hard se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
1536 questions tagged with hard.
If \(\alpha\) and \(\beta\) are zeroes of \(x^2-7x+10\), what is \(\alpha^2+\beta^2\)?
A. (29)
(\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) होता है। (72-2(10)=29) है। / (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta). Thus (72-2(10)=29).
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A. \(-\frac{3}{5}\)
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\) है। यहां \(\frac{3}{2}\div-\frac{5}{2}=-\frac{3}{5}\)। / \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here \(\frac{3}{2}\div-\frac{5}{2}=-\frac{3}{5}\).
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A. \(x^2-5x+6\)
मोनिक बहुपद (x-2-\(\alpha+\beta\)x+\alpha\beta) होता है। इसलिए \(x^2-5x+6\) सही है। / The monic polynomial is (x-2-\(\alpha+\beta\)x+\alpha\beta). Hence \(x^2-5x+6\) is correct.
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If (p(x)=x-2-2x+5), what is (p(1)), and is (1) a zero?
A. (4), नहीं(4), no
(p(1)=1-2+5=4) है इसलिए (1) शून्यक नहीं है। शून्यक तभी होगा जब मान (0) हो। / (p(1)=1-2+5=4), so (1) is not a zero. A zero must make the value (0).
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Which one is not a linear factor of \(x^4-1\)?
C. \(x^2+1\)
\(x^2+1\) रैखिक नहीं है क्योंकि उसकी घात (2) है। वास्तविक संख्याओं में इसे रैखिक गुणनखंड नहीं माना जाता। / \(x^2+1\) is not linear because its degree is (2). Over real numbers, it is not a linear factor.
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If the remainder when (p(x)) is divided by (x-4) is (7), what is (p(4))?
B. (7)
शेषफल प्रमेय के अनुसार (x-4) से भाग देने पर शेषफल (p(4)) होता है। इसलिए (p(4)=7)। / By the remainder theorem, the remainder on division by (x-4) is (p(4)). Therefore, (p(4)=7).
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What is the remainder when \(3x^3-2x^2+x+7\) is divided by (x+2)?
A. (-27)
शेषफल (p(-2)) होगा। (3(-8)-2(4)-2+7=-27) है। / The remainder is (p(-2)). (3(-8)-2(4)-2+7=-27).
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What is the remainder when \(x^3-5x^2+8x-4\) is divided by (x-1)?
A. (0)
शेषफल प्रमेय से शेषफल (p(1)) है। (1-5+8-4=0) मिलता है। / By the remainder theorem, the remainder is (p(1)). We get (1-5+8-4=0).
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If (x+1) is a factor of \(2x^3+kx^2-5x+2\), what is the correct value of (k)?
A. (-5)
(p(-1)=0) रखने पर (-2+k+5+2=0) मिलता है। इसलिए (k=-5) सही है। / Putting (p(-1)=0) gives (-2+k+5+2=0). Therefore, (k=-5) is correct.
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If (x+1) is a factor of \(2x^3+kx^2-5x+2\), what is (k)?
A. (5)
(x+1) गुणनखंड होने पर (p(-1)=0) होगा। (-2+k+5+2=0) से (k=-5) नहीं बल्कि (k=-5) मिलता है। / If (x+1) is a factor, then (p(-1)=0). From (-2+k+5+2=0), we get (k=-5).
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If (p(x)=x-3-3x-2-4x+12), which is a factor of (p(x))?
A. (x-2)
(p(2)=8-12-8+12=0) है इसलिए (x-2) गुणनखंड है। गुणनखंड प्रमेय का प्रयोग करें। / Since (p(2)=8-12-8+12=0), (x-2) is a factor. Use the factor theorem.
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What are the zeroes of \(x^3-6x^2+11x-6\)?
A. (1,2,3)
यह ((x-1)(x-2)(x-3)) के बराबर है। अतः शून्यक (1), (2) और (3) हैं। / It equals ((x-1)(x-2)(x-3)). Therefore, the zeroes are (1), (2), and (3).
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If (x-a), (x-b), and (x-c) are factors of a cubic polynomial, which is a possible polynomial?
A. (x-3-(a+b+c)x-2+(ab+bc+ca)x-abc)
((x-a)(x-b)(x-c)) फैलाने पर पहला रूप मिलता है। शून्यकों और गुणनखंडों का संबंध याद रखें। / Expanding ((x-a)(x-b)(x-c)) gives the first form. Remember the link between zeroes and factors.
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Which statement about the zeroes of \(x^2-6x+9\) is correct?
B. दोनों शून्यक बराबर हैंBoth zeroes are equal
(x-2-6x+9=(x-3)2) है इसलिए दोनों शून्यक (3) हैं। वर्ग रूप को पहचानना परीक्षा में तेज तरीका है। / (x-2-6x+9=(x-3)2), so both zeroes are (3). Recognizing square form is a fast exam method.
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If the product of zeroes of (p(x)=3x-2-10x+m) is \(\frac{4}{3}\), what is (m)?
A. (4)
गुणनफल \(\frac{m}{3}\) है। \(\frac{m}{3}=\frac{4}{3}\) से (m=4) मिलता है। / The product is \(\frac{m}{3}\). From \(\frac{m}{3}=\frac{4}{3}\), we get (m=4).
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If the sum of zeroes of (p(x)=2x-2+kx+8) is (3), what is (k)?
B. (-6)
योग \(-\frac{k}{2}\) होता है और यह (3) है। इसलिए (k=-6) है। / The sum is \(-\frac{k}{2}\), and it equals (3). Hence (k=-6).
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If (p(x)=x-2-5x+6), what is the value of (p(3)-p(2))?
A. (0)
(2) और (3) दोनों शून्यक हैं, इसलिए (p(3)=0) और (p(2)=0)। अतः अंतर (0) है। / Both (2) and (3) are zeroes, so (p(3)=0) and (p(2)=0). Therefore, the difference is (0).
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Which polynomial has degree (4)?
A. \(7x^4+0x^5-3x+1\)
\(0x^5\) पद घात नहीं बढ़ाता क्योंकि उसका गुणांक (0) है। सबसे बड़ा शून्येतर घात (4) है। / The term \(0x^5\) does not affect degree because its coefficient is (0). The highest non-zero power is (4).
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If the zeroes of \(x^2+ax+b\) are (4) and (-7), what is (a-b)?
B. (31)
योग (-3) है इसलिए (a=3) और गुणनफल (-28) है इसलिए (b=-28)। अतः (a-b=31) है। / The sum is (-3), so (a=3), and the product is (-28), so (b=-28). Hence (a-b=31).
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Which can be a quadratic polynomial whose zeroes are \(\frac{2}{3}\) and \(-\frac{5}{4}\)?
A. \(12x^2+7x-10\)
शून्यकों का योग \(-\frac{7}{12}\) और गुणनफल \(-\frac{5}{6}\) है। इसलिए \(12x^2+7x-10\) सही है। / The sum is \(-\frac{7}{12}\) and the product is \(-\frac{5}{6}\). So \(12x^2+7x-10\) is correct.
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If the zeroes of (p(x)=x-3+px-2+qx-6) are (1), (2), and (3), what is the correct value of (p+q)?
A. (5)
((x-1)(x-2)(x-3)=x-3-6x-2+11x-6) से (p=-6) और (q=11)। अतः (p+q=5) है। / From ((x-1)(x-2)(x-3)=x-3-6x-2+11x-6), (p=-6) and (q=11). Hence (p+q=5).
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If the zeroes of (p(x)=x-3+px-2+qx-6) are (1), (2), and (3), what is (p+q)?
A. (-5)
बहुपद ((x-1)(x-2)(x-3)=x-3-6x-2+11x-6) है। इसलिए (p=-6), (q=11) और (p+q=5) नहीं बल्कि (5) है। / The polynomial is ((x-1)(x-2)(x-3)=x-3-6x-2+11x-6). Hence (p=-6), (q=11), so (p+q=5).
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What conclusion follows when (x=2) is substituted in \(x^3-4x^2+x+6\)?
A. (2) शून्यक है(2) is a zero
(p(2)=8-16+2+6=0) है इसलिए (2) शून्यक है। सीधे प्रतिस्थापन से जल्दी जांच होती है। / (p(2)=8-16+2+6=0), so (2) is a zero. Direct substitution is the quickest check.
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If the zeroes of \(x^2-9x+k\) are equal, what is the value of (k)?
C. \(\frac{81}{4}\)
बराबर शून्यकों के लिए विविक्तकर \(b^2-4ac=0\) होता है। इसलिए (81-4k=0) से \(k=\frac{81}{4}\) मिलता है। / For equal zeroes, the discriminant \(b^2-4ac=0\). Thus (81-4k=0), giving \(k=\frac{81}{4}\).
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What is the product of the zeroes of the quadratic polynomial \(2x^2-7x+3\)?
B. \(\frac{3}{2}\)
द्विघात \(ax^2+bx+c\) में गुणनफल \(\frac{c}{a}\) होता है। यहां \(\frac{3}{2}\) सही है। / For \(ax^2+bx+c\), the product is \(\frac{c}{a}\). Here, \(\frac{3}{2}\) is correct.
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If the zeroes of (p(x)=x-2-(m+3)x+12) are (3) and (4), what is (m)?
B. (4)
शून्यकों का योग (7) है और यह (m+3) के बराबर है। इसलिए (m=4) मिलेगा। / The sum of zeroes is (7), which equals (m+3). Therefore, (m=4).
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If (2) is a zero of (p(x)=kx-2-5x+6), what is the value of (k)?
A. (1)
(p(2)=0) रखने पर (4k-10+6=0) से (k=1) मिलता है। शून्यक के लिए (p\(\alpha\)=0) लगाएं। / Putting (p(2)=0) gives (4k-10+6=0), so (k=1). For a zero, use (p\(\alpha\)=0).
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C. क्योंकि घात (3) हैBecause degree is (3)
सबसे बड़ी घात (3) है इसलिए बहुपद घन है। परीक्षा में हमेशा सबसे बड़े शून्येतर घातांक को देखें। / The highest power is (3), so the polynomial is cubic. In exams, always check the highest non-zero exponent.
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If the roots of \(x^2-17x+70=0\) are \(\alpha\) and \(\beta\), what is \(\alpha^2+\beta^2\)?
A. (149)
\(\alpha+\beta=17\) और \(\alpha\beta=70\), इसलिए (\alpha-2+\beta-2=172-2(70)=149) है। परीक्षा में (\(\alpha+\beta\)2-2\alpha\beta) याद रखें। / \(\alpha+\beta=17\) and \(\alpha\beta=70\), so (\alpha-2+\beta-2=172-2(70)=149). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).
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Which root is common to \(5x^2-16x+12=0\) and \(6x^2-17x+12=0\)?
A. \(x=\frac{3}{2}\)
पहले समीकरण के मूल \(\frac{6}{5},2\) नहीं बल्कि \(\frac{6}{5}\) और (2) हैं, इसलिए यह विकल्प नहीं है। सही जांच में दोनों समीकरणों के मूल क्रमशः \(2,\frac{6}{5}\) और \(\frac{3}{2},\frac{4}{3}\) हैं। / The first equation has roots \(2,\frac{6}{5}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so there is no common root among the given values. In exams, solve both equations before comparing.
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