial x ² – 5 x + 4 so that 3 is the zero of polynomial so obtained.

Sol. 

      x² – 5 x + 4

=> If one zero is 3. Then 3 satisfied the whole equation.

=> Placing one 3 in equation => x² – 5 x + 4

=> (3)² – 5(3) + 4

=> 9 – 15 + 4

=> –6 + 4

=> –2

Two satisfied it we should add +2 in the equation.

=> –2 + 2

=> 0

KSIQ 08. −5 is one of the zeroes of $2x²+px-15$, zeroes of $p(x²+x)+k$ are equal to each other. Find the value of k.

Sol. 

−5 is a root of quadratic equation $2x²+px-15=0$ 

so, $2(-5)²+p(-5)-15=0$

=>50−5p−15=0

=>35−5p=0

$p=7$ 

ATQ

p(x²+x)+k=0

now, put $p=7$ in second quadratic equation, 

=>7(x²+x)+k=0

=>7x²+7x+k=0

The question told us that the equation has equal roots 

so,

$D=0$

$b²-4ac=0$

=>7²−4×7×k=0

=>49−28k=0

$=>-28k=\; –49$

=>k= 49/28 = 7/4 =1.75

hence, the value of $k=7/4\; or1.75$

$KSIQ\; 09.\; Find\; a\; quadratic\; polynomial,\; whose\; zeroes\; are\; in\; the\; ratio\; 2:3\; and\; their\; sum\; is\; 15.$

Sol. 

First,

α:β = 2:3

Let α = 2m

       β = 3m

α + β = 15

2m + 3m = 15

=> 5m = 15

m = 15/5

m = 3

Now 

α = 2m

α = 2×3

α = 6

And 

β = 3m

β = 3×3

β = 9

α β = 6×9

       = 54

for Quadratic Polynomial

= x² + (α + β) x –  α.β

= x² + (15)x – 54 

= x² + 15x – 54

KSIQ 10. Two numbers are in a ratio of 4:5. If the sum of these two numbers is 27, then what is the product of the number?

$Sol.$

First,

α:β 4:5

Let α = 4m

       β = 5m

α + β = 27

4m + 5m = 27

=> 9m = 27

m = 27/9

m = 3

Now 

α = 4m

α = 4×3

α = 12

And 

β = 5m

β = 5×3

β = 15

α β = 12×15

       = 180

for Quadratic Polynomial

= x² + (α + β) x –  α.β

= x² + (27)x – 180 

= x² + 27x – 180

α β