7 results found for "specialised-input" in Class 10.
ग्राफ में बिंदु ((-8,0)) दिखने पर बहुपद का कौन-सा मान शून्य होगा?
If the point ((-8,0)) appears on the graph, which value of the polynomial input gives zero?
#negative-zero
#coordinate-reading
#x-intercept
#polynomials
A (x=-8)
B (x=8)
C (x=0)
D (x=-1)
Explanation opens after your attempt
Step 1
Concept
The point ((-8,0)) means (y=0) at (x=-8). So (x=-8) makes the polynomial zero.
Step 2
Why this answer is correct
The correct answer is A. (x=-8). The point ((-8,0)) means (y=0) at (x=-8). So (x=-8) makes the polynomial zero.
Step 3
Exam Tip
((-8,0)) का अर्थ है (x=-8) पर (y=0)। इसलिए (x=-8) बहुपद को शून्य बनाता है।
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(q(x)=x-3 +2x-2 -5x-6) के लिए (q(-2)) क्या होगा?
For (q(x)=x-3 +2x-2 -5x-6), what is (q(-2))?
#evaluation
#negative input
#cubic
A (0)
B (2)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
(q(-2)=-8+8+10-6=4). Check the sign of every term separately when substituting a negative value.
Step 2
Why this answer is correct
The correct answer is C. (4). (q(-2)=-8+8+10-6=4). Check the sign of every term separately when substituting a negative value.
Step 3
Exam Tip
(q(-2)=-8+8+10-6=4)। ऋणात्मक मान रखते समय प्रत्येक पद का संकेत अलग जांचें।
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यदि (p(x)=x-2 +x) है, तो (p(-1)) क्या होगा?
If (p(x)=x-2 +x), what is (p(-1))?
#polynomials
#value
#negative-input
#easy
A (0)
B (-1)
C (1)
D (2)
Explanation opens after your attempt
Step 1
Concept
(p(-1)=(-1)2 +(-1)=1-1=0). The square of a negative number is positive.
Step 2
Why this answer is correct
The correct answer is A. (0). (p(-1)=(-1)2 +(-1)=1-1=0). The square of a negative number is positive.
Step 3
Exam Tip
(p(-1)=(-1)2 +(-1)=1-1=0) है। ऋणात्मक संख्या का वर्ग धनात्मक होता है।
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(p(x)=x-2 +2x+1) में (p(0)) का मान क्या है?
For (p(x)=x-2 +2x+1), what is (p(0))?
#polynomials
#value
#easy
#zero-input
A (0)
B (1)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
(p(0)=02 +2\cdot0+1=1). When (x=0), only the constant term remains.
Step 2
Why this answer is correct
The correct answer is B. (1). (p(0)=02 +2\cdot0+1=1). When (x=0), only the constant term remains.
Step 3
Exam Tip
(p(0)=02 +2\cdot0+1=1) है। (x=0) रखने पर केवल नियत पद बचता है।
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यदि (p(x)=x-2 -2x-2) है, तो (p\(1+\sqrt{3}\)) क्या है?
If (p(x)=x-2 -2x-2), what is (p\(1+\sqrt{3}\))?
#polynomial-evaluation
#irrational-input
#zero
A (0)
B \(2\sqrt{3}\)
C (-2)
D (3)
Explanation opens after your attempt
Step 1
Concept
(\(1+\sqrt{3}\)2 -2\(1+\sqrt{3}\)-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0). Do not forget the middle term while expanding the square.
Step 2
Why this answer is correct
The correct answer is A. (0). (\(1+\sqrt{3}\)2 -2\(1+\sqrt{3}\)-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0). Do not forget the middle term while expanding the square.
Step 3
Exam Tip
(\(1+\sqrt{3}\)2 -2\(1+\sqrt{3}\)-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0)। वर्ग खोलते समय बीच का पद न भूलें।
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यदि (p\(\sqrt{5}\)=0) और (p(x)=x-2 +ax-5) है, तो (a) का मान क्या है?
If (p\(\sqrt{5}\)=0) and (p(x)=x-2 +ax-5), what is the value of (a)?
#polynomial-value
#parameter
#irrational-input
A (0)
B \(\sqrt{5}\)
C \(-\sqrt{5}\)
D (5)
Explanation opens after your attempt
Step 1
Concept
Substitution gives \(5+a\sqrt{5}-5=0\), so \(a\sqrt{5}=0\) and (a=0). Simplify like terms first while substituting.
Step 2
Why this answer is correct
The correct answer is A. (0). Substitution gives \(5+a\sqrt{5}-5=0\), so \(a\sqrt{5}=0\) and (a=0). Simplify like terms first while substituting.
Step 3
Exam Tip
रखने पर \(5+a\sqrt{5}-5=0\), इसलिए \(a\sqrt{5}=0\) और (a=0)। मान रखते समय समान पद पहले सरल करें।
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यदि (p(x)=x-2 -2x-2) है, तो (p\(\sqrt{2}\)) का मान क्या है?
If (p(x)=x-2 -2x-2), what is the value of (p\(\sqrt{2}\))?
#evaluation
#irrational-input
#polynomial
A \(-2\sqrt{2}\)
B (0)
C \(2\sqrt{2}\)
D (-4)
Explanation opens after your attempt
Correct Answer
A. \(-2\sqrt{2}\)
Step 1
Concept
(p\(\sqrt{2}\)=\(\sqrt{2}\)2 -2\sqrt{2}-2=2-2\sqrt{2}-2=-2\sqrt{2}). When substituting, write (\(\sqrt{2}\)2 =2).
Step 2
Why this answer is correct
The correct answer is A. \(-2\sqrt{2}\). (p\(\sqrt{2}\)=\(\sqrt{2}\)2 -2\sqrt{2}-2=2-2\sqrt{2}-2=-2\sqrt{2}). When substituting, write (\(\sqrt{2}\)2 =2).
Step 3
Exam Tip
(p\(\sqrt{2}\)=\(\sqrt{2}\)2 -2\sqrt{2}-2=2-2\sqrt{2}-2=-2\sqrt{2}) है। मान रखते समय (\(\sqrt{2}\)2 =2) लिखें।
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