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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 651. |
Factorise `x^(2)+6sqrt(6)x+48`. |
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Answer» Here, `a=1, b=6sqrt(6), c=48` `therefore a xx c=48` `48=2xx2xx2xx2xx3` `=sqrt(2)xxsqrt(2)xx2xx2xx2xxsqrt(3)xxsqrt(3)` `=2sqrt(6)xx4sqrt(6)` Now, we take two factors of 48 whose sum is `6sqrt(6)` Such factors are `2sqrt(6)` and `4sqrt(6)` `therefore x^(2)+6sqrt(6)x+48=x^(2)+2sqrt(6)x+4sqrt(6)x+48` `=x(x+2sqrt(6))+4sqrt(6)(x+2sqrt(6))` `=(x+2sqrt(6))(x+4sqrt(6))` |
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| 652. |
Write (p + 2q + r)2 in expanded form. |
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Answer» (p + 2q + r)2 = (p)2 + (2q)2 + (r)2 + 2 (P) (2q) + 2 (2q) (r) + 2(r) (p) = p2 + 4q2 + r2 + 4pq + 4qr + 2rp |
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| 653. |
Classify the following as linear, quadratic and cubic polynomials : (i) x2 + x (ii) x – x3 (iii) y + y2 + 4 (iv) 1 + x (v) 3t (vi) r2(vii) 7x3 |
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| 654. |
Write the coefficient of x3 in each of the following.i) x3 + x + 1 ii) 2 – x3 + x2iii)√2x3 + 5iv) 2x3 + 5v) π/2 x3 + xvi) - 2/3 x3vii) 2x2 + 5viii) 4 |
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Answer» Write the coefficient of x3 in each of the following. i) x3 + x + 1 : co-efficient of x3 is 1. ii) 2 – x3 + x2 : co-efficient of x3 is – 1. iii)√2x3 + 5 : co-efficient of x3 is √2 iv) 2x3 + 5 : co-efficient of x3 is 2 v) π/2 x3 + x : co-efficient of x3 is π/2 vi) - 2/3 x3 : co-efficient of x3 is -2/3 vii) 2x2 + 5 : co-efficient of x3 is '0'. viii) 4 : co-efficient of x3 is '0'. |
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| 655. |
Write whether the following statements are True or False. Justify your answer. i) A binomial can have at the most two terms ii) Every polynomial is a binomial iii) A binomial may have degree 3 iv) Degree of zero polynomial is zero v) The degree of x2 + 2xy + y2 is 2 vi) πr2 is monomial |
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Answer» i) A binomial can have at the most two terms - True ii) Every polynomial is a binomial – False [∵ A polynomial can have more than two terms] iii) A binomial may have degree 3 – True iv) Degree of zero polynomial is zero – False v) The degree of x2 + 2xy + y2 is 2 – True vi) πr2 is monomial – True |
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| 656. |
Find the value of each of the following : `(i) p(x)=3x+7 " at" x=1 " " (ii) q(y)=y^(3)-3y^(2)+sqrt(3) " at" y=1` (iii)`p(a)=a^(4)+6a^(2)-6a+3` at a=m |
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Answer» (i) p(x)=3x+7 `implies p(1)=3xx1+7=10` (ii) `q(y)=y^(3)=3y^(2)+sqrt(3)` `implies q(1)=1^(3)-3xx1^(2)+sqrt(3)=1-3+sqrt(3)=-2+sqrt(3)` (iii) `p(a)=a^(4)+6a^(2)-6a+3` `implies p(m)=m^(4)+6m^(2)-6m+3` |
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| 657. |
Find the coefficient of `x^(2)` in the polynomial : `2x^(3)-3x^(2)+5x(1-(x)/(2))+2x^(2)(x+7)-13` |
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Answer» Given polynomial is `2x^(3)-3x^(2)+5x(1-(x)/(2))+2x^(2)(x+7)-13` `=2x^(3)-3x^(2)+5x-(5)/(2)x^(2)+2x^(3)+14x^(2)-13` `x^(3)(2+2)+x^(2)(-3-(5)/(2)+14)+x(5)-13=4x^(3)+(17)/(2)x^(2)+5x-13` `therefore` Coefficient of `x^(2) " is" (17)/(2)`. |
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| 658. |
Evaluate : `(i) (a+6b)^(2) " " (ii) (3x-4y)^(2) " " (iii) (2a-b+c)^(2)` |
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Answer» (i) `(a+6b)^(2)=a^(2)+2xxaxx6b+(6b)^(2)=a^(2)+12ab+36b^(2)` `(ii) (3x-4y)^(2)=(3x)^(2)-2xx3x xx4y+(4y)^(2)=9x^(2)-24xy+16y^(2)` `(iii) " We know that " (x+y+z)^(2)=x^(2)+y^(2)+z^(2)+2xy+2yz+2zx` Now, `(2a-b+c)^(2)=(2a)^(2)+(-b)^(2)+(c )^(2)+2{(2a)(-b)xxc+cxx(2a)}` `=4a^(2)+b^(2)+c^(2)-4ab-2bc+4ac` |
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| 659. |
Factorise using appropriate identities : x2 + 3x + 2 |
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Answer» x2 + 3x + 2 = x2 + (2 + 1) x + (2 x 1) (x + 2) (x + 1) |
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| 660. |
Determine the degree of each of the following polynomials:(i) 2x – 1(ii) –10(iii) x3 – 9x + 3x5(iv) y3 (1 – y4) |
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Answer» Degree of a polynomial in one variable = highest power of the variable in algebraic expression (i) 2x – 1 Power of x = 1 Highest power of the variable x in the given expression = 1 Hence, degree of the polynomial 2x – 1 = 1 (ii) –10 There is no variable in the given term. Let us assume that the variable in the given expression is x. – 10 = –10x0 Power of x = 0 Highest power of the variable x in the given expression = 0 Hence, degree of the polynomial – 10 = 0 (iii) x3 – 9x + 3x5 Powers of x = 3, 1 and 5 respectively. Highest power of the variable x in the given expression = 5 Hence, degree of the polynomial x3 – 9x + 3x5= 5 (iv) y3 (1 – y4) The equation can be written as, y3 (1 – y4) = y3 – y7 Powers of y = 3 and 7 respectively. Highest power of the variable y in the given expression = 7 Hence, degree of the polynomial y3 (1 – y4) = 7 |
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| 661. |
Write the coefficient of x2 in each of the following(i) (π/6)x + x2 – 1(ii) 3x – 5(iii) (x –1) (3x – 4)(iv) (2x – 5) (2x2 – 3x + 1) |
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Answer» (i) (π/6) x + x2−1 (π/6) x + x2−1 = (π/6) x + (1) x2−1 The coefficient of x2 in the polynomial (π/6) x + x2−1 = 1. (ii) 3x – 5 3x – 5 = 0x2 + 3x – 5 The coefficient of x2 in the polynomial 3x – 5 = 0, zero. (iii) (x – 1) (3x – 4) (x – 1)(3x – 4) = 3x2 – 4x – 3x + 4 = 3x2 – 7x + 4 The coefficient of x2 in the polynomial 3x2 – 7x + 4 = 3. (iv) (2x – 5) (2x2 – 3x + 1) (2x – 5) (2x2 – 3x + 1) = 4x3 – 6x2 + 2x – 10x2 + 15x– 5 = 4x3 – 16x2 + 17x – 5 The coefficient of x2 in the polynomial (2x – 5) (2x2 – 3x + 1) = – 16 |
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| 662. |
Factorise : `27x^(3)+8y^(3)+8z^(3)-36xyz` |
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Answer» We can write it `(3x)^(3)+(2y)^(3)+(2z)^(3)-3xx3x xx2yxx2z` `=a^(3)+b^(3)+c^(3)-3abc " " ("where" a=3x,b=2y,c=2z)` `=(3x+2y+2z)(9x^(2)+4y^(2)+4z^(2)-5xy-4yz-6zx)` |
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| 663. |
Give possible expressions for the length and breadth of the rectangle whose area is given by 25a2 – 35a + 12 |
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Answer» Given 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a (5a – 4) – 3 (5a – 4) = (5a – 4) (5a – 3) ∴ (5a – 4) (5a – 3) are the length and breadth. |
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| 664. |
Give possible expressions for the length and breadth of the rectangle whose area is given by 4a2 + 4a – 3 |
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Answer» Given that area = 4a2 + 4a – 3 = 4a2 + 6a – 2a – 3 = 2a (2a + 3) – 1 (2a + 3) = (2a – 1) (2a + 3) ∴ Length = (2a + 3); breadth = (2a – 1). |
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| 665. |
Find product using appropriate identities : (t + 2) (t + 4) |
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Answer» (t + 2) (t + 4) = t2 + t(2 + 4) + 2 x 4 = t2 + 6t + 8 |
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| 666. |
Find product using appropriate identities : (y – 1) (y – 1) |
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Answer» (y – 1) (y – 1) = (y – 1)2 = y2 – 2y + 1 |
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| 667. |
Factorise : `(a) x^(6)+y^(6) " " (b) x^(6)-y^(6)` |
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Answer» (a) We have `x^(6)+y^(6)=(x^(2))^(3)+(y^(2))^(3)` `=(x^(2)+y^(2))(x^(4)-x^(2)y^(2)+y^(4))` `[because a^(3)+b^(3)=(A+b)(a^(2)-ab+b^(2))]` Most of students write factors till here, which is not proper. We shall factorise further, as `x^(6)+y^(6)=(x^(2)+y^(2))[underset("making perfect square")(ubrace((x^(2))^(2)+2x^(2).y^(2)+(y^(2))^(2)))-2x^(2)y^(2)-x^(2)y^(2)]` `implies x^(6)+y^(6)=(x^(2)+y^(2))[(x^(2)+y^(2))^(2)-3x^(2)y^(2)]=(x^(2)+y^(2))[(x^(2)+y^(2))^(2)-(sqrt(3)xy)^(2)]` `=(x^(2)+y^(2))[(x^(2)+y^(2)+sqrt(3)xy)(x^(2)+y^(2)-sqrt(3)xy)]` `implies underset("polynomial")(ubrace(x^(6)+y^(6)))=underset("polynomial")(ubrace((x^(2)+y^(2))))underset("polynomial")(ubrace((x^(2)+sqrt(3)xy+y^(2))))underset("polynomial")(ubrace((x^(2)-sqrt(3)xy+y^(2))))` `(b) x^(6)-y^(6)=(x^(2))^(3)-(y^(2))3)=(x^(2)-y^(2))(x^(4)+x^(2)y^(2)+y^(4))` `=(x+y)(x-y)[underset("making perfect square")ubrace((x^(2))^(2)+2(x^(2))(y^(2))+(y^(2))^(2))-2x^(2)y^(2)+x^(2)y^(2)]` `=(x+y)(x-y)[(x^(2)+y^(2))^(2)-(xy)^(2)]` `=(x+y)(x-y)[(x^(2)+y^(2)+xy)(x^(2)+y^(2)-xy)]` `=(x+y)(x-y)(x^(2)+xy+y^(2))(x^(2)-xy+y^(2))` Alternative Method : `x^(6)-y^(6)=(x^(3))^(2)-(y^(3))^(2)=(x^(3)+y^(3))(x^(3)-y^(3))` `=(x+y)(x^(2)-xy)+y^(2))(x-y)(x^(2)+xy+y^(2))` `=(x+y)(x-y)(x^(2)+xy+y^(2))(x^(2)-xy+y^(2))` |
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| 668. |
Find the remainder when `x^3+3x^2+3x+1`is divided by(i) `x+1` (ii) `x-1/2` (iii) `x` (iv) `x+pi` (v) `5+2x` |
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Answer» From remainder theorem, we know that if `P(x)` is divided by `(x-a)`, then, ` P(a)` will be the remainder. In this question, `P(x) = x^3+3x^2+3x+1` (i) Here, `a = -1` `P(-1) = (-1)^3+3(-1)^2+3(-1)+1 = -1+3-3+1=0` Remainder is `0`. (ii) Here, `a = 1/2` `P(1/2) = (1/2)^3+3(1/2)^2+3(1/2)+1 =1/8+3/4+3/2+1=27/8` Remainder is `27/8`. (iii) Here, `a = 0` `P(0) = (0)^3+3(0)^2+3(0)+1 =1` Remainder is `1`. (iv)Here, `a = -pi` `P(-pi) = (-pi)^3+3(-pi)^2+3(-pi)+1 =-pi^3+3pi^2-3pi+1` Remainder is `-pi^3+3pi^2-3pi+1`. (v)Here, `a = -5/2` `P(-5/2) = (-5/2)^3+3(-5/2)^2+3(-5/2)+1 =-125/8+75/4-15/2+1` ` =1/8(-125+150-60+8=-27/8` Remainder is `-27/8`. |
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| 669. |
Factorise : `(a) x^(4)+4x^(2)+16 " " (ii) x^(4)+4` |
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Answer» `(a) " We have" , x^(4)+4x^(2)+16=(x^(2))^(2)+2(x^(2))(4)+(4)^(2)-8x^(2)+4x^(2)` [we are trying to make the formula `a^(2)+2ab+b^(2)=(a+b)^(2)]` `=(x^(2)+4)^(2)-4x^(2)=(x^(2)+4)^(2)-(2x)^(2)` `=(x^(2)+4+2x)(x^(2)+4-2x) " " [because a^(2)-b^(2)=(a+b)(a-b)]` `=(x^(2)+2x+4)(x^(2)-2x+4)` `(b) x^(4)+4=(x^(2))^(2)+(2)^(2)` `=ubrace((x^(2))^(2)+2(x^(2))(2)+(2)^(2))-2(x^(2))(2)` (adding and subtracting the same quantity) `(x^(2)+2)^(2)-4x^(2)` `=(x^(2)+2)^(2)-(2x)^(2)=(x^(2)+2+2x)(x^(2)+2-2x)=(x^(2)+2x+2)(x^(2)-2x+2)` |
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| 670. |
What are the possible expressions for the dimensions of the cuboids whose volumes are given below : (i) Volume: 3x2 – 12x (ii) Volume: 12ky2 + 8ky – 20k |
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Answer» (i) length × breadth × height = Volume of Cuboid = 3x2 – 12x = 3x2 – 12x = 3x(x – 4) = 3 × x × (x – 4) l × b × h ∴ Length = 3 Breadth = x Height = (x – 4) (ii) length × breadth × height = Volume of Cuboid = 12ky2 + 8ky – 20k 12ky2 + 8ky – 20k = 4k(3y2 + 2y – 5) = 4k (3y2 + 5y – 3y – 5) = 4k (y(3y + 5) – 1 (3y + 5)) = 4k × (3y + 5) × (y – 1) l × b × h ∴ Length = 4k Breadth = (3y + 5) Height = (y – 1) |
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| 671. |
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given : (i) Area: 25a2 – 35a +12 (ii) Area : 35y2 + 13y – 12 |
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Answer» (i) Area: 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a -4) (5a – 3) length × breadth = Area of rectangle (5a – 4) (5a – 3) = 25a2 – 35a + 12 ∴ Length = (5a – 4) Breadth = (5a – 3) (ii) Area : 35y2 + 13y – 12 Area of Rectangle – Length × Breadth 35y2 + 13y – 12 35y2 + 28y – 15y 7y(5y + 4) – 3(5y + 4) (7y – 3) (5y + 4) ∴ Length = (7y – 3) Breadth = (5y + 4) |
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| 672. |
Find the zero of the polynomial in each of the following eases:(i) `p(x)=x+5` (ii) `p(x)=x-5` (iii) `p(x)=2x+5` (iv) `p(x)=3x-2`(v) `p(x)=3x` (vi) `p(x)=a x , a!=0`(vii) `p(x)=c x+d , c!=0, c , d` are real numbers. |
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Answer» (i)` p(x) = x +5 = 0` `x=-5` (ii) `p(x) = x-5=0` `x=5` (iii) `p(x)= 2x+5 = 0` `x=-5/2` (iv) `p(x)= 3x-2=0` `x=2/3` (v) `p(x)=3x=0` `x=0` (vi) `p(x)= ax=0` `x=0` (vii) `p(x) = cx+d=0` `x=-d/c` answers |
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| 673. |
Factorise: (i) `49 a^2+70 a b+25 b^2` (ii) `(25)/4x^2-(y^2)/9` |
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Answer» `(i) 49a^2+70ab+25b^2` `=(7a)^2+2**7a**5b+(5b)^2` As, we know,` (x+y)^2 = x^2+y^2+2xy` So, our expression becomes, `=(7a+5b)^2 = (7a+5b)(7a+5b)` `(ii)(25/4)x^2 - y^2/9` `=(5x/2)^2 - (y/3)^2` As, we know, `a^2 - b^2 = (a+b)(a-b)` So, our expression becomes, `=(5/2x+y/3)(5/2x-y/3)` |
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| 674. |
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:Area: `25 a^2-35 a+12` Area: `35 y^2+13 y-12`(i) (ii) |
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Answer» (i) `25a^2 - 35a +12` `= 25a^2 - 15a -20a +12` `= 5a(5a-3) - 4(5a-3)` `= (5a-4)(5a-3)` so, `l = 5a-3` `b= 5a-4` (ii) `35y^2 +13y - 12` `35y^2 - 15y +28y -12` `= 5y(7y-3)+4(7y-3)` `= (5y+4)(7y-3)` `l = 5y+4` `b= 7y-3` answer |
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| 675. |
Verify : (i) `x^3+y^3=(x+y)(x^2-x y+y^2)` (ii) `x^3-y^3=(x-y)(x^2+x y+y^2)` |
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Answer» (i) RHS= `(x+y)(x^2-xy+y^2)` `= x(x^2-xy+y^2) + y(x^2-xy+y^2)` `= x^3 cancel(- x^2y) + cancel(xy^2) + cancel(yx^2) cancel(- xy^2)+y^3` `=x^3 + y^3` = LHS hence proved (ii) RHS = `(x-y)(x^2+xy+y^2)` `= x(x^2+xy+y^2) - y(x^2+xy+y^2)` `=x^3 +cancel(x^2y)+cancel(xy^2) cancel( - yx^2) cancel(-xy^2 )- y^3` `=x^3- y^3` =LHS hence proved |
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| 676. |
Find the remainder when `x^3-a x^2+6x-a` is divided by `x-a`. |
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Answer» From remainder theorem, we know that if `p(x)` is divided by `(x-a)`, then remainder will be `p(a).` Here, `p(x) = x^3-ax^2+6x-a` `p(a) = a^3 - a*a^2+6a -a = a^3 -a^3 +6a - a = 5a` So, remainder will be `5a`. |
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| 677. |
Show that if 2 (a2 + b2 ) = (a + b)2 then a = b |
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Answer» Given that 2 (a2 + b2 ) = (a + b)2 To prove a = b As 2 (a2 + b2 ) = (a + b)2 We have 2a2 + 2b2 = a2 + 2ab + b2 2a2 – a2 + 2b2 – b2 = 2ab a2 + b2 = 2ab This is possible only when a = b ∴ a = b |
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| 678. |
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?(i) Volume: `3x^2-12 x` (ii) Volume: `12 k y^2+8k x-20 k` |
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Answer» 1) Volume=`3x^2-12x` =3x(x-4) Possible outcomes=3,x,(x-4) =1,3x,(x-4) =1,x,3(x-4) =1,3,x(x-4) 2) Volume=`12ky^2+8ky-20k` =k(12y^2+8y-20) =4k(y-1)(3y+5) one possible outcome=4k,(y-1)(3y+5) |
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| 679. |
What are the possible polynomial expressions for the dimensions of the cuboids whose volumes are given below ?3x3 – 12x |
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Answer» Volume = 3x3 – 12x = 3x (x2 – 4) = 3x (x + 2) (x – 2) are the dimensions. |
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| 680. |
What are the possible polynomial expressions for the dimensions of the cuboids whose volumes are given below ?12y2 + 8y – 20 |
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Answer» Given that volume = 12y2 + 8y – 20 = 4 (3y2 + 2y – 5) = 4 [3y2 + 5y – 3y – 5] = 4 [y (3y + 5) – 1 (3y + 5)] = 4 (3y + 5) (y – 1) Hence 4, (3y + 5) and (y – 1) are the dimensions. |
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| 681. |
Write the zeros of the polynomial x2 – x – 6. |
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Answer» Given, Polynomial = X2 – x – 6 = 0 x2 – 3x + 2x – 6 = 0 x(x – 3) + 2(x – 3) = 0 (x + 2)(x – 3) = 0 So, – 2 and 3 are the zeros of the given polynomial |
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| 682. |
If the degree of a polynomial AB is 15 and the degree of polynomial A is 5, then find the degree of polynomial B. |
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Answer» Correct Answer - 10 |
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| 683. |
Factorise `a^(6)-b^(6)`. |
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Answer» Correct Answer - `(a-b)(a+b)(a^(2)+ab+b^(2))(a^(2)-ab+b^(2))` |
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| 684. |
If fourth degree polynomial is divided by a quadratic polynomial, write the degree of the remainder. |
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Answer» Degree of reminder is less than the degree of divisor so the degree of reminder could be one or zero depending on the quadratic polynomial. |
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| 685. |
If `p(x)=x^(3)+2x^(2)+x` find : (i) p(0) (ii)p(2) |
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Answer» Correct Answer - (i) 0, (ii) 18 |
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| 686. |
Factorise `(a+b)^(3)-a-b`. |
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Answer» Correct Answer - `(a+b)(a+b+1)(a+b-1)` |
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| 687. |
If f(x) = x3 + x2 – ax + b is divisible by x2 – x write the values of a and b. |
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Answer» a = 2, b = 0 Given, A polynomial f(x) = x3 + x2 – ax + b Which is divisible by x2 – x Now, x 2 – x = (x – 1) = (x – 0)(x – 1) (x – 0) and (x – 1) are factors of polynomial f(x) ⇒ f(0) = 0 ⇒ 0 3 + 0 – a × 0 + b = 0 ⇒ b = 0 And f(1) 1 3 + 1 – a × 1 + b = 0 ⇒ 2 – a = 0 ⇒ a = 2 ∴ a = 2 and b = 0 |
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| 688. |
Write the coefficients of the polynomial p(z) = z5 – 2z2 + 4. |
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Answer» Given, Polynomial p(z) = z5 – 2z2 + 4 Now first break the given equation, We get, Coefficient of z5 = 1 Coefficient of z4 = 0 Coefficient of z3 = 0 Coefficient of z2 = – 2 Coefficient of z = 0 Constant term = 4 Therefore 1, 0, 0, – 2, 0, 4 are coefficients of the polynomial p(z) |
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| 689. |
Check whether p(x) is a multiple of g(x) or not(i) p(x) = x3 – 5x2 + 4x – 3, g(x) = x – 2.(ii) p(x) = 2x3 – 11x2 – 4x+ 5, g(x) = 2x + l |
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Answer» (i) p(x) = x3 – 5x2 + 4x – 3, g(x) = x – 2. According to the question, g(x)=x – 2, Then, zero of g(x), g(x) = 0 x – 2 = 0 x = 2 Therefore, zero of g(x) = 2 So, substituting the value of x in p(x), we get, p(2) =(2)3 – 5(2)2 + 4(2) – 3 = 8 – 20 + 8 – 3 = – 7 ≠ 0 Hence, p(x) is not the multiple of g(x) since the remainder ≠ 0. (ii) p(x) = 2x3 – 11x2 – 4x+ 5, g(x) = 2x + l According to the question, g(x)= 2x + 1 Then, zero of g(x), g(x) = 0 2x + 1 = 0 2x = – 1 x = – ½ Therefore, zero of g(x) = – ½ So, substituting the value of x in p(x), we get, p(–½) = 2 × ( – ½ )3 – 11 × ( – ½ )2 – 4 × ( – 1/2) + 5 = – ¼ – 11/4 + 7 = 16/4 = 4 ≠ 0 Hence, p(x) is not the multiple of g(x) since the remainder ≠ 0. |
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| 690. |
If `p(y)=y^(3)+y^(2)+y+1` find : (i) y(1) (ii) y(-1) |
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Answer» Correct Answer - (i) 4, (ii) 0 |
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| 691. |
Simplify `(4a+2b)^(3)+(4a-2b)^(3)`. |
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Answer» Correct Answer - `128a^(3)+96ab^(2)` |
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| 692. |
Show that (i) `x+3 ` is factor of `69+11x-x^(2)+x^(3)` (ii) `2x-3` is factor of `x+2x^(3)-9x^(2)+12` |
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Answer» (I) let `p(x) =x^(3)-x^(2)+11 x+69` we have to show that , x+3 is a factor of p(x) . `i.e., p(-3)=0` Now `p(-3)=(-3)^(3)-(-3)^(2)+11(-3)+69` `=-2-9-33+69=-69+69=0` hence ,(x+3) is a factor of p(x) . (ii) Let `p(x) =2x^(3)-9x^(2)+x+12` we have to show that ,2x -3 is a factor of p(x). i.e., `p((3)/(2))=0` Now , `p((3)/(2))=2((3)/(2))^(3)-9((3)/(2))^(2)+(3)/(2)+12` `=2xx(27)/(8) -9xx(9)/(4)+(3)/(2)+12` ` =(27)/(4)-(81)/(4) +(3)/(2)+12` `=(27-81+6+48)/(4)=(81-81)/(4)=0` Hence ,`(2x-3)` is a factor of p(x). |
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| 693. |
If a – b, a and a + b are zeros of the polynomial f (x) = 2x3 – 6x2 + 5x – 7, write the value of a. |
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Answer» Given: A polynomial f(x) = 2x3 – 6x2 + 5x – 7 And zeroes are a – b, a and a + b, Let’s take a = α b = β c = y As we know that, α + β + y = – b/a (a – b) + a + (a + b) = – ( – 6)/2 3a = 3 a = 1 So, the value of a = 1 |
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| 694. |
If x3 + x2 - ax + b is divisible by (x2 -x),write the value of a and b. |
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Answer» Equating x2 – x to 0 to find the zeroes, we will get x(x – 1) = 0 ⇒ x = 0 or x – 1 = 0 ⇒ x = 0 or x = 1 Since, x3 + x2 – ax + b is divisible by x2 – x. Hence, the zeroes of x2 – x will satisfy x3 + x2 – ax + b ∴ (0)3 + 02 – a(0) + b = 0 ⇒ b = 0 And (1)3 + 12 – a(1) + 0 = 0 [∵b = 0] ⇒ a = 2 |
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| 695. |
Show that p - 1 is a factor of p10 -1 and also of p11 -1. |
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Answer» Let g (p) = p10 -1 and h(p) = p11 -1. On putting p=1 in Eq. (i), we get g(1)=110-1 = 1-1=0 Hence, p-1 is a factor of g(p). Again, putting p = 1 in Eq. (ii), we get h(1) = (1)11 - 1 = 1 - 1 = 0 Hence, p -1 is a factor of h(p) |
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| 696. |
Show that,(i) x + 3 is a factor of 69 + 11c – x2 + x3(ii) 2x – 3 is a factor of x + 2x3 -9x2 +12 |
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Answer» (i) According to the question, Let p(x) = 69 + 11x − x2 + x3 and g(x) = x + 3 g(x) = x + 3 zero of g(x) ⇒ g(x) = 0 x + 3 = 0 x = – 3 Therefore, zero of g(x) = – 3 So, substituting the value of x in p(x), we get, p( – 3) = 69 + 11( – 3) –( – 3)2 + ( – 3)3 = 69 – 69 = 0 Since, the remainder = zero, We can say that, g(x) = x + 3 is factor of p(x) = 69 + 11x − x2 + x3 (ii) According to the question, Let p(x) = x + 2x3 – 9x2 + 12 and g(x) =2x−3 g(x) = 2x – 3 zero of g(x) ⇒ g(x) = 0 2x – 3 = 0 x = 3/2 Therefore, zero of g(x) = 3/2 So, substituting the value of x in p(x), we get, P(3/2) = 3/2 + 2(3/2)3 – 9(3/2)2 + 12 = (81 – 81) / 4 = 0 Since, the remainder = zero, We can say that, g(x) = 2x – 3 is factor of p(x) = x + 2x3 – 9x2 + 12 |
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| 697. |
Evaluate `(2x-3y+5)^(3)`. |
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Answer» Correct Answer - `8x^(3)-27y^(2)-36x^(2)y+54xy^(2)+125+60x^(2)+135y^(2)-180xy+150x-225y` |
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| 698. |
Evaluate `((2)/(3)x-(3)/(4)y+(4)/(5)z)^(2)`. |
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Answer» Correct Answer - `(4)/(9)x^(2)+(9)/(16)y^(2)+(16)/(25)z^(2)-xy-(6)/(5)yz+(16)/(15)zx` |
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| 699. |
If (a - b) , a and (a + b) are zeros of the polynomial 2x3 - 6x2 + 5x - 7 write the value of a |
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Answer» By using the relationship between the zeroes of the quadratic polynomial. We have Sum of zeroes = \(\frac{-(coefficient\,of\,x^2)}{coefficient\,of\,x^3}\) ⇒ a – b + a + a + b = \(\frac{-(-6)}2\) ⇒ 3a = 3 ⇒ a = 1 |
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| 700. |
If `alpha,beta,gamma` are the roots of equation `x^3+ qx + r =0` then find thene find the value of `1/6 (sum alpha^3)^2`. |
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Answer» `x^3 + qx + r =0` roots of the equation are`alpha , beta, gamma` `alpha + beta + gamma= 0` `gamma beta + beta alpha + gamma alpha = - q` `alpha beta gamma = r` we have to find : `[1/3 sum alpha^3][1/2 sum alpha^3]` `= 1/6 [ sum alpha^3]^2` `= 1/6[alpha^3 + beta^3 + gamma ^3]^2` if `a+b+c = 0` then `a^3+ b^3+c^3 = 3abc` `= 1/6(3 alpha beta gamma)^2` `= (9r^2)/6 = 3/2r^2` answer |
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