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NCERT solutions for class 10

10th class is the foundation of a student's academic career and their future will be based on the marks they will get in class 10, so It is really mandatory to get the best study material for Scoring good marks in the most challenging subject Maths. You can see that so many websites are providing good NCERT Solutions for Class 10 Maths and students can download them free of cost very easily from: https://www.entrancei.com/ncert-solutions-class-10

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NCERT solutions for class 10

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  1. https://www.entrancei.com/ https://www.entrancei.com/ncert-solutions-class-10-maths

  2. https://www.entrancei.com/ POLYNOMIALS 1. Polynomial: Algebraic expressions in which the variables involved with only non- negative integral exponents are called polynomials. 2. Terms of a polynomial and their coefficient: If x a x a x p + + = 1 0 ) ( n n a x a x a x a , ,.... , 1 1 0 − a a a a ,...., , 2 1 , 0 are known as their coefficients. − − n n 1 n 2 + + + is a polynomial in variable x, then a are known as the terms of polynomial x ... a x a − 2 n 1 n − 1 and p (x ) n n n 3. Quadratic polynomial: A polynomial of degree 2 is called quadratic polynomial or any polynomial of the form c bx ax + + where a polynomial. 2  are real and a 0 is called quadratic , b , c 4. Cubic polynomial: A polynomial of degree 3 is called cubic polynomial. A polynomial of the form d cx bx ax + + + where b a , polynomial. 3 2  are real and is called cubic , c , d a 0 5. 6. Zeroes of a polynomial. k is said to be zero of a polynomial p(x) if p(k) = 0 Graph of polynomial. (i) Graph of a linear polynomial (ii) A linear polynomial has exactly one zero. ax + is a straight line. b 2 = + + p ( x ) ax bx c (iii) Graph of a quadratic polynomial upwards like  if is a parabola open  a 0 . 2 = + + p ( x ) ax bx c (iv) Graph of a quadratic polynomial downwards like  if is a parabola open  a 0 . (v) A quadratic polynomial has at most two zeros. (vi) In general a polynomial of degree n crosses the x-axis at atmost n points. p (x ) 7. Discriminate of a quadratic polynomial. For bx ax x p ) ( + + = 2 2 − is known as its discriminate ‘D’. c , b 4 ac 2− 0  (i) If = D D b  4 ac 2 = + + , graph of p ( x ) ax bx c will intersect the x-axis at two distinct points, x-coordinates of points of intersection with x-axis are known as ‘zeroes’ of . p (x ) https://www.entrancei.com/ncert-solutions-class-10-maths

  3. https://www.entrancei.com/ y Q x x O y=ax2+bx+c y 2 = + + = (x p (ii) If , graph of will have only one zero. will touch the x-axis at one point only. p ( x ) ax bx c D 0  ) y x x O Q y y=ax2+bx+c ax x = ) ( 2 + +  (x p (iii) If , graph of will not have any real zeroes. will neither touch nor intersect the x-axis. p bx c D 0  ) y x x O Q y y=ax2+bx+c 8. Relationship between the zeroes and the coefficients of a Polynomial. (i) For quadratic polynomial (a) If ,  are zeroes of bx ax x p + = ) ( b − 2 + , then c Sum of zeroes =  +  = a − (coefficie nt of x )   +  = 2 Coefficien c t of x Product of zeroes =  = a ( Constant term )   = 2 Coefficien t of x For Example: To find the zeros of the quadratic polynomial between the zeros and their coefficients. Given quadratic polynomial is 2 − x − and verify the relationship x 2 8 https://www.entrancei.com/ncert-solutions-class-10-maths

  4. https://www.entrancei.com/ 2 2 = − x + − ) = − 4 + ( x − )( y = = y = 0 gives x = –2, 4. These are two zeros such that  = –2,  = 4. x 2 − 2 x 4 )( 8 ( 2 + − x x 4 ) x 2 + x 8 x − = − x ( ( x ) 2 4 ) x 4 − ( 2 ) b  Sum of zeros ( + ) = –2 + 4 = 2 = − = − 1 a 8 c Product of zeros () = – 2 × 4 = – 8 = − = a 1 (b) If ,  are roots of quadratic polynomial  (Sum ) ( − = x k x p or p(x) = k[(x–) (x–)] For Example: , then  of p + (x  )  2 = of −  +  p ( x ) k x ( ) x   2 + zeroes) x Product zeroes 1 and –1 To find a quadratic polynomial, the sum and product of whose zeros are 2 respectively. 1 =  +  ,  = –1 Here 2   f Required polynomial is ) ( [  +  − x k  − − 1 2 2 = +  ( x ) x  ( )] 1 2 = where k is a non-zero constant. k x x     (ii) For cubic polynomial (a) If ,  and  are zeroes of 3 2 = + (Coefficie + + p ( x ) ax bx cx d of 2 − − nt x ) b ,  +  +  = Sum of roots = 3 a Coefficien t of c x Sum of product of roots taken two at a time = a Coefficien t of x  +  +  = 3 Coefficien d − t of x Product of roots = a − (Constant term) x  = 3 Coefficien t of For Example: To verify the relationship between the zeros and co-efficient of cubic polynomial 2 5 4 − + − x x x whose zeros are 2, 1 and 1. Here,  = 2,  = 1,  = 1 b − = − = = + + =  +  +  1 ) 2 (− − = a 3 2 − ( 4 ) 2 1 1 4 a d −  =   = 2 1 1 1 https://www.entrancei.com/ncert-solutions-class-10-maths

  5. https://www.entrancei.com/ 5 c 5 =  +  +  =  +  +  = 2 + 1 +2 = = a 2 1 1 1 1 2 1 (b) If ,  and  are zeroes of a cubic polynomial ( x p   − = ) ( k x p Hence the result. , then  + p ( (x  )   ) 3 2 = −  +  +  + +  −  ) k x ( ) x ) x (  3 2 + x (Sum of zeroes) x (Sum of product of zeroes taken two at a time) x     − (Product of zeroes) or p(x) = k[(x–) (x–) (x–)] 9. For Example: To form a cubic polynomial with zeros  = 3,  = 2,  = –1. Here,  = 3,  = 2,  = –1 Required polynomial = − −  3 − − − where k is a non-zero constant. − − + + +  2 6 6 − + +  1 k k k k k [( [( [( [ [ x x x )( )( 5 5 4 x x x )( )( )( x x x x x + + )] )] )] 1 x ] 6 = = = = 2 3 2 2 − + x x x x 5 x ] 6 3 2 Division Algorithm for polynomials. If p (x ) and g (x ) are any two polynomials with g (x )  0, then we can find polynomials q (x ) and r (x ) such that p (x ) = q (x ) × g (x ) + r (x ), where r (x ) = 0 or degree of r (x < degree of g (x ). ) Step 1: Divide the highest degree term of the dividend by the highest degree term of the divisor and obtain the remainder. Step 2: If the remainder is zero or degree of remainder is less than divisor, then we cannot continue the division any further. If degree of remainder is equal to or more than divisor repeats step 1. For Example: Apply the division algorithm to find the quotient and remainder on dividing ) (x g as given below. 3 5 3 ) ( − + − = x x x x p , ) ( x g 2 x 5 3 + − x x – + –3x2 + 7x–3 –3x2 +6 + 7x – 9  3 ) ( − = x x q  ( r by p (x ) 3 2 2− = x 2 x – 3 2− 3 2 − x 3 x3– 2x – 9 = x − x ) 7 https://www.entrancei.com/ncert-solutions-class-10-maths

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