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SubjectsMathematics
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
Up to 20 questions from this page. Select your focus, then start.
Given \(p(x)=rx-12\) and \(p(6)=30\), we have \(6r-12=30\). Thus, \(6r=42\) and \(r=7\). Now, \(p(-2)=7(-2)-12=-14-12=-26\). Hence, \(-26\) is the correct answer. \(-14\) is only the value of \(7(-2)\); the constant term \(-12\) must also be included. Exam tip: First use the given function value to find the unknown coefficient, then substitute the required input.
Given \(p(x)=ax+15\), we have \(p(2)=2a+15\) and \(p(8)=8a+15\). Substituting these into \(p(2)=p(8)-30\) gives \(2a+15=8a+15-30\), or \(2a+15=8a-15\). Hence, \(6a=30\), so \(a=5\). For example, \(a=6\) does not satisfy the condition. Exam tip: Write the required polynomial values separately before substituting them into the given relation.
Given \(p(x)=7x-30\), we have \(p(a)=7a-30\). Using the condition \(p(a)=a\), \(7a-30=a\). Hence, \(6a=30\), so \(a=5\). Therefore, option B is correct. For example, if \(a=6\), then \(p(6)=12\), not 6. Exam tip: In a condition such as \(p(a)=a\), first substitute \(a\) for \(x\), then solve the resulting equation.
Given \(p(x)=mx+n\). Substituting \(x=0\) gives \(p(0)=n=-8\). Next, \(p(5)=27\) gives \(5m-8=27\), so \(5m=35\) and \(m=7\). Therefore, \(m+n=7+(-8)=-1\). The option \(7\) is only the value of \(m\), not of \(m+n\). Exam tip: for a linear polynomial, \(p(0)\) directly gives the constant term \(n\).
Take the difference of the given values: \(p(2)-p(-4)=27-3=24\). For the linear polynomial \(p(x)=mx+n\), \(p(2)-p(-4)=m[2-(-4)]=6m\). Hence, \(6m=24\), so \(m=4\). The value \(24\) is the difference of the function values, not the value of \(m\). Exam tip: when values of a linear polynomial at two points are given, divide the difference in polynomial values by the difference in the corresponding \(x\)-values to find \(m\).
In option D, \(p(x)=x-9\). On substituting \(x=9\), we get \(p(9)=9-9=0\). Its degree is 1, so it is a linear polynomial. Option C also gives \(p(9)=0\), but its degree is 2, so it is quadratic rather than linear. Exam tip: a linear polynomial always has degree 1.
Given \(p(x)=8x-15\). Substituting \(-x\) for \(x\), \(p(-x)=8(-x)-15=-8x-15\). Therefore, \(p(x)+p(-x)=(8x-15)+(-8x-15)=-30\). Option \(0\) is incorrect because only the terms containing \(x\) cancel; the constant terms \(-15\) and \(-15\) add to \(-30\). Exam tip: while finding \(p(-x)\), replace every \(x\) with \((-x)\) using brackets.
Given \(p(x)=8x-15\). Replacing \(x\) with \(-x\) gives \(p(-x)=8(-x)-15=-8x-15\). Therefore, \(p(x)-p(-x)=(8x-15)-(-8x-15)=16x\). Hence, \(16x\) is correct. \(-30\) may result from incorrectly handling only the constant terms. Exam tip: while finding \(p(-x)\), replace \(x\) by \(-x\) in every term containing \(x\).
For the zero to be (0), the constant term (6s=0). At (s=0), the coefficient of (x) is (9), so the polynomial is linear.
Adding the given polynomials gives (p(x)+q(x))=(7x+a)+(9x-a)=16x. The terms containing a cancel, so the sum does not depend on a. The only zero of 16x is 0; at x=2, its value is 32, not 0. Therefore, no value of a can make 2 a zero of the sum. Exam tip: To check a zero, substitute the given x-value and verify that the polynomial becomes 0.
\(q(x)-p(x)=(9x-a)-(7x+a)\). Since a minus sign precedes the second bracket, the signs of all its terms change: \(9x-a-7x-a=2x-2a\). Therefore, the correct answer is \(2x-2a\). The option \(2x\) incorrectly omits the constant terms \(-a-a\). Exam tip: while subtracting polynomials, keep the second polynomial in brackets and change the sign of every term.
First, \(p(0)=a\cdot0+7=7\). Therefore, \(p(p(0))=p(7)=7a+7\). Given \(7a+7=63\), we get \(7a=56\), so \(a=8\). Option 7 is the value of \(p(0)\), not the value of \(a\). Exam tip: In a composite function, evaluate the inner function first.
Given \(p(x)=x+a\), we have \(p(5)=5+a\). Therefore, \(p(p(5))=p(5+a)=(5+a)+a=5+2a\). Hence \(5+2a=19\), so \(2a=14\) and \(a=7\). Thus, option C is correct. \(14\) is the value of \(2a\), not of \(a\). Exam tip: In a composite function, evaluate the inner function first and substitute its result into the outer function.
Given \(p(-5)=p(3)\), we get \(-5m+n=3m+n\). Thus \(-8m=0\), so \(m=0\). Then \(p(x)=n\), which is a constant polynomial, not a linear polynomial. Option A satisfies the given equality, but it makes the coefficient of \(x\) zero, so the degree is not 1. Exam tip: For \(mx+n\) to be linear, \(m\ne0\) is necessary.
Writing \(q(x)=-\frac{1}{3}x+\frac{5}{2}\), the highest power of \(x\) is 1, so it is linear. Fractions are valid coefficients; denominators do not decide degree. Exam tip: check powers of variables only.
Given \(p(x)=13x-7\), we get \(p(x+2)=13(x+2)-7=13x+19\) and \(p(x-1)=13(x-1)-7=13x-20\). Hence, \(p(x+2)-p(x-1)=(13x+19)-(13x-20)=39\). The value 26 would correspond to an input difference of 2, whereas \(x+2\) and \(x-1\) differ by 3. Exam tip: for a linear polynomial \(p(x)=ax+b\), use \(p(u)-p(v)=a(u-v)\).
First evaluate the inner function: p(3)=11×3-32=1. Now substitute this value into p again: p(p(3))=p(1)=11×1-32=-21. Option 1 is only the value of p(3), not of p(p(3)). In exams, evaluate a composite function from inside to outside.
First, \(p(0)=15(0)+a=a\). Therefore, \(p(p(0))=p(a)=15a+a=16a\). Given \(16a=240\), we get \(a=240/16=15\). Option 16 is the coefficient in \(p(a)=16a\), not the value of \(a\). Exam tip: In a composite function, evaluate the inner function first and then substitute its value into the outer function.
Given \(p(x)=16x-11\), \(p(x+2)=16(x+2)-11=16x+21\) and \(p(x-3)=16(x-3)-11=16x-59\). Therefore, \(p(x+2)+p(x-3)=(16x+21)+(16x-59)=32x-38\). Option A results from an incorrect addition of the constant terms. Exam tip: always keep \(x-3\) in brackets while substituting it into a polynomial.
The direct answer is option A: \(r=5\) or \(r=6\). A linear polynomial in x has the form \(ax+b\), where the coefficient a must be nonzero. Here the coefficient of x is \(r^2-11r+30\). The expression stops being linear when this coefficient becomes zero, leaving the constant polynomial 8. Solve \(r^2-11r+30=0\). Factoring gives \((r-5)(r-6)=0\), so \(r=5\) or \(r=6\). Option A is correct. Option B, \(r=0\) or \(r=30\), does not make the coefficient zero: it gives 30 or 600, respectively. Option C, \(r=11\), gives \(121-121+30=30\), so the polynomial remains linear. Option D, “no value”, is wrong because two values have been found. At either 5 or 6, the expression becomes 8, a constant polynomial, not a linear polynomial. Exam cue: for \(ax+b\), check whether the coefficient of x is zero; do not confuse the coefficient with the constant term.
QUIZ COMPLETE
