For which value will the zero of ((s-8)x+5s) be (0)?
Correct answer: A
For the zero to be (0), the constant term (5s=0). At (s=0), the coefficient of (x) is (-8), so the polynomial is linear.
View question detailsMathematics
Linear polynomials
रैखिक बहुपद
In this Class 9 Mathematics topic from “Introduction to Polynomials,” students learn how to recognize and work with linear polynomials, whose degree is one and which are commonly written as ax + b, where a is non-zero. They identify the variable, coefficient, constant term, and degree, distinguish linear polynomials from other types, evaluate them for given values, and understand how to find their zero. These ideas build a foundation for simplifying expressions and studying polynomial relationships.
TOPIC PRACTICE
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Expert · Level 33 · polynomial,linear,zero_at_origin
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- (s=0)
- (s=8)
- (s=-8)
- No value
Expert · Level 33 · polynomials,linear polynomials,zero of polynomial,algebraic simplification,class 9 mathematics
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- No value
- 0
- 1
- 14
Expert · Level 33 · polynomials,linear polynomials,subtraction of polynomials,algebraic expressions,grade 9 mathematics
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- \(14x\)
- \(2x-2a\)
- \(2x\)
- \(-2x-2a\)
Expert · Level 33 · polynomials,linear polynomial,composite functions,substitution,algebraic equations
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- 6
- 7
- 8
- 48
Expert · Level 33 · mathematics,polynomials,linear polynomials,function composition,parameter solving,class 9
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- 4
- 5
- 6
- 8
Expert · Level 33 · polynomials,linear polynomials,constant polynomial,algebraic conditions,class 9 mathematics
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- जब \(m=0\)
- जब \(n=0\)
- Never
- जब \(m=n\)
Expert · Level 33 · linear polynomials, degree of polynomial, constant term, polynomial classification, mathematics class 9
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- It is a linear polynomial because the highest power of \(x\) is 1.
- It is a constant polynomial because its constant term is 120.
- It is a quadratic polynomial because it has two terms.
- It is not a polynomial because \(x\) represents the number of coupons.
Expert · Level 33 · polynomials,linear polynomials,function evaluation,algebraic substitution,shifted arguments
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- 12
- 24
- 36
- 48
Expert · Level 33 · polynomials,linear polynomials,composite functions,function evaluation,class 9 mathematics
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- 1
- -17
- 3
- 9
Expert · Level 33 · polynomials,linear polynomials,function composition,constant term,algebra
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- \(13\)
- \(14\)
- \(169\)
- \(182\)
Expert · Level 33 · polynomial,linear,coefficients
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- (10)
- (15)
- (20)
- (25)
Expert · Level 33 · polynomials,linear polynomials,function substitution,algebraic simplification,class 9 mathematics
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- \(28x-46\)
- \(28x-18\)
- \(14x-46\)
- \(28x+46\)
Medium · Level 33 · polynomial,linear polynomial,degree,parameter,Introduction to Polynomials,Class 9 MCQ,Linear polynomials,Mathematics
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- r = 4 or r = 5
- r = 0 or r = 20
- r = 9 only
- No value
Expert · Level 34 · polynomials,linear polynomials,algebraic simplification,distributive property,grade 9 mathematics
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- \(x-20\)
- \(31x-20\)
- \(x+4\)
- \(16x-12\)
Expert · Level 34 · linear polynomials, zeroes of polynomial, real numbers, algebra, class 9 mathematics
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- It has exactly one real zero.
- It has no real zero.
- It has exactly two real zeroes.
- It has infinitely many real zeroes.
Expert · Level 34 · linear polynomials,zero of polynomial,error analysis,algebra,grade 9 mathematics
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- The sign of the constant term has been handled incorrectly; the correct zero is \(-5\).
- The coefficient of \(x\) has been read incorrectly; the correct zero is \(3\).
- This polynomial has no zero because its constant term is positive.
- The zero is \(5\), because \(15\div 3=5\).
Expert · Level 34 · polynomials,linear polynomials,algebraic expressions,like terms,subtraction of polynomials
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- x-65
- 31x+45
- x+45
- 16x-55
Expert · Level 34 · polynomials,linear polynomials,zero of polynomial,parameter substitution,algebra
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- \(a=5\)
- \(a=4\)
- \(a=0\)
- \(a=\frac{5}{4}\)
Expert · Level 34 · polynomials,linear polynomials,zeros of polynomials,coefficient,algebra
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- 7
- -28
- 28
- -7
Expert · Level 34 · polynomials,linear polynomials,algebraic substitution,constant term,grade 9 mathematics
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- Only \(c=0\)
- Every real value of \(c\)
- No real value of \(c\)
- Only \(c=33\)
If (p(x)=6x+a) and (q(x)=8x-a), what is (a) if the zero of (p(x)+q(x)) is (-1)?
Correct answer: A
We have \(p(x)+q(x)=(6x+a)+(8x-a)=14x\). The terms containing \(a\) cancel, so the sum polynomial does not depend on any value of \(a\). Its zero is \(x=0\), since \(14x=0\) gives \(x=0\). Therefore, no value of \(a\) can make \(-1\) a zero. Exam tip: simplify the sum first; variable terms may cancel before you solve for a zero.
View question detailsIf (p(x)=6x+a) and (q(x)=8x-a), what is (q(x)-p(x))?
Correct answer: B
\(q(x)-p(x)=(8x-a)-(6x+a)\). Since there is a minus sign before the second bracket, the signs of both its terms change: \(8x-a-6x-a=2x-2a\). Hence, the correct answer is \(2x-2a\). The option \(2x\) incorrectly omits \(-a-a=-2a\). Exam tip: While subtracting polynomials, keep the second polynomial in brackets and change all its signs.
View question detailsIf (p(x)=ax+6) and (p(p(0))=48), what is (a)?
Correct answer: B
First, \(p(0)=a\cdot0+6=6\). Hence, \(p(p(0))=p(6)=6a+6\). Given \(6a+6=48\), we get \(6a=42\) and therefore \(a=7\). If \(a=8\), then \(p(6)=54\), not 48. Exam tip: In a composite expression such as \(p(p(0))\), evaluate the inner function first.
View question detailsIf (p(x)=x+a) and (p(p(4))=16), what is the value of (a)?
Correct answer: C
Given p(x)=x+a, we get p(4)=4+a. Therefore, p(p(4))=p(4+a)=(4+a)+a=4+2a. Hence, 4+2a=16, so 2a=12 and a=6. If a=8, then p(p(4)) would be 20, so it is not correct. Exam tip: For p(p(4)), first find p(4) and then substitute that result into p again.
View question detailsIf (p(x)=mx+n) and (p(-4)=p(2)), when can (p(x)) be linear?
Correct answer: C
Given \(p(-4)=p(2)\), we get \(-4m+n=2m+n\). Thus \(-6m=0\), so \(m=0\). Then \(p(x)=n\), which is a constant polynomial, not a linear polynomial. Hence \(p(x)\) can never be linear under the given condition. Exam tip: \(ax+b\) is linear only when \(a\ne0\).
View question detailsIn a shop, if \(x\) represents the number of discount coupons, the net amount is \(B(x)=120-8x\). A student calls it a constant polynomial because 120 is a constant number. Which is the correct correction of the student's statement?
Correct answer: A
In \(B(x)=120-8x\), the highest power of \(x\) is 1, so its degree is 1 and it is a linear polynomial. Only 120 is the constant term; the whole polynomial is not constant. Exam tip: use the highest power, not the number of terms.
View question detailsIf (p(x)=12x-5), what is (p(x+2)-p(x-1))?
Correct answer: C
Given \(p(x)=12x-5\), we get \(p(x+2)=12(x+2)-5=12x+19\) and \(p(x-1)=12(x-1)-5=12x-17\). Hence, \(p(x+2)-p(x-1)=(12x+19)-(12x-17)=36\). Option 24 would correspond to an input difference of 2, whereas \(x+2\) and \(x-1\) differ by 3. Exam tip: for a linear polynomial \(ax+b\), an input change of \(h\) produces an output change of \(ah\).
View question detailsIf (p(x)=9x-26), what is the correct value of (p(p(3)))?
Correct answer: B
First evaluate the inner function: \(p(3)=9\times3-26=1\). Then \(p(p(3))=p(1)=9\times1-26=-17\). Hence, the correct answer is \(-17\). Option \(1\) is only the value of \(p(3)\); it must be substituted into \(p\) once more. Exam tip: In a composite function, always evaluate the innermost function first.
View question detailsIf (p(x)=13x+a) and (p(p(0))=182), what is the value of (a)?
Correct answer: A
Given \(p(x)=13x+a\), we first get \(p(0)=a\). Therefore, \(p(p(0))=p(a)=13a+a=14a\). Since \(14a=182\), \(a=\frac{182}{14}=13\). The value \(14\) arises after adding the constant term \(a\) to the coefficient expression, but it is not the value of \(a\). Exam tip: In composition questions, find the inner function value first and then substitute it into the outer function.
View question detailsIf (p(x)=mx+n), (p(-2)=5) and (p(3)=30), what is (m+n)?
Correct answer: B
Subtracting the two values gives (5m=25), so (m=5). From (-2m+n=5), (n=15), so (m+n=20), not (15).
View question detailsIf (p(x)=14x-9), what is (p(x+1)+p(x-3))?
Correct answer: A
Substitute x+1 and x-3 for x separately. \(p(x+1)=14(x+1)-9=14x+5\), while \(p(x-3)=14(x-3)-9=14x-51\). Therefore, \(p(x+1)+p(x-3)=(14x+5)+(14x-51)=28x-46\). The option \(28x-18\) results from adding the constant terms incorrectly. Exam tip: use brackets whenever substituting an expression for x, especially when it contains a minus sign.
View question detailsFor which value of r will ((r²−9r+20)x+6) not remain a linear polynomial?
Correct answer: A
For an expression ax+b to be a linear polynomial, the coefficient a of x must be nonzero. Here the coefficient is r²−9r+20. The expression fails to be linear when this coefficient equals zero. Factorising gives r²−9r+20=(r−4)(r−5). Thus (r−4)(r−5)=0, so r=4 or r=5. At either value, the x-term vanishes and the expression becomes the constant polynomial 6, which has degree 0 rather than degree 1. Therefore option A is correct. Option B contains values that do not solve the coefficient equation. Option C gives neither of the roots, and option D is false because two valid values exist.
View question detailsAfter simplifying (8(2x-1)-3(5x+4)), which linear polynomial is obtained?
Correct answer: A
On opening the brackets, \(8(2x-1)=16x-8\) and \(-3(5x+4)=-15x-12\). Therefore, \(16x-8-15x-12=x-20\). Hence, the required linear polynomial is \(x-20\). The option \(16x-12\) results from not distributing \(-3\) correctly to both terms in the second bracket. Exam tip: When a negative factor precedes a bracket, multiply every term inside by it.
View question detailsIf \(p(x)=ax+b\), where \(a\neq 0\) and \(a,b\) are real numbers, is a linear polynomial, which statement about its zeroes is correct?
Correct answer: A
Since \(a\neq0\), solving \(ax+b=0\) gives \(x=-b/a\). This is only one value, so a linear polynomial has exactly one real zero. Exam tip: always check that \(a\neq0\).
View question detailsA student writes that the zero of the polynomial \(p(x)=3x+15\) is \(5\). What is the error in the student's solution?
Correct answer: A
For a zero, set \(p(x)=0\): \(3x+15=0\Rightarrow3x=-15\Rightarrow x=-5\). The student dropped the negative sign after division. Exam tip: substitute the value back to verify that the polynomial becomes zero.
View question detailsIf (p(x)=8x-5) and (q(x)=3x+11), what is (2p(x)-5q(x))?
Correct answer: A
Here, 2p(x)=2(8x-5)=16x-10 and 5q(x)=5(3x+11)=15x+55. Therefore, 2p(x)-5q(x)=(16x-10)-(15x+55)=16x-10-15x-55=x-65. The option x+45 results from not applying the negative sign correctly to the constant term of q(x). Exam tip: When subtracting a bracket, change the sign of every term inside it.
View question detailsFor which value will the zero of ((4a-5)x+30) be (-2)?
Correct answer: A
If \(x=-2\) is a zero of \((4a-5)x+30\), the polynomial must equal zero at \(x=-2\): \((4a-5)(-2)+30=0\). Thus, \(-8a+10+30=0\), or \(-8a+40=0\), giving \(a=5\). For \(a=\frac{5}{4}\), the coefficient of \(x\) becomes zero and the expression is \(30\), so it cannot have a zero. Exam tip: substitute the given zero for \(x\) and equate the polynomial to zero.
View question detailsIf the zero of (p(x)=ax+b) is (7) and (a=-4), what is (b)?
Correct answer: C
Since 7 is a zero, p(7)=0. Therefore, 7a+b=0. Substituting a=-4 gives 7(-4)+b=0, or -28+b=0. Hence, b=28. The close distractor -28 is the value of 7a, not of b. Exam tip: if r is a zero of the linear polynomial ax+b, use ar+b=0 to find an unknown coefficient.
View question detailsIf (p(x)=11x+c) and (p(4)=p(1)+33), what is correct about (c)?
Correct answer: B
\(p(4)=44+c\) and \(p(1)=11+c\). Hence, \(p(1)+33=11+c+33=44+c=p(4)\). The term \(c\) occurs equally on both sides, so it cancels for every real value of \(c\). Therefore, every real value of \(c\) is valid. \(c=0\) and \(c=33\) are only particular values, not necessary conditions. Exam tip: Evaluate the polynomial at the given inputs first, then simplify the constant terms.
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