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## Wednesday, June 4, 2014

### X maths Chapter: 02 Polynomials CBSE Test paper Trend Setter problems

10th Polynomials Ch-02-Mathematics [Key points] Check point [Formative Assessment]
Þ  Any algebraic expression having non zero integral power (whole number) is called polynomial.

Þ If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x).

Þ A polynomial of degree 1 is called a linear polynomial. For example, 2x – 3, √3x+1,y +√3

Þ A polynomial of degree 2 is called a quadratic polynomial. The name ‘quadratic’ has been derived from the word ‘quadrate’, which means ‘square’.       2x2 +  3x + 2 ,y2 + 2

Þ Any quadratic polynomial in x is of the form ax2 + bx + c, where a, b, c are real numbers and a ≠ 0.

Þ A polynomial of degree 3 is called a cubic polynomial e.g. 2 – x3x3, √2 x3.

Þ General form of a cubic polynomial is  ax3 + bx2 + c x + d, where, a, b, c, d are real numbers and a ≠ 0.

Þ If p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k). A real number k is said to be a zero of a polynomial p(x), if p (k) = 0.

Þ Thus, the zero of a linear polynomial is related to its coefficients because if is a zero of p(x) = ax b, then (k) = a k + b = 0, i.e., k  = -b/a

Þ  Zero of a linear polynomial ax + b is −b/a

ÞThe graph of y = ax + b is a straight line like the graph of y = ax + b is a straight line passing through the points (– 2, –1) and (2, 7) and straight line straight line intersects the x-axis at exactly one point

ÞThe linear polynomial ax + b, a ≠ 0, has exactly one zero, namely, the x-coordinate of the point (-b/a , 0) where the graph of y = ax + b intersects the x-axis.

Þ The graph of equation y = ax+ + b x + c has one of the two shapes either open upwards like È (a > 0 ) or open downwards like Ç ( a < 0) . These curves are called parabolas.

Þ The zeroes of a quadratic polynomial ax2 + bx + c, a ≠ 0, are precisely the x-coordinates of the points where the parabola representing y = ax2 + bx + c intersects the x-axis.

Þ In general, given a polynomial p(x) of degree n, the graph of y = p(x) intersects the x- axis at atmost n[n or less than n]  points.

Þ A polynomial p(x) of degree n has at most n zeroes.

Þ if α and β are the zeroes of the quadratic polynomial p(x) = ax2 + b x + c, a ≠ 0, then you know that x – α and x – β are the factors of p(x).
Therefore, ax2 + bx + c = a(x – α) (x – β)=  ax2 – a(α + β)x + a α β]

Comparing the coefficients of x2, x and constant terms on both the sides, we get  , a = kb = – k(α + β) and c = kαβ.

This gives

α + β = -b/a ; α β = c/a

Þ Relationship between the zeroes of a cubic polynomial and its coefficients of
ax3 + bx2 + c x + d= a(x-a)(x-b)(x-g)  = ax3 – a(a+b+g)x+ a(ab +bg+ga) - aabg

Comparing the coefficients of terms on both the sides, we get,  α + β + γ = –b/a; α β + β γ + γ α =c/a; α β γ =– d/a
Þ Division algorithm states that given any polynomial p(x) and any non-zero polynomial g(x), there are polynomials q(x) and r(x) such that p(x) = g(xq(x) + r(x), where r(x) = 0 or degree r(x) < degree g(x).

Check point [Formative Assessment]
Q. 1. Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.

Q.2. Find the zeroes of the polynomial x2 – 3 and verify the relationship between the zeroes and the coefficients.

Q.3. Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively.

Q.4. Verify that 3, –1, -1/3 are the zeroes of the cubic   polynomial p(x) = 3x– 5x2 – 11x – 3, and then verify the relationship between the zeroes and the coefficients.

Q.5. Find a quadratic polynomial if the sum and product of its zeroes respectively √2, 1/3

Q.6. Find all the zeroes of 2x– 3x3 – 3x2 + 6x – 2, if you know that two of its zeroes are √2 and − √2

Q.7. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are   √(5/3) , -√(5/3)

Q.8. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively

Q.9. If the zeroes of the polynomial x3 – 3x2 + + 1 are – bab, find and b.

Q.10. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138– 35 are 2 ± √3 , find other zeroes.

Q.11.  If the polynomial x4 – 6x3 + 16x2 – 25+ 10 is divided by another polynomial x2 – 2k, the remainder comes out to be a, find and a.

Q.12. If α and β are the zeros of the quadratic polynomial f(x) = x2 - p (x+1) - c, Show that (α+ 1) (β + 1) = 1- c.
Q.13. 1.      For which values of a and b , are the zeros of g(x) = x3+2x2+a, also the zeros of the polynomial f(x) =x5-x4-4x3+3x2+3x+b ? Which zeros of f(x) are not the zeros of g(x)?

(Ans. 1 and 2 are the zeros of g(x) which are not the zeros f(x) and this happens    when    a= -2 , b= -2)
Q.14. 1.      If one zero of the polynomial 3x2- 8x –( 2k + 1) is seven times the other , find both zeroes of the polynomial and the value of k .

( Ans. The zeroes of the given polynomial are 1/3  , 7/3  and the value of k is (-5/3)

Q.14. if (x+a) is a factor of 2x2+2ax+5x+10then find the value of a. [Ans : 2]

Q.15. If two zeros of the polynomials   x4 – 6x3 – 26x2 +138x -35 are 2- Root3  and 2+ root3 , find all the zeros  .

( Ans. Zeros p(x) are 2 - root3  and 2 + root3 , -5 and 7)

Q.16. It is true to say that for k= 2 the pair of linear equation 3x+y = 1;(2k-1)x + (k-1)y = 2k+1 has no solution. Justify

Ans: Given system of equations is:
3x+y=1 and
(2k-1)x+(k-1)y=2k+1
They can be rewrite as:
3x + y - 1 = 0 and
(2k-1)x + (k-1)y - (2k+1) = 0
We know that, for the given system to have no solution, a1/a2 = b1/b2 ¹ c1/c2
Þ 3/(2k-1) = 1/(k-1)
3k - 3 = 2k - 1
k = 3-1
k = 2
Hence, for k = 2, the given system of equation will have no solution.

Q.17. In a competition, one mark is awarded for every correct answer and half mark is deducted for every wrong answer. Jayanthi answered 120 questions and got 90 marks. How many questions did she answer correctly?

Ans: Let the number of questions correctly answered be x

Therefore, number of questions incorrectly answered will be 120 - x.

According to the given condition,

1× x – (1/2)(120-x) = 90

⇒ x - 60 +(1/2)( x = 90

⇒ (3/2) x = 150
⇒ x= 100

Class X Polynomial Test Paper-1
Class X Polynomial Test Paper-2
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