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If (C(t)=28+kt) and (C(10)-C(4)=72), what is (k)?यदि (C(t)=28+kt) और (C(10)-C(4)=72), तो (k) क्या है?
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Correct answer:सही उत्तर: 1212
Given \(C(t)=28+kt\), we have \(C(10)=28+10k\) and \(C(4)=28+4k\). Hence, \(C(10)-C(4)=(28+10k)-(28+4k)=6k\). As the difference is 72, \(6k=72\), so \(k=12\). For example, choosing 14 would give a difference of \(6\times14=84\), not 72. Exam tip: when subtracting two values of a linear expression, the constant term cancels out.दिया है कि \(C(t)=28+kt\)। अतः \(C(10)=28+10k\) और \(C(4)=28+4k\)। इसलिए \(C(10)-C(4)=(28+10k)-(28+4k)=6k\)। प्रश्नानुसार \(6k=72\), इसलिए \(k=12\)। विकल्प 14 लेने पर अंतर \(6\times14=84\) आता, जो दिया गया 72 नहीं है। परीक्षा टिप: रैखिक व्यंजक के दो मानों का अंतर लेने पर स्थिर पद 28 कट जाता है।
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