Question : If $l+m+n=9$ and $l^{2}+m^{2}+n^{2}=31$, then the value of $(lm+mn+nl )$ will be:
Option 1: 22
Option 2: 50
Option 3: 25
Option 4: –25
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Correct Answer: 25
Solution : Given: $l^{2}+m^{2}+n^{2}=31$ And $l+m+n=9$ Squaring both sides, $(l+m+n)^2=81$ ⇒ $l^2+m^2+n^2+2(lm+mn+nl)= 81$ ⇒ $31+2(lm+mn+nl) = 81$ ⇒ $2(lm+mn+nl) = 81 – 31 = 50$ Thus, $(lm+mn+nl)=\frac{50}{2}=25$ Hence, the correct answer is 25.
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Question : If $m-n=2$ and $mn=15,(m,n>0)$ , then the value of $(m^2-n^2)(m^3-n^3)$ is:
Option 1: 1856
Option 2: 1658
Option 3: 1586
Option 4: 1568
Question : If $m$ and $n$ are two positive real numbers such that $9 m^2+n^2=40$ and $mn = 4$, then the value of $3m + n$ is:
Option 1: 160
Option 2: 64
Option 3: 10
Option 4: 8
Question : Directions: If 1 * 2 = 1, 2 * 3 = –1 and 3 * 4 = –5, then find the value of 7 * 9 = ?
Option 1: –47
Option 2: –29
Option 3: –2
Option 4: –9
Question : $\triangle$LMN is right angled at M. If $\angle$N = 45$^\circ$, what is the length of MN (in cm), if NL = 9$\sqrt2$cm?
Option 1: $9\sqrt2$
Option 2: $\frac{9}{\sqrt2}$
Option 3: $18$
Option 4: $9$
Question : If $\frac{\cos\alpha}{\cos\beta}= m$ and $\frac{\cos\alpha}{\sin\beta}= n$, then the value of $(m^{2}+n^{2})\cos^{2}\beta$ is:
Option 1: n2
Option 2: m2
Option 3: mn
Option 4: 1
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