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If \(p(x)=2x^3+rx^2+s\), \(p(0)=-3\), and \(p(1)=4\), what is the value of \(r\)?यदि \(p(x)=2x^3+rx^2+s\), \(p(0)=-3\) और \(p(1)=4\) है, तो \(r\) का मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: 55
Since \(p(0)=s\), we get \(s=-3\). Substituting \(x=1\) and \(s=-3\) gives \(p(1)=2+r-3=4\), so \(r=5\). Hence, option C is correct. Exam tip: use the value at \(x=0\) first to determine the constant term, then apply the second condition.\(p(0)=s\), इसलिए \(s=-3\)। अब \(p(1)=2+r+s=4\) में \(s=-3\) रखने पर \(2+r-3=4\), अर्थात् \(r=5\)। इसलिए सही उत्तर विकल्प C है। परीक्षा-युक्ति: पहले \(x=0\) रखकर अचर पद ज्ञात करें, फिर दूसरे दिए गए मान का उपयोग करें।
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