A quadratic polynomial, the sum and product of whose zeroes are 0 and , respectively is
(i) x2 + (ii) x2 - (iii) x2 – 5 (iv) None of these.
2. Which of the following is a terminating decimal?
(i) 35/14(ii) 2/35(iii) 1/7(iv) 8/7
3. If p, q are two co-prime numbers, then HCF (p,q) is:
(i) p(ii) q(iii) pq(iv) 1
4. If 1 is zero of the polynomial p(x)= ax2 - 3(a-1)x - 1, then the value of 'a' is:
(i) 1 (ii) -1(iii) 2(iv) -2
5. Write the name of solution for the given condition .
(i) Infinite solution(ii) Unique solution(iii) No solution(iv) None of these.
6. If α and β are zeroes of x2+5x+8, then the value of (α+β).
(i) 5 (ii) -5 (iii) 8 (iv) -8
7. Polynomial 2x4+3x3-5x2+9x+1 is a
(i) linear polynomial (ii) quadratic polynomial (iii) cubic polynomial (iv) Bi-quadratic polynomial
8. Check graphically whether the pair of equations is consistent. If so, solve them graphically.
x-2y=0 and 3x+4y=20.
9. Show that any positive even integer is of the form 6q , 6q+2 or 6q+4, where q is some integer.
10. Use Euclid's division algorithm to find the H.C.F of 4052 and 12576.2
Section-C3×4=12
11. Find all the zeroes of 2x4-3x3-3x2+6x-2, if you know that two of its zeroes are .
12. A fraction becomes 1/3 when 1 is subtracted from numerator and it becomes 1/4 when 8 is added to its
denominator. Find the fraction.
13. On dividing x3-3x2+x+2 by a polynomial g(x), the quotient and remainder were (x-2) and (-2x+4), respectively. Find g(x).
14. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing
in still water and the speed of the current.
15. If the zeroes of the polynomial x3-3x2+x+1 are (a-b),(a),(a+b) , find a and b.
Answers
Step-by-step explanation:
2.in photos ok
3.p and q are know to be co-prime meaning the only common factor that they have is 1 and no other number. Therefore, irrespective of how large the numbers are, if the numbers are co-prime pairs, then their HCF will always be equal to 1.
4. ax2+bx+c=0⇒α+β=−ab,αβ=ac
(i)
x2−2x−8=0
⇒a=1,b=−2,c=−8
x2−2x−8=0
(x−4)(x+2)=0
α=−2,β=4
α+β=−ab→−2+4=−1−2⇒2=2
αβ=ac→(−2)(4)=1−8⇒−8=−8
Hence the relationship between zeros and coefficients is verified.
(ii)
4s2−4s+1=0
⇒a=4,b=−4,c=1
4s2−4s+1=0
(2s−1)(2s−1)=0
α
5.sorry I don't know
6. I don't know sorry
7. in photos ok
8. Since we have given that
\begin{gathered}x-2y=0\\\\and\\\\3x+4y=20\end{gathered}x−2y=0and3x+4y=20
We need to solve it graphically.
In the graph given below:
Red line represents x-2y=0x−2y=0
and
Blue line represents 3x+4y=203x+4y=20
When they intersect, we get asolutions:
So, At (4,2), they get intersected.
So, x = 4 and y= 2
9. in photos ok
10. According to Euclid's Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r ≤ b.
11. in photos ok
12. in photos ok
13. in photos ok
14.Let, speed of Ritu in still water be x km/hr
and speed of current be y km/hr
Then speed downstream =(x+y) km/hr
speed upstream =(x−y) km/hr
we know, Time=SpeedDistance
Then x+y20=2 and x−y4=2
⇒x+y=10 ...(1)
& x−y=2 ...(2)
adding both we get, 2x=12⇒x=6
Thus using (1) y=10−6=4
Hence the speed of Ritu in still water is 6 km/hr and the speed of the current is 4 km/hr.
15.
\begin{gathered}p(x) = x^3 - 3x^2 + x + 1\\\\\text{A standard cubic equation is of the form $ax^3 + bx^2 + cx + d$}\\\\\text{On comparing the two equations, we get}\\\\a = 1\\\\b = -3\\\\c= 1\\\\\\d=1\\\\\text{Given zeroes are:}\\\\\alpha = a-b\\\\\beta = a\\\\\gamma = a + b\\\\\text{We know that in a cubic polynomial,}\\\\\text{sum of zeroes = $-\dfrac{b}{a}$}\\\\\\\implies \alpha + \beta + \gamma = -\dfrac{(-3)}{1}\\\\\\\implies a - b + a + a + b = 3\\\\\implies 3a = 3\\\end{gathered}p(x)=x3−3x2+x+1A standard cubic equation is of the form ax3+bx2+cx+dOn comparing the two equations, we geta=1b=−3c=1d=1Given zeroes are:α=a−bβ=aγ=a+bWe know that in a cubic polynomial,sum of zeroes = −ab⟹α+β+γ=−1(−3)⟹a−b+a+a+b=3⟹3a=3
\begin{gathered}\implies a = 3 \div 3 = 1\\\\\implies \boxed{\underline{\boxed{\bold{a = 1}}}}\\\\\\\text{We also know that in a cubic polynomial,}\\\\\text{Product of zeroes = $-\dfrac{c}{a}$}\\\\\\\implies \alpha \times \beta \times \gamma = -\dfrac{c}{a}\\\\\\\implies (a-b)(a)(a+b) = -\dfrac{1}{1}\\\\\\\text{Put a = 1}\\\\(1-b)(1)(1+b) = -1\\\\\implies 1^2 - b^2 = -1\\\\\implies 1 - b^2 = -1\\\\\implies b^2 = 1 + 1\\\\\implies b^2 = 2\\\end{gathered}⟹a=3÷3=1⟹a=1We also know that in a cubic polynomial,Product of zeroes = −ac⟹α×β×γ=−ac⟹(a−b)(a)(a+b)=−11Put a = 1(1−b)(1)(1+b)=−1⟹12−b2=−1⟹1−b2=−1⟹b2=1+1⟹b2=2
\implies \boxed{\underline{\boxed{\bold{b = \pm \sqrt{2}}}}}⟹
