<\/span><\/h2>\nMark the correct alternative in each of the following :<\/strong>
Question 1.<\/strong>
If \u03b1, \u03b2 are the zeros of the polynomial f(x) = x2<\/sup>\u00a0+ x + 1, then\u00a01\u03b1+1\u03b2\u00a0=
(a) 1
(b) -1
(c) 0
(d) None of these
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/q.png\")
<\/p>\nQuestion 2.<\/strong>
If \u03b1, \u03b2 are the zeros of the polynomial p(x) = 4x2<\/sup>\u00a0+ 3x + 7, then\u00a01\u03b1+1\u03b2\u00a0is equal to
(a)\u00a073
(b) \u2013\u00a073
(c)\u00a037
(d) \u2013\u00a037
Solution:<\/strong>
(d)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/q1.png\")
<\/p>\nQuestion 3.<\/strong>
If one zero of the polynomial f(x) = (k2<\/sup>\u00a0+ 4) x2<\/sup>\u00a0+ 13x + 4k is reciprocal of the other, then k =
(a) 2
(b) -2
(c) 1
(d) -1
Solution:<\/strong>
(a)<\/strong>\u00a0f (x) = (k2<\/sup>\u00a0+ 4) x2<\/sup>\u00a0+ 13x + 4k
Here a = k2<\/sup>\u00a0+ 4, b = 13, c = 4k
One zero is reciprocal of the other
Let first zero = \u03b1
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/q3.png\")
k = 2<\/p>\nQuestion 4.<\/strong>
If the sum of the zeros of the polynomial f(x) = 2x3<\/sup>\u00a0\u2013 3kx2<\/sup>\u00a0+ 4x \u2013 5 is 6, then value of k is
(a) 2
(b) 4
(c) -2
(d) -4
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/q5.png\")
<\/p>\nQuestion 5.<\/strong>
If \u03b1 and \u03b2 are the zeros of the polynomial f(x) = x2<\/sup>\u00a0+ px + q, then a polynomial having \u03b1 and \u03b2 is its zeros is
(a) x2<\/sup>\u00a0+ qx + p
(b) x2<\/sup>\u00a0\u2013 px + q
(c) qx2<\/sup>\u00a0+ px + 1
(d) px2<\/sup>\u00a0+ qx + 1
Solution:<\/strong>
(c)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/q6.png\")
<\/p>\nQuestion 6.<\/strong>
If \u03b1, \u03b2 are the zeros of polynomial f(x) = x2<\/sup>\u00a0\u2013 p (x + 1) \u2013 c, then (\u03b1 + 1) (\u03b2 + 1) =
(a) c \u2013 1
(b) 1 \u2013 c
(c) c
(d) 1 + c
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/q7.png\")
<\/p>\nQuestion 7.<\/strong>
If \u03b1, \u03b2 are the zeros of the polynomial f(x) = x2<\/sup>\u00a0\u2013 p(x + 1) \u2013 c such that (\u03b1 + 1) (\u03b2 + 1) = 0, then c =
(a) 1
(b) 0
(c) -1
(d) 2
Solution:<\/strong>
(a)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/aa-1.png\")
<\/p>\nQuestion 8.<\/strong>
If f(x) = ax2<\/sup>\u00a0+ bx + c has no real zeros and a + b + c < 0, then
(a) c = 0
(b) c > 0
(c) c < 0
(d) None of these
Solution:<\/strong>
(d)<\/strong>\u00a0f(x) = ax2<\/sup>\u00a0+ bx + c
Zeros are not real
b2<\/sup>\u00a0\u2013 4ac < 0 \u2026.(i)
but a + b + c < 0
b < \u2013 (a + c)
Squaring both sides b2<\/sup>\u00a0< (a + c)2<\/sup>
=> (a + c)2<\/sup>\u00a0\u2013 4ac < 0 {From (i)}
=> (a \u2013 c)2<\/sup>\u00a0< 0
=> a \u2013 c < 0
=> a < c<\/p>\nQuestion 9.<\/strong>
If the diagram in figure shows the graph of the polynomial f(x) = ax2<\/sup>\u00a0+ bx + c, then
(a) a > 0, b < 0 and c > 0
(b) a < 0, b < 0 and c < 0
(c) a < 0, b > 0 and c > 0
(d) a < 0, b > 0 and c < 0
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/a-2.png\")
Solution:<\/strong>
(a)<\/strong>\u00a0Curve ax2<\/sup>\u00a0+ bx + c intersects x-axis at two points and curve is upward.
a > 0, b < 0 and c> 0<\/p>\nQuestion 10.<\/strong>
Figure shows the graph of the polynomial f(x) = ax2<\/sup>\u00a0+ bx + c for which
(a) a < 0, b > 0 and c > 0
(b) a < 0, b < 0 and c > 0
(c) a < 0, b < 0 and c < 0
(d) a > 0, b > 0 and c < 0
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/1-2.png\")
Solution:<\/strong>
(b)<\/strong>\u00a0Curve ax2<\/sup>\u00a0+ bx + c intersects x-axis at two points and curve is downward.
a < 0, b < 0 and c > 0<\/p>\nQuestion 11.<\/strong>
If the product of zeros of the polynomial f(x) = ax3<\/sup>\u00a0\u2013 6x2<\/sup>\u00a0+ 11x \u2013 6 is 4, then a =
(a)\u00a032
(b) \u2013\u00a032
(c)\u00a023
(d) \u2013\u00a023
Solution:<\/strong>
(a)<\/strong>\u00a0f(x) = ax3<\/sup>\u00a0\u2013 6x2<\/sup>\u00a0+ 11x \u2013 6
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/2-2.png\")
<\/p>\nQuestion 12.<\/strong>
If zeros of the polynomial f(x) = x3<\/sup>\u00a0\u2013 3px2<\/sup>\u00a0+ qx \u2013 r are in AP, then
(a) 2p3<\/sup>\u00a0= pq \u2013 r
(b) 2p3<\/sup>\u00a0= pq + r
(c) p3<\/sup>\u00a0= pq \u2013 r
(d) None of these
Solution:<\/strong>
(a)<\/strong>\u00a0f(x) = x3<\/sup>\u00a0\u2013 3px2<\/sup>\u00a0+ qx \u2013 r
Here a = 1, b = -3p, c = q, d= -r
Zeros are in AP
Let the zeros be \u03b1 \u2013 d, \u03b1, \u03b1 + d
<\/p>\n![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/3-2.png\")
<\/p>\n
Question 13.<\/strong>
If the product of two zeros of the polynomial f(x) = 2x3<\/sup>\u00a0+ 6x2<\/sup>\u00a0\u2013 4x + 9 is 3, then its third zero is
(a)\u00a032
(b) \u2013\u00a032
(c)\u00a092
(d) \u2013\u00a092
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/4-2.png\")
<\/p>\nQuestion 14.<\/strong>
If the polynomial f(x) = ax2<\/sup>\u00a0+ bx \u2013 c is divisible by the polynomial g(x) = ax2<\/sup>\u00a0+ bx + c, then ab =
(a) 1
(b)\u00a01c
(c) \u2013 1
(d) \u2013\u00a01c
Solution:<\/strong>
(a)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/5a.png\")
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/5b.png\")
<\/p>\nQuestion 15.<\/strong>
In\u00a0Q. No. 14<\/strong>, ac =
(a) b
(b) 2b
(c) 2b2
(d) -2b
Solution:<\/strong>
(b)<\/strong>\u00a0In the previous questions
Remainder = 0
(b \u2013 ac + ab2<\/sup>) = 0
b + ab2<\/sup>\u00a0= ac
=> ac = b (1 + ab) = b (1 + 1) = 2b<\/p>\nQuestion 16.<\/strong>
If one root of the polynomial f(x) = 5x2<\/sup>\u00a0+ 13x + k is reciprocal of the other, then the value of k is
(a) 0
(b) 5
(c)\u00a016
(d) 6
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/5c.png\")
<\/p>\nQuestion 17.<\/strong>
If \u03b1, \u03b2, \u03b3 are the zeros of the polynomial f(x) = ax3<\/sup>\u00a0+ bx2<\/sup>\u00a0+ cx + d, then\u00a01\u03b1+1\u03b2+1\u03b3\u00a0=
(a) \u2013\u00a0bd
(b)\u00a0cd
(c) \u2013\u00a0cd
(d)\u00a0ca
Solution:<\/strong>
(c)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/5d.png\")
<\/p>\nQuestion 18.<\/strong>
If \u03b1, \u03b2, \u03b3 are the zeros of the polynomial f(x) = ax3<\/sup>\u00a0+ bx2<\/sup>\u00a0+ cx + d, then \u03b12<\/sup>\u00a0+ \u03b22<\/sup>\u00a0+ \u03b32<\/sup>\u00a0=
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/6.png\")
Solution:<\/strong>
(d)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/6a.png\")
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/6d.png\")
<\/p>\nQuestion 19.<\/strong>
If \u03b1, \u03b2, \u03b3 are the zeros of the polynomial f(x) = x3<\/sup>\u00a0\u2013 px2<\/sup>\u00a0+ qx \u2013 r, then\u00a01\u03b1\u03b2+1\u03b2\u03b3+1\u03b3\u03b1\u00a0=
(a)\u00a0rp
(b)\u00a0pr
(c) \u2013\u00a0pr
(d) \u2013\u00a0rp
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/7.png\")
<\/p>\nQuestion 20.<\/strong>
If \u03b1, \u03b2 are the zeros of the polynomial f(x) = ax2<\/sup>\u00a0+ bx + c, then\u00a01\u03b12+1\u03b22\u00a0=
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/8.png\")
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/05\/RD-Sharma-Class-10-Solutions-Chapter-2-Polynomials-MCQS-24.png\")
<\/p>\nQuestion 21.<\/strong>
If two of the zeros of the cubic polynomial ax3<\/sup>\u00a0+ bx2<\/sup>\u00a0+ cx + d are each equal to zero, then the third zero is
(a)\u00a0\u2212da
(b)\u00a0ca
(c)\u00a0\u2212ba
(d)\u00a0ba
Solution:<\/strong>
(c)<\/strong>\u00a0Two of the zeros of the cubic polynomial ax3<\/sup>\u00a0+ bx2<\/sup>\u00a0+ cx + d are each equal to zero
Let \u03b1, \u03b2 and \u03b3 are its zeros, then
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/05\/RD-Sharma-Class-10-Solutions-Chapter-2-Polynomials-MCQS-25.png\")
Third zero will be\u00a0\u2212ba<\/p>\nQuestion 22.<\/strong>
If two zeros of x3<\/sup>\u00a0+ x2<\/sup>\u00a0\u2013 5x \u2013 5 are \u221a5 and \u2013 \u221a5 then its third zero is
(a) 1
(b) -1
(c) 2
(d) -2
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/05\/RD-Sharma-Class-10-Solutions-Chapter-2-Polynomials-MCQS-26.png\")
<\/p>\nQuestion 23.<\/strong>
The product of the zeros of x3<\/sup>\u00a0+ 4x2<\/sup>\u00a0+ x \u2013 6 is
(a) \u2013 4
(b) 4
(c) 6
(d) \u2013 6
Solution:<\/strong>
(c)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/05\/RD-Sharma-Class-10-Solutions-Chapter-2-Polynomials-MCQS-27.png\")
<\/p>\nQuestion 24.<\/strong>
What should be added to the polynomial x2<\/sup>\u00a0\u2013 5x + 4, so that 3 is the zero of the resulting polynomial ?
(a) 1
(b) 2
(c) 4
(d) 5
Solution:<\/strong>
(b)<\/strong>\u00a03 is the zero of the polynomial f(x) = x2<\/sup>\u00a0\u2013 5x + 4
x \u2013 3 is a factor of f(x)
Now f(3) = (3)2<\/sup>\u00a0\u2013 5 x 3 + 4 = 9 \u2013 15 + 4 = 13 \u2013 15 = -2
-2 is to be subtracting or 2 is added<\/p>\nQuestion 25.<\/strong>
What should be subtracted to the polynomial x2<\/sup>\u00a0\u2013 16x + 30, so that 15 is the zero of the resulting polynomial ?
(a) 30
(b) 14
(b) 15
(d) 16
Solution:<\/strong>
(c)<\/strong>\u00a015 is the zero of polynomial f(x) = x2<\/sup>\u00a0\u2013 16x + 30
Then f(15) = 0
f(15) = (15)2<\/sup>\u00a0\u2013 16 x 15 + 30 = 225 \u2013 240 + 30 = 255 \u2013 240 = 15
15 is to be subtracted<\/p>\nQuestion 26.<\/strong>
A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is
(a) x2<\/sup>\u00a0\u2013 9
(b) x2<\/sup>\u00a0+ 9
(c) x2<\/sup>\u00a0+ 3
(d) x2<\/sup>\u00a0\u2013 3
Solution:<\/strong>
(a)<\/strong>\u00a0In a quadratic polynomial
Let \u03b1 and \u03b2 be its zeros
and \u03b1 + \u03b2 = 0
and one zero = 3
3 + \u03b2 = 0 \u21d2 \u03b2 = -3 .
Second zero = -3
Quadratic polynomial will be
(x \u2013 3) (x + 3) \u21d2 x2<\/sup>\u00a0\u2013 9<\/p>\nQuestion 27.<\/strong>
If two zeroes of the polynomial x3<\/sup>\u00a0+ x2<\/sup>\u00a0\u2013 9x \u2013 9 are 3 and -3, then its third zero is
(a) -1
(b) 1
(c) -9
(d) 9
Solution:<\/strong>
(a)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/05\/RD-Sharma-Class-10-Solutions-Chapter-2-Polynomials-MCQS-28.png\")
=> \u03b3 = -1
Third zero = -1<\/p>\nQuestion 28.<\/strong>
If \u221a5 and \u2013 \u221a5 are two zeroes of the polynomial x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0\u2013 5x \u2013 15, then its third zero is
(a) 3
(b) \u2013 3
(c) 5
(d) \u2013 5
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/05\/RD-Sharma-Class-10-Solutions-Chapter-2-Polynomials-MCQS-29.png\")
<\/p>\nQuestion 29.<\/strong>
If x + 2 is a factor x2<\/sup>\u00a0+ ax + 2b and a + b = 4, then
(a) a = 1, b = 3
(b) a = 3, b = 1
(c) a = -1, b = 5
(d) a = 5, b = -1
Solution:<\/strong>
(b)<\/strong>\u00a0x + 2 is a factor of x2<\/sup>\u00a0+ ax + 2b and a + b = 4
x + 2 is one of the factor
x = \u2013 2 is its one zero
f(-2) = 0
=> (-2)2<\/sup>\u00a0+ a (-2) + 2b = 0
=> 4 \u2013 2a + 2b = 0
=> 2a \u2013 2b = 4
=> a \u2013 b = 2
But a + b = 4
Adding we get, 2a = 6 => a = 3
and a + b = 4 => 3 + b = 4 => b = 4 \u2013 3 = 1
a = 3, b = 1<\/p>\nQuestion 30.<\/strong>
The polynomial which when divided by \u2013 x2<\/sup>\u00a0+ x \u2013 1 gives a quotient x \u2013 2 and remainder 3, is
(a) x3<\/sup>\u00a0\u2013 3x2<\/sup>\u00a0+ 3x \u2013 5
(b) \u2013 x3<\/sup>\u00a0\u2013 3x2<\/sup>\u00a0\u2013 3x \u2013 5
(c) \u2013 x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0\u2013 3x + 5
(d) x3<\/sup>\u00a0\u2013 3x2<\/sup>\u00a0\u2013 3x + 5
Solution:<\/strong>
(c)<\/strong>\u00a0Divisor = \u2013 x2<\/sup>\u00a0+ x \u2013 1, Quotient = x \u2013 2 and
Remainder = 3, Therefore
Polynomial = Divisor x Quotient+Remainder
= (-x2<\/sup>\u00a0+ x \u2013 1) (x \u2013 2) + 3
= \u2013 x3<\/sup>\u00a0+ x2<\/sup>\u00a0\u2013 x + 2x2<\/sup>\u00a0\u2013 2x + 2 + 3
= \u2013 x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0\u2013 3x + 5<\/p>\nQuestion 31.<\/strong>
The number of polynomials having zeroes -2 and 5 is
(a) 1
(b) 2
(c) 3
(d) more than 3
Solution:<\/strong>
(d)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/05\/RD-Sharma-Class-10-Solutions-Chapter-2-Polynomials-MCQS-30.png\")
Hence, the required number of polynomials are infinite i.e., more than 3.<\/p>\n
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-2-MCQs.jpg\")
<\/p>\n