116907
If \(f(x)=2 x^2\), find \(\frac{f(3.8)-f(4)}{3.8-4}\)
1 156
2 0.156
3 1.56
4 15.6
Explanation:
D Given, \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^2\) Then find \(\frac{f(3.8)-f(4)}{(3.8-4)}=\) ? So, \(\frac{\mathrm{f}(3.8)-\mathrm{f}(4)}{3.8-4}=\frac{2 \times(3.8)^2-4^2 \times 2}{3.8-4}\) \(=\frac{2\left[(3.8)^2-4^2\right]}{(3.8-4)}\) \(=\frac{-2\left[4^2-(3.8)^2\right]}{-(4-3.8)}\) \(=\frac{2(4-3.8)(4+3.8)}{(4-3.8)}=2 \times 7.8=15.6\)
Karnataka CET-2015
Sets, Relation and Function
116975
If \(y=3^{x-1}+3^{-x-1}\) (x real), then the least value of \(y\) is
1 2
2 6
3 \(2 / 3\)
4 None of these
Explanation:
C Given that, \(y=3^{x-1}+3^{-x-1}\) Now we know that- \(\text { A.M } \geq \mathrm{G} \cdot \mathrm{M}\) \(\frac{3^{\mathrm{x}-1}+3^{-\mathrm{x}-1}}{2} \geq\left(3^{\mathrm{x}-1} \cdot 3^{-\mathrm{x}-1}\right)^{\frac{1}{2}}\) \(3^{\mathrm{x}-1}+3^{-\mathrm{x}-1} \geq 2\left(3^{\mathrm{x}-1} \cdot 3^{-\mathrm{x}-1}\right)^{\frac{1}{2}}\) \(3^{\mathrm{x}-1}+3^{-\mathrm{x}-1} \geq \frac{2}{3}\)
Jamia Millia Islamia-2006
Sets, Relation and Function
117024
If \(p\) and \(q\) are positive real numbers such that \(p^2+q^2=1\), then the maximum value of \((p+q)\) is
1 2
2 \(\frac{1}{2}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\sqrt{2}\)
Explanation:
D Given, \(\mathrm{p}^2+\mathrm{q}^2=1\) Applying \(\mathrm{AM} \geq \mathrm{GM}\) inequality \(\frac{p^2+q^2}{2} \geq \sqrt{p^2 q^2}\) \(p^2+q^2 \geq 2 p q\) \(\frac{1}{2} \geq p q\) \(\mathrm{pq} \leq \frac{1}{2}\) Now we know that - \((p+q)^2=p^2+q^2=p q\) \((p+q)^2=1+2 p q\) \((p+q)^2 \leq 1+1\) \((p+q) \leq \sqrt{2}\) Hence maximum value of \(\mathrm{p}+\mathrm{q}=\sqrt{2}\)
AIEEE-2007
Sets, Relation and Function
116859
If \(p=\frac{1}{\log _3 \pi}+\frac{1}{\log _4 \pi}+1\), then
116907
If \(f(x)=2 x^2\), find \(\frac{f(3.8)-f(4)}{3.8-4}\)
1 156
2 0.156
3 1.56
4 15.6
Explanation:
D Given, \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^2\) Then find \(\frac{f(3.8)-f(4)}{(3.8-4)}=\) ? So, \(\frac{\mathrm{f}(3.8)-\mathrm{f}(4)}{3.8-4}=\frac{2 \times(3.8)^2-4^2 \times 2}{3.8-4}\) \(=\frac{2\left[(3.8)^2-4^2\right]}{(3.8-4)}\) \(=\frac{-2\left[4^2-(3.8)^2\right]}{-(4-3.8)}\) \(=\frac{2(4-3.8)(4+3.8)}{(4-3.8)}=2 \times 7.8=15.6\)
Karnataka CET-2015
Sets, Relation and Function
116975
If \(y=3^{x-1}+3^{-x-1}\) (x real), then the least value of \(y\) is
1 2
2 6
3 \(2 / 3\)
4 None of these
Explanation:
C Given that, \(y=3^{x-1}+3^{-x-1}\) Now we know that- \(\text { A.M } \geq \mathrm{G} \cdot \mathrm{M}\) \(\frac{3^{\mathrm{x}-1}+3^{-\mathrm{x}-1}}{2} \geq\left(3^{\mathrm{x}-1} \cdot 3^{-\mathrm{x}-1}\right)^{\frac{1}{2}}\) \(3^{\mathrm{x}-1}+3^{-\mathrm{x}-1} \geq 2\left(3^{\mathrm{x}-1} \cdot 3^{-\mathrm{x}-1}\right)^{\frac{1}{2}}\) \(3^{\mathrm{x}-1}+3^{-\mathrm{x}-1} \geq \frac{2}{3}\)
Jamia Millia Islamia-2006
Sets, Relation and Function
117024
If \(p\) and \(q\) are positive real numbers such that \(p^2+q^2=1\), then the maximum value of \((p+q)\) is
1 2
2 \(\frac{1}{2}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\sqrt{2}\)
Explanation:
D Given, \(\mathrm{p}^2+\mathrm{q}^2=1\) Applying \(\mathrm{AM} \geq \mathrm{GM}\) inequality \(\frac{p^2+q^2}{2} \geq \sqrt{p^2 q^2}\) \(p^2+q^2 \geq 2 p q\) \(\frac{1}{2} \geq p q\) \(\mathrm{pq} \leq \frac{1}{2}\) Now we know that - \((p+q)^2=p^2+q^2=p q\) \((p+q)^2=1+2 p q\) \((p+q)^2 \leq 1+1\) \((p+q) \leq \sqrt{2}\) Hence maximum value of \(\mathrm{p}+\mathrm{q}=\sqrt{2}\)
AIEEE-2007
Sets, Relation and Function
116859
If \(p=\frac{1}{\log _3 \pi}+\frac{1}{\log _4 \pi}+1\), then
116907
If \(f(x)=2 x^2\), find \(\frac{f(3.8)-f(4)}{3.8-4}\)
1 156
2 0.156
3 1.56
4 15.6
Explanation:
D Given, \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^2\) Then find \(\frac{f(3.8)-f(4)}{(3.8-4)}=\) ? So, \(\frac{\mathrm{f}(3.8)-\mathrm{f}(4)}{3.8-4}=\frac{2 \times(3.8)^2-4^2 \times 2}{3.8-4}\) \(=\frac{2\left[(3.8)^2-4^2\right]}{(3.8-4)}\) \(=\frac{-2\left[4^2-(3.8)^2\right]}{-(4-3.8)}\) \(=\frac{2(4-3.8)(4+3.8)}{(4-3.8)}=2 \times 7.8=15.6\)
Karnataka CET-2015
Sets, Relation and Function
116975
If \(y=3^{x-1}+3^{-x-1}\) (x real), then the least value of \(y\) is
1 2
2 6
3 \(2 / 3\)
4 None of these
Explanation:
C Given that, \(y=3^{x-1}+3^{-x-1}\) Now we know that- \(\text { A.M } \geq \mathrm{G} \cdot \mathrm{M}\) \(\frac{3^{\mathrm{x}-1}+3^{-\mathrm{x}-1}}{2} \geq\left(3^{\mathrm{x}-1} \cdot 3^{-\mathrm{x}-1}\right)^{\frac{1}{2}}\) \(3^{\mathrm{x}-1}+3^{-\mathrm{x}-1} \geq 2\left(3^{\mathrm{x}-1} \cdot 3^{-\mathrm{x}-1}\right)^{\frac{1}{2}}\) \(3^{\mathrm{x}-1}+3^{-\mathrm{x}-1} \geq \frac{2}{3}\)
Jamia Millia Islamia-2006
Sets, Relation and Function
117024
If \(p\) and \(q\) are positive real numbers such that \(p^2+q^2=1\), then the maximum value of \((p+q)\) is
1 2
2 \(\frac{1}{2}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\sqrt{2}\)
Explanation:
D Given, \(\mathrm{p}^2+\mathrm{q}^2=1\) Applying \(\mathrm{AM} \geq \mathrm{GM}\) inequality \(\frac{p^2+q^2}{2} \geq \sqrt{p^2 q^2}\) \(p^2+q^2 \geq 2 p q\) \(\frac{1}{2} \geq p q\) \(\mathrm{pq} \leq \frac{1}{2}\) Now we know that - \((p+q)^2=p^2+q^2=p q\) \((p+q)^2=1+2 p q\) \((p+q)^2 \leq 1+1\) \((p+q) \leq \sqrt{2}\) Hence maximum value of \(\mathrm{p}+\mathrm{q}=\sqrt{2}\)
AIEEE-2007
Sets, Relation and Function
116859
If \(p=\frac{1}{\log _3 \pi}+\frac{1}{\log _4 \pi}+1\), then
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sets, Relation and Function
116907
If \(f(x)=2 x^2\), find \(\frac{f(3.8)-f(4)}{3.8-4}\)
1 156
2 0.156
3 1.56
4 15.6
Explanation:
D Given, \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^2\) Then find \(\frac{f(3.8)-f(4)}{(3.8-4)}=\) ? So, \(\frac{\mathrm{f}(3.8)-\mathrm{f}(4)}{3.8-4}=\frac{2 \times(3.8)^2-4^2 \times 2}{3.8-4}\) \(=\frac{2\left[(3.8)^2-4^2\right]}{(3.8-4)}\) \(=\frac{-2\left[4^2-(3.8)^2\right]}{-(4-3.8)}\) \(=\frac{2(4-3.8)(4+3.8)}{(4-3.8)}=2 \times 7.8=15.6\)
Karnataka CET-2015
Sets, Relation and Function
116975
If \(y=3^{x-1}+3^{-x-1}\) (x real), then the least value of \(y\) is
1 2
2 6
3 \(2 / 3\)
4 None of these
Explanation:
C Given that, \(y=3^{x-1}+3^{-x-1}\) Now we know that- \(\text { A.M } \geq \mathrm{G} \cdot \mathrm{M}\) \(\frac{3^{\mathrm{x}-1}+3^{-\mathrm{x}-1}}{2} \geq\left(3^{\mathrm{x}-1} \cdot 3^{-\mathrm{x}-1}\right)^{\frac{1}{2}}\) \(3^{\mathrm{x}-1}+3^{-\mathrm{x}-1} \geq 2\left(3^{\mathrm{x}-1} \cdot 3^{-\mathrm{x}-1}\right)^{\frac{1}{2}}\) \(3^{\mathrm{x}-1}+3^{-\mathrm{x}-1} \geq \frac{2}{3}\)
Jamia Millia Islamia-2006
Sets, Relation and Function
117024
If \(p\) and \(q\) are positive real numbers such that \(p^2+q^2=1\), then the maximum value of \((p+q)\) is
1 2
2 \(\frac{1}{2}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\sqrt{2}\)
Explanation:
D Given, \(\mathrm{p}^2+\mathrm{q}^2=1\) Applying \(\mathrm{AM} \geq \mathrm{GM}\) inequality \(\frac{p^2+q^2}{2} \geq \sqrt{p^2 q^2}\) \(p^2+q^2 \geq 2 p q\) \(\frac{1}{2} \geq p q\) \(\mathrm{pq} \leq \frac{1}{2}\) Now we know that - \((p+q)^2=p^2+q^2=p q\) \((p+q)^2=1+2 p q\) \((p+q)^2 \leq 1+1\) \((p+q) \leq \sqrt{2}\) Hence maximum value of \(\mathrm{p}+\mathrm{q}=\sqrt{2}\)
AIEEE-2007
Sets, Relation and Function
116859
If \(p=\frac{1}{\log _3 \pi}+\frac{1}{\log _4 \pi}+1\), then
116907
If \(f(x)=2 x^2\), find \(\frac{f(3.8)-f(4)}{3.8-4}\)
1 156
2 0.156
3 1.56
4 15.6
Explanation:
D Given, \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}^2\) Then find \(\frac{f(3.8)-f(4)}{(3.8-4)}=\) ? So, \(\frac{\mathrm{f}(3.8)-\mathrm{f}(4)}{3.8-4}=\frac{2 \times(3.8)^2-4^2 \times 2}{3.8-4}\) \(=\frac{2\left[(3.8)^2-4^2\right]}{(3.8-4)}\) \(=\frac{-2\left[4^2-(3.8)^2\right]}{-(4-3.8)}\) \(=\frac{2(4-3.8)(4+3.8)}{(4-3.8)}=2 \times 7.8=15.6\)
Karnataka CET-2015
Sets, Relation and Function
116975
If \(y=3^{x-1}+3^{-x-1}\) (x real), then the least value of \(y\) is
1 2
2 6
3 \(2 / 3\)
4 None of these
Explanation:
C Given that, \(y=3^{x-1}+3^{-x-1}\) Now we know that- \(\text { A.M } \geq \mathrm{G} \cdot \mathrm{M}\) \(\frac{3^{\mathrm{x}-1}+3^{-\mathrm{x}-1}}{2} \geq\left(3^{\mathrm{x}-1} \cdot 3^{-\mathrm{x}-1}\right)^{\frac{1}{2}}\) \(3^{\mathrm{x}-1}+3^{-\mathrm{x}-1} \geq 2\left(3^{\mathrm{x}-1} \cdot 3^{-\mathrm{x}-1}\right)^{\frac{1}{2}}\) \(3^{\mathrm{x}-1}+3^{-\mathrm{x}-1} \geq \frac{2}{3}\)
Jamia Millia Islamia-2006
Sets, Relation and Function
117024
If \(p\) and \(q\) are positive real numbers such that \(p^2+q^2=1\), then the maximum value of \((p+q)\) is
1 2
2 \(\frac{1}{2}\)
3 \(\frac{1}{\sqrt{2}}\)
4 \(\sqrt{2}\)
Explanation:
D Given, \(\mathrm{p}^2+\mathrm{q}^2=1\) Applying \(\mathrm{AM} \geq \mathrm{GM}\) inequality \(\frac{p^2+q^2}{2} \geq \sqrt{p^2 q^2}\) \(p^2+q^2 \geq 2 p q\) \(\frac{1}{2} \geq p q\) \(\mathrm{pq} \leq \frac{1}{2}\) Now we know that - \((p+q)^2=p^2+q^2=p q\) \((p+q)^2=1+2 p q\) \((p+q)^2 \leq 1+1\) \((p+q) \leq \sqrt{2}\) Hence maximum value of \(\mathrm{p}+\mathrm{q}=\sqrt{2}\)
AIEEE-2007
Sets, Relation and Function
116859
If \(p=\frac{1}{\log _3 \pi}+\frac{1}{\log _4 \pi}+1\), then