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Page No 64:
Question 1:
Which one of the following is a polynomial?
(a)
(b)
(c)
(4)
Answer:
(a)
Exponent of x cannot be negative.
Therefore, it is not a polynomial.
(b)
Exponent of x must be a whole number.
Therefore, it is not a polynomial.
(c)
It is a polynomial.
(d)
It is not a polynomial.
Hence, the correct option is (c).
Page No 64:
Question 2:
Degree of the polynomial f(x) = 4x4 + 0x3 + 0x5 + 5x + 7 is
(a) 4
(b) 5
(c) 3
(d) 7
Answer:
Given: f(x) = 4x4 + 0x3 + 0x5 + 5x + 7 = 4x4 + 5x + 7
Degree is the highest power of x in the polynomial.
Here, the highest power of x is 4.
Thus, degree of the polynomial is 4.
Hence, the correct option is (a).
Page No 64:
Question 3:
Degree of the zero polynomial is
(a) 0
(b) 1
(c) any natural number
(d) not defined
Answer:
The zero polynomial is a polynomial in which all the coefficients are zero.
We can't say about powers of x.
Thus, degree of the zero polynomial is not defined.
Hence, the correct option is (d).
Page No 64:
Question 4:
is a polynomial of degree
(a) 2
(b) 0
(c) 1
(d)
Answer:
Given: f(x) =
It can also be written as .
Degree is the highest power of x in the polynomial.
Here, the highest power of x is 0.
Thus, degree of the polynomial is 0.
Hence, the correct option is (b).
Page No 64:
Question 5:
Zero of the zero polynomial is
(a) 0
(b) 1
(c) any real number
(d) not defined
Answer:
Given: zero polynomial i.e., f(x) = 0
Zero of a polynomial is the value of x at which the polynomial becomes zero.
Here, any value of x can be a zero of the zero polynomial.
Hence, the correct option is (c).
Page No 64:
Question 6:
If f(x) = x + 3, then f(x) + f(–x) is equal to
(a) 3
(b) 2x
(c) 0
(d) 6
Answer:
Given: f(x) = x + 3
f(–x) = –x + 3
Thus,
f(x) + f(–x) = x + 3 + (–x + 3)
= x + 3 – x + 3
= 6
Hence, the correct option is (d).
Page No 65:
Question 7:
Zero of the polynomial f(x) = 3x + 7 is
(a)
(b)
(c)
(d) –7
Answer:
Given: f(x) = 3x + 7
Zero of a polynomial is the value of x at which the polynomial becomes zero.
Therefore, zero of the polynomial f(x) = 3x + 7 is .
Hence, the correct option is (c).
Page No 65:
Question 8:
On of the zeros of the polynomial f(x) = 2x2 + 7x – 4 is
(a) 2
(b)
(c)
(d) –2
Answer:
Given: f(x) = 2x2 + 7x – 4
Zero of a polynomial is the value of x at which the polynomial becomes zero.
Therefore, zeroes of the polynomial f(x) = 2x2 + 7x – 4 are
â
Hence, the correct option is (b).
Page No 65:
Question 9:
If is equal to
(a) 0
(b) 1
(c)
(d)
Answer:
Given: f(x) =
â
Hence, the correct option is (b).
Page No 65:
Question 10:
x + 1 is a factor of the polynomial
(a) x3 + x2 – x + 1
(b) x3 + x2 + x + 1
(c) x4 + x3 + x2 + 1
(d) x4 + 3x3 + 3x2 + x + 1
Answer:
The polynomial f(x) which becomes zero when x = −1 is the required polynomial whose factor is x + 1.
(a) Let f(x) = x3 + x2 – x + 1
f(−1) = (−1)3 + (−1)2 − (−1) + 1
= −1 + 1 + 1 + 1 = 2 ≠ 0
Therefore, x + 1 is not a factor of the polynomial,
(b) Let f(x) = x3 + x2 + x + 1
f(−1) = (−1)3 + (−1)2 + (−1) + 1
= −1 + 1 − 1 + 1 = 0
Therefore, x + 1 is a factor of the polynomial,
(c) Let f(x) = x4 + x3 + x2 + 1
f(−1) = (−1)4 + (−1)3 + (−1)2 + 1
= 1 − 1 + 1 + 1 = 2 ≠ 0
Therefore, x + 1 is not a factor of the polynomial,
(d) Let f(x) = x4 + 3x3 + 3x2 + x + 1
f(−1) = (−1)4 + 3(−1)3 + 3(−1)2 + (−1) + 1
= 1 − 3 + 3 − 1 + 1 = 1 ≠ 0
Therefore, x + 1 is not a factor of the polynomial,
Hence, the correct option is (b).
Page No 65:
Question 11:
If x2 + kx + 6 = (x + 2) (x + 3) for all x, then the value of k is
(a) 1
(b) –1
(c) 5
(d) 3
Answer:
Given: x2 + kx + 6 = (x + 2) (x + 3)
Hence, the correct option is (c).
Page No 65:
Question 12:
If x − 2 is a factor of x2 + 3ax − 2a, then a =
(a) 2
(b) − 2
(c) 1
(d) −1
Answer:
As is a factor of
i.e.
Hence, correct option is (d).
Page No 65:
Question 13:
If x3 + 6x2 + 4x + k is exactly divisible by x + 2, then k =
(a) −6
(b) −7
(c) −8
(d) −10
Answer:
As is exactly divisible by
i.e., is a factor of f(x).
Therefore,
Hence, the correct option is (c).
Page No 65:
Question 14:
If x − a is a factor of x3 −3x2a + 2a2x + b, then the value of b is
(a) 0
(b) 2
(c) 1
(d) 3
Answer:
As is a factor of
i.e.
Thus, b = 0
Thus, the correct option is (a).
Page No 65:
Question 15:
If x140 + 2x151 + k is divisible by x + 1, then the value of k is
(a) 1
(b) −3
(c) 2
(d) −2
Answer:
As is divisible by
i.e., is a factor of f(x).
Therefore,
Hence, the correct option is (a).
Page No 65:
Question 16:
If x + 2 is a factor of x2 + mx + 14, then m =
(a) 7
(b) 2
(c) 9
(d) 14
Answer:
As is a factor of
Therefore,
Hence, the correct option is (c).
Page No 65:
Question 17:
If x − 3 is a factor of x2 − ax − 15, then a =
(a) −2
(b) 5
(c) −5
(d) 3
Answer:
As is a factor of polynomial
i.e.
Therefore,
Hence, the correct option is (a)
Page No 65:
Question 18:
If x51 + 51 is divided by x + 1, the remainder is
(a) 0
(b) 1
(c) 49
(d) 50
Answer:
As is divided by
The remainder will be
Hence, the correct option is (d).
Page No 65:
Question 19:
If x + 1 is a factor of the polynomial 2x2 + kx, then k =
(a) −2
(b) −3
(c) 4
(d) 2
Answer:
As is a factor of polynomial Therefore,
So,
Hence, the correct option is (d).
Page No 65:
Question 20:
If x + a is a factor of x4 − a2x2 + 3x − 6a, then a =
(a) 0
(b) −1
(c) 1
(d) 2
Answer:
Asis a factor of polynomial
Therefore,
Hence, the correct option is (a).
Page No 66:
Question 21:
The value of k for which x − 1 is a factor of 4x3 + 3x2 − 4x + k, is
(a) 3
(b) 1
(c) −2
(d) −3
Answer:
As is a factor of polynomial f(x) = 4x3 + 3x2 − 4x + k
Therefore,
Hence, the correct option is (d).
Page No 66:
Question 22:
If x + 2 and x − 1 are the factors of x3 + 10x2 + mx + n, then the values of m and n are respectively
(a) 5 and −3
(b) 17 and −8
(c) 7 and −18
(d) 23 and −19
Answer:
It is given and are the factors of the polynomial
i.e., and
Now
And
Solving equation (i) and (ii) we get
m = 7 and n = − 18
Hence, the correct option is (c)
Page No 66:
Question 23:
Let f(x) be a polynomial such that = 0, then a factor of f(x) is
(a) 2x − 1
(b) 2x + 1
(c) x − 1
(d) x + 1
Answer:
Let f(x) be a polynomial such that
i.e., is a factor.
On rearranging can be written as
Thus, is a factor of f(x).
Hence, the correct option is (b).
Page No 66:
Question 24:
When x3 − 2x2 + ax − b is divided by x2 − 2x − 3, the remainder is x − 6. The values of a and b are respectively
(a) −2, −6
(b) 2 and −6
(c) −2 and 6
(d) 2 and 6
Answer:
If the reminder (x −6) is subtracted from the given polynomial then rest of part of this polynomial is exactly divisible by x2 − 2x − 3.
Therefore,
Now,
Therefore, are factors of polynomial p(x).
Now,
And
and
Solving (i) and (ii) we get
Hence, the correct option is (c).
Page No 66:
Question 25:
One factor of x4 + x2 − 20 is x2 + 5. The other factor is
(a) x2 − 4
(b) x − 4
(c) x2 − 5
(d) x + 2
Answer:
It is given that is a factor of the polynomial
Here, reminder is zero. Therefore, is a factor of polynomial.
Thus, the correct option is (a).
Page No 66:
Question 26:
If (x − 1) is a factor of polynomial f(x) but not of g(x) , then it must be a factor of
(a) f(x) g(x)
(b) −f(x) + g(x)
(c) f(x) − g(x)
(d)
Answer:
Asis a factor of polynomial f(x) but not of g(x)
Therefore
Now,
Let
Now
Therefore (x − 1) is also a factor of f(x).g(x).
Hence, the correct option is (a).
Page No 66:
Question 27:
(x+1) is a factor of xn + 1 only if
(i) n is an odd integer
(ii) n is an even integer
(iii) n is a negative integer
(iv) n is a positive integer
Answer:
The linear polynomial is a factor of only if
If n is odd integer, then
Hence, the correct option is (a).
Page No 66:
Question 28:
If x2 + x + 1 is a factor of the polynomial 3x3 + 8x2+ 8x + 3 + 5k, then the value of k is
(a) 0
(b) 2/5
(c) 5/2
(d) −1
Answer:
Let be the given polynomial,
Since is the factor of f(x). Therefore, re`maider will be zero.
Now,
Now,
Hence, the correct option is (b).
Page No 66:
Question 29:
If (3x − 1)7 = a7x7 + a6x6 + a5x5 +...+ a1x + a0, then a7 + a5+ ...+a1 + a0 =
(a) 0
(b) 1
(c) 128
(d) 64
Answer:
Given that,
Putting
We get
Hence, the correct option is (c).
Page No 66:
Question 30:
If both x − 2 and are factors of px2 + 5x + r, then
(a) p = r
(b) p + r = 0
(c) 2p + r = 0
(d) p + 2r = 0
Answer:
As and are the factors of the polynomial
i.e., and
Now,
And
From equation (i) and (ii), we get
Hence, the correct option is (a).
Page No 66:
Question 31:
If x2 − 1 is a factor of ax4 + bx3 + cx2 + dx + e, then
(a) a + c + e = b + d
(b) a + b +e = c + d
(c) a + b + c = d + e
(d) b + c + d = a + e
Answer:
Asis a factor of polynomial
Therefore,
And
f(1) = 0
And
Hence,
The correct option is (a).
Page No 66:
Question 32:
If f(x + 3) = x2 − 7x + 2, then the remainder when f(x) is divided by (x + 1), is
(a) 8
(b) –4
(c) 20
(d) 46
Answer:
Given: f(x + 3) = x2 − 7x + 2 and divisor is (x + 1).
f(x + 3) = x2 − 7x + 2
Replace x be x − 3,
⇒f(x) = (x − 3)2 − 7(x − 3) + 2 .....(1)
Considering (x + 1) = 0
Thus, the remainder, when f(x) is divided by (x + 1) is f(−1).
f(−1) = (−1 − 3)2 − 7(−1 − 3) + 2
⇒ f(−1) = 16 + 28 + 2
⇒ f(−1) = 46
Hence, the correct answer is option (d).
Page No 66:
Question 33:
If then the remainder when f(x) is divided by (2x + 1), is
(a)
(b)
(c)
(d)
Answer:
Given:
Considering (2x + 1) = 0
Hence, the correct answer is option (a).
Page No 66:
Question 34:
If then the remainder when f(x) is divided by (x – 3), is
(a) 10
(b) 11
(c) 7
(d)
Answer:
Given:
Considering (x − 3) = 0
Thus, the remainder, when f(x) is divided by (x − 3) is f(3).
Hence, the correct answer is option (b).
Page No 67:
Question 35:
If f(x – 2) = 2x2 – 3x + 4, then the remainder when f(x) is divided by (x – 1), is
(a) 3
(b) 9
(c) 13
(c) –13
Answer:
Given: f(x – 2) = 2x2 – 3x + 4, and the divisor is (x – 1).
f(x – 2) = 2x2 – 3x + 4
Let t = x – 2 ⇒ t + 2 = x
f(t + 2– 2) = 2(t + 2)2 – 3(t + 2) + 4
⇒ f(t) = 2(t + 2)2 – 3(t + 2) + 4
Now, replacing t by x, we get
f(x) = 2(x + 2)2 – 3(x + 2) + 4 .....(1)
Considering (x – 1) = 0
Thus, the remainder, when f(x) is divided by (x – 1) is f(1).
f(1) = 2(1 + 2)2 – 3(1 + 2) + 4
⇒ f(1) = 2(3)2 – 3(3) + 4
⇒ f(1) = 2(9) – 9 + 4
⇒ f(1) = 18 – 9 + 4
⇒ f(1) = 9 + 4
⇒ f(1) = 13
Hence, the correct answer is option (c).
Page No 67:
Question 36:
When the polynomial p(x) = ax2 + bx + c is divided by x, x – 2 and x + 3, the remainders obtained are 7, 9 and 49 respectively. The value of 3a + 5b + 2c is
(a) –5
(b) 5
(c) 2
(d) –2
Answer:
Given: polynomial p(x) = ax2 + bx + c is divided by x, x – 2 and x + 3, the remainders obtained are 7, 9 and 49 respectively.
According to remainder theorem,
Considering x = 0
⇒ p(0) = c .....(1)
Also, p(0) = 7 .....(2)
From (1) and (2), we get
⇒ c = 7 .....(3)
Considering x – 2 = 0 ⇒ x = 2
⇒ p(2) = 4a + 2b + 7 .....(4)
Also, p(2) = 9 .....(5)
From (4) and (5), we get
⇒ 9 = 4a + 2b + 7
⇒ 2 = 4a + 2b .....(6)
Considering x + 3 = 0 ⇒ x = –3
⇒ p(–3) = 9a – 3b + 7 .....(7)
Also, p(–3) = 49 .....(8)
From (7) and (8), we get
⇒ 49 = 9a – 3âb + 7
⇒ 42 = 9a – 3âb â .....(9)
3 × (6) ⇒ 6 = 12a + 6b .....(10)
Adding (10) and (11), we get
90 = 30a
⇒ a = 3 .....(12)
From (11) and (6), we get
⇒ 2 = 4(3) + 2b
⇒ 2 = 12 + 2b
⇒ –10 = 2b
⇒ –5 = b .....(13)
3a + 5b + 2c = 3(3) + 5(–5) + 2(7) [From (3), (12) and (13)]
= 9 – 25 + 14
= – 2
Hence, the correct answer is option (d).
Page No 67:
Question 37:
If (x – a) and (x – b) are factors of x2 + ax + b, then
(a) a = 1, b = –2
(b) a = –2, b = 1
(c) a = 2, b = –3
(d)
Answer:
Given: (x – a) and (x – b) are factors of x2 + ax + b
Let p(x) = x2 + ax + b
According to the Factor theorem, if (x – a) and (x – b) are factors of x2 + ax + b then a and b are roots of the equation x2 + ax + b.
i.e., p(a) = 0 and p(b) = 0
Also,
Substituting (2) in (1),
Hence, the correct answer is option (a).
Page No 67:
Question 38:
If (x – a) and (x – b) are factors of x2 + ax – b, then
(a) a = –1, b = –2
(b) a = 0, b = 1
(c)
(d) a = –1, b = 2
Answer:
Given: (x – a) and (x – b) are factors of x2 + ax – b
Let p(x) = x2 + ax – b
According to the Factor theorem, if (x – a) and (x – b) are factors of x2 + ax – b then a and b are roots of the equation x2 + ax – b.
i.e., p(a) = 0 and p(b) = 0
Also,
Substituting (2) in (1),
Hence, the correct answer is option (d).
Page No 67:
Question 39:
The ratio of remainders when f(x) = x2 + ax + b is divided by (x – 2) and (x – 3) respectively is 5 : 4. If (x – 1) is a factor of f(x), then
(a)
(b)
(c)
(d)
Answer:
Given: The ratio of remainders when f(x) = x2 + ax + b is divided by (x – 2) and (x – 3) respectively is 5 : 4.
According to the Factor theorem, if (x – a) is a factor of f(x) then a is a root of the equation.
i.e., f(a) = 0.
As, (x – 1) is a factor of f(x) then f(1) = 0
⇒ f(1) = (1)2 + a(1) + b
⇒ â0 = 1 + a + b
⇒ −1 − a = b .....(1)
According to remainder theorem, if is divided by , the remainder will be .
Thus, remainders of f(x) = x2 + ax + b when divided by (x – 2) is f(2).
∴ f(2) = (2)2 + a(2) + b
⇒ f(2) = 4 + 2a + b .....(2)
Thus, remainders of f(x) = x2 + ax + b when divided by (x – 3) is f(3).
∴ f(3) = (3)2 + a(3) + b
⇒ f(3) = 9 + 3a + b .....(2)
Given,
Thus,
Hence, the correct answer is option (b).
Page No 67:
Question 40:
The remainder when f(x) = x45 + x25 + x14 + x9 + x is divided by g(x) = x2 – 1, is
(a) 4x – 1
(b) 4x + 2
(c) 4x + 1
(d) 4x – 2
Answer:
Given: f(x) = x45 + x25 + x14 + x9 + x and divisor g(x) = x2 – 1
Let the remainder is in the form of ax+b.
The function is divided by x2 – 1 so it is a factor of the function.
Thus,
x2 – 1 = 0
⇒ x2 – 12 = 0
â⇒ (x – 1) (x + 1) = 0
ââ⇒ (x – 1) = 0 or (x + 1) = 0
âââ⇒ x = 1 or x = –1
Now, by division algorithm,
Dividend = Divisor × Quotient + Remainder
Since f(x) = (x – 1) (x + 1) × q(x) + ax+b
At x = –1
∴ f(–1) = (–1 – 1) (–1 + 1) × q(–1) – a+b
âââ⇒ f(–1) = –a+b .....(1)
Putting,x= –1 in f(x)
f(–1) = (–1)45 + (–1)25 + (–1)14 + (–1)9 + (–1)
⇒ f(–1) = –1 – 1 + 1 – 1 – 1
⇒ f(–1) = –3 .....(2)
From (1) and (2)
⇒ –a+b = –3 .....(3)
f(x) = (x – 1) (x + 1) × q(x) + ax+b
∴ f(1) = (1 – 1) (1 + 1) × q(1) + a+b
âââ⇒ f(1) = a+b .....(4)
Putting,x= 1 in f(x)
f(1) = (1)45 + (1)25 + (1)14 + (1)9 + (1)
⇒ f(1) = 1 + 1 + 1 + 1 + 1
⇒ f(1) = 5 .....(5)
From (4) and (4)
⇒ a+b = 5 .....(6)
Adding (3) and (4),
b = 1
⇒ a = 4
∴ The remainder = ax+b
= 4x + 1
Hence, the correct answer is option (c).
Page No 67:
Question 41:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) âStatement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If the polynomial p(x) = x3 + ax2 – 2x + a + 4 has (x + a) as one of its factors, then .
Statement-2 (Reason): If f(x) = ax2 + b + c is exactly divisible by 2x – 3 then 4a + 6b + 9c = 0.
Answer:
Statement-2 (Reason): If f(x) = ax2 + b + c is exactly divisible by 2x – 3 then 4a + 6b + 9c = 0.
According to the Factor theorem, if (x – a) is a factor of f(x) then a is a root of the equation.
i.e., f(a) = 0.
As, (2x – 3) is a factor of f(x) then f() = 0
Thus, Statement-2 is false.
Statement-1 (Assertion): If the polynomial p(x) = x3 + ax2 – 2x + a + 4 has (x + a) as one of its factors, then .
Given that, p(x) = x3 + ax2 – 2x + a + 4 has (x + a) as one of its factors.
According to the Factor theorem, if (x – a) is a factor of f(x) then a is a root of the equation.
i.e., f(a) = 0.
As, (x + a) is a factor of p(x) then p(−a) = 0
⇒ p(−a) = (−a)3 + a(−a)2 – 2(−a) + a + 4
⇒ 0 = −a3 + a3 + 2a + a + 4
⇒ 0 = 3a + 4
Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.
Hence, the correct answer is option (c).
Page No 67:
Question 42:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) âStatement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If the polynomial f(x) = 3x4 – 11x2 + 6x + k when divided by (x – 3) leaves remainder 7, then k = –155.
Statement-2 (Reason): If a polynomial is divided by (x – a), the remainder is f(a).
Answer:
Statement-2 (Reason): If a polynomial is divided by (x – a), the remainder is f(a).
According to the remainder theorem, if is divided by , the remainder will be .
Thus, Statement-2 is true.
Statement-1 (Assertion): If the polynomial f(x) = 3x4 – 11x2 + 6x + k when divided by (x – 3) leaves remainder 7, then k = –155.
Given that, f(x) = 3x4 – 11x2 + 6x + k when divided by (x – 3) leaves the remainder 7.
According to the remainder theorem, if is divided by , the remainder will be .
As, (x – 3) leaves remainder 7 when dividing f(x) ⇒ f(3) = 7
⇒ f(3) = 3(3)4 – 11(3)2 + 6(3) + k
⇒ 7 = 243 – 99 + 18 + k
⇒ 7 = 162 + k
⇒ −155 = k
Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Hence, the correct answer is option (a).
Page No 67:
Question 43:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) âStatement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If f(x + 2) = 2x2 + 7x + 5, then the remainder when f(x) is divided by (x – 1) is 0.
Statement-2 (Reason): If a polynomial f(x) is divided by (ax + b), then the remainder is f
Answer:
Statement-2 (Reason):
If a polynomial f(x) is divided by (ax + b), then the remainder is f
Thus, Statement-2 is false.
Statement-1 (Assertion): If f(x + 2) = 2x2 + 7x + 5, then the remainder when f(x) is divided by (x – 1) is 0.
Given that, f(x + 2) = 2x2 + 7x + 5 and divisor is (x – 1).
At x = x − 2 in f(x + 2) = 2x2 + 7x + 5
⇒ f(x − 2 + 2) = 2(x − 2)2 + 7(x − 2) + 5
⇒ f(x) = 2(x2 − 4x + 4) + 7x − 14 + 5
⇒ f(x) = 2x2 − 8x + 8 + 7x − 9
⇒ f(x) = 2x2 − x − 1
Here, f(x) = 2x2 − x − 1 is divided by by (x – 1) so the the remainder will be f(1).
So, f(1) = 2(1)2 − (1) − 1
⇒f(1) = 2− 2
⇒f(1) = 0
Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.
Page No 67:
Question 44:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) âStatement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If x +1 is a factor of f(x) = px2 + 5x + r, then p + r + 5 = 0.
Statement-2 (Reason): If x – 2 and 2x – 1 are factors of f(x) = px2 + 5x + r, then p = r.
Answer:
Statement-2 (Reason): If x – 2 and 2x – 1 are factors of f(x) = px2 + 5x + r, then p = r.
Given that, x – 2 and 2x – 1 are factors of f(x) = px2 + 5x + r
According to the Factor theorem, if (x – a) is a factor of f(x) then a is the root of the equation.
i.e., f(a) = 0.
So, f(2) = p(2)2 + 5(2) + r
⇒ 0 = 4p + 10 + r .....(1)
Multiplying (1) by 4, we get
0 = 16p + 40 + 4r .....(3)
Subtracting (2) from (3),
0 − 0 = 16p + 40 + 4r − (p + 10 + 4r )
⇒ 0 = 16p + 40 + 4r − p − 10 − 4r
⇒ 0 = 15p + 30
⇒ p = −2 .....(4)
From (4) and (1), we get
⇒ 0 = 4(−2 ) + 10 + r
⇒ 0 = −8 + 10 + r
⇒ 0 = 2 + r
⇒ r = −2
⇒ p = r = −2
Thus, Statement-2 is true.
Statement-1 (Assertion): If x +1 is a factor of f(x) = px2 + 5x + r, then p + r + 5 = 0.
Given that, x +1 is a factor of f(x) = px2 + 5x + r
According to the Factor theorem, if (x – a) is a factor of f(x) then a is the root of the equation.
i.e., f(a) = 0.
So, f(−1) = p(−1)2 + 5(−1) + r
⇒ 0 = p − 5 + r
Thus, Statement-1 is false.
So, Statement-1 is false, Statement-2 is true.
Hence, the correct answer is option (d).
Page No 67:
Question 45:
Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.
(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) âStatement-1 is false, Statement-2 is true.
Statement-1 (Assertion): If x + 2a is a factor of f(x) = x5 – 4a2x3 + 2x + 2a + 3, then 2a – 3 = 0.
Statement-2 (Reason): If f(x) is divisible by (ax + b), then
Answer:
Statement-2 (Reason): If f(x) is divisible by (ax + b), then
According to the Factor theorem, if (x – a) is a factor of f(x) then a is a root of the equation.
i.e., f(a) = 0.
So, if (ax + b) is a factor of f(x) then is a root of the equation.
Thus, Statement-2 is true.
Statement-1 (Assertion): If x + 2a is a factor of f(x) = x5 – 4a2x3 + 2x + 2a + 3, then 2a – 3 = 0.
Given that, x + 2a is a factor of f(x) = x5 – 4a2x3 + 2x + 2a + 3.
According to the Factor theorem, if (x – a) is a factor of f(x) then a is a root of the equation.
i.e., f(a) = 0.
Since x + 2a is a factor of f(x) = x5 – 4a2x3 + 2x + 2a + 3 then –2a is the root of f(x), i.e., f(–2a) = 0.
⇒ f(–2âa) = (–2a)5 – 4a2(–2a)3 + 2(–2a) + 2a + 3
⇒ 0 = –32a5 + 32a5 – 4a + 2a + 3
⇒ 0 = –2a + 3
⇒ 2a – 3 = 0
Thus, Statement-2 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Hence, the correct answer is option (a).
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