86617
The integral \(\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x\) is equal to
1 2
2 4
3 1
4 6
Explanation:
(C) : Let, \(I=\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x\) \(=\int_{2}^{4} \frac{2 \log x}{2 \log x+\log (6-x)^{2}} d x\) \(=\int_{2}^{4} \frac{2 \log x d x}{2[\log x+\log (6-x)]} d x\) \(I=\int_{2}^{4} \frac{\log x}{\log x+\log (6-x)} d x\) \(I=\int_{2}^{4} \frac{\log (6-x)}{\log (6-x)+\log x} d x \ldots . \text { (ii) } \tag{ii}\) On adding equation (i) and (ii), we get - \(2 I=\int_{2}^{4} \frac{\log x+\log (6-x)}{\log x+\log (6-x)} d x\) \(2 I=\int_{2}^{4} d x\) \(2 I=[x]_{2}^{4}\) \(2 I=2\) \(I=1\)
[JCECE-2018]
Integral Calculus
86625
Let \(I=\int_{a}^{b}\left(x^{4}-2 x^{2}\right) d x\). If \(l\) is minimum, then the ordered pair \((a, b)\) is
1 \((-\sqrt{2}, 0)\)
2 \((0, \sqrt{2})\)
3 \((\sqrt{2},-\sqrt{2})\)
4 \((-\sqrt{2}, \sqrt{2})\)
Explanation:
(D) : \(I=\int_{a}^{b}\left(x^{4}-2 x^{2}\right) d x\) \(y=x^{4}-2 x^{2}=0\) \(x=0, \pm \sqrt{2} \tag{i}\) Therefore, the graph of above curve Hence, from the graph we can conduce that the internal value I in minimum when the curve is founded between \((-\sqrt{2}, 0),(\sqrt{2} ., 0)\) \(\therefore\) The ordered pair \((a, b)\) is \((-\sqrt{2}, \sqrt{2})\)
JEE Main-2019-10.01.2019
Integral Calculus
86655
If \(f(x)=\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x\), then \(\int_{0}^{\pi / 4} \frac{\sin ^{2} 2 x}{f(x)} d x=\)
1 2
2 \(\frac{2}{3}\)
3 \(\frac{-2}{3}\)
4 \(\frac{1}{6}\)
Explanation:
(B) : Given, \(f(x)=\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x\) and \(I=\int_{0}^{\pi / 4} \frac{\sin ^{2} 2 x}{f(x)} d x\) \(=\int_{0}^{\pi / 4} \frac{4 \sin ^{2} x \cos ^{2} x}{\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x} d x\) \(=\int_{0}^{\pi / 4} \frac{4 \tan ^{2} x \sec ^{2} x}{\tan ^{6} x+2 \tan ^{3} x+1} d x\) Now, put \(\tan ^{3} x=t\), \(3 \tan ^{2} x \sec ^{2} x d x=d t\) So at \(\mathrm{x}=0, \mathrm{t}=0\) and at \(\mathrm{x}=\frac{\pi}{4}, \mathrm{t}=1\) So, \(I=\frac{4}{3} \int_{0}^{1} \frac{d t}{t^{2}+2 t+1}=\frac{4}{3} \int_{0}^{1} \frac{d t}{(t+1)^{2}}\) \(=\frac{4}{3}\left[\frac{-1}{(t+1)}\right]_{0}^{1}=\frac{-4}{3}\left[\frac{1}{2}-1\right]=\frac{-4}{3} \times \frac{-1}{2}=\frac{2}{3}\)
TS EAMCET-2020-11.09.2020
Integral Calculus
86663
Let \(\frac{d}{d x} F(x)=\left(\frac{e^{\sin x}}{x}\right), x>0\) if \(\int_{1}^{4} \frac{3}{x} e^{\sin x^{3}} d x=f(k)-f(1)\), then one of the possible value of \(k\), is
1 15
2 16
3 63
4 64
Explanation:
(D) : Given that, \(\frac{d}{d x} F(x)=\left(\frac{e^{\sin x}}{x}\right), x>0\) On integrating both sides we get - \(F(x)=\int \frac{e^{\sin x}}{x} d x\) \(\int_{1}^{4} \frac{3}{x} e^{\sin x^{3}} d x =\int_{1}^{4} \frac{3 x^{2}}{x^{3}} \cdot e^{\sin x^{3}} d x\) \(=F(k)-F(1)\) Let us consider, \(x^{3}=z\) \(3 d^{2} d x=d z\) \(\therefore \quad \int_{1}^{64} \frac{\mathrm{e}^{\sin z}}{\mathrm{z}} \mathrm{dz}=\mathrm{F}(\mathrm{k})-\mathrm{F}(1)\) \({[F(z)]_{1}^{64}=F(F)-F(1) \quad[\because \text { from equation }(i)]}\) \(F(64)-F(1)=F(k)-f(1)\) Comparing on both sides we get- \(\mathrm{k}=64\)
86617
The integral \(\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x\) is equal to
1 2
2 4
3 1
4 6
Explanation:
(C) : Let, \(I=\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x\) \(=\int_{2}^{4} \frac{2 \log x}{2 \log x+\log (6-x)^{2}} d x\) \(=\int_{2}^{4} \frac{2 \log x d x}{2[\log x+\log (6-x)]} d x\) \(I=\int_{2}^{4} \frac{\log x}{\log x+\log (6-x)} d x\) \(I=\int_{2}^{4} \frac{\log (6-x)}{\log (6-x)+\log x} d x \ldots . \text { (ii) } \tag{ii}\) On adding equation (i) and (ii), we get - \(2 I=\int_{2}^{4} \frac{\log x+\log (6-x)}{\log x+\log (6-x)} d x\) \(2 I=\int_{2}^{4} d x\) \(2 I=[x]_{2}^{4}\) \(2 I=2\) \(I=1\)
[JCECE-2018]
Integral Calculus
86625
Let \(I=\int_{a}^{b}\left(x^{4}-2 x^{2}\right) d x\). If \(l\) is minimum, then the ordered pair \((a, b)\) is
1 \((-\sqrt{2}, 0)\)
2 \((0, \sqrt{2})\)
3 \((\sqrt{2},-\sqrt{2})\)
4 \((-\sqrt{2}, \sqrt{2})\)
Explanation:
(D) : \(I=\int_{a}^{b}\left(x^{4}-2 x^{2}\right) d x\) \(y=x^{4}-2 x^{2}=0\) \(x=0, \pm \sqrt{2} \tag{i}\) Therefore, the graph of above curve Hence, from the graph we can conduce that the internal value I in minimum when the curve is founded between \((-\sqrt{2}, 0),(\sqrt{2} ., 0)\) \(\therefore\) The ordered pair \((a, b)\) is \((-\sqrt{2}, \sqrt{2})\)
JEE Main-2019-10.01.2019
Integral Calculus
86655
If \(f(x)=\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x\), then \(\int_{0}^{\pi / 4} \frac{\sin ^{2} 2 x}{f(x)} d x=\)
1 2
2 \(\frac{2}{3}\)
3 \(\frac{-2}{3}\)
4 \(\frac{1}{6}\)
Explanation:
(B) : Given, \(f(x)=\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x\) and \(I=\int_{0}^{\pi / 4} \frac{\sin ^{2} 2 x}{f(x)} d x\) \(=\int_{0}^{\pi / 4} \frac{4 \sin ^{2} x \cos ^{2} x}{\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x} d x\) \(=\int_{0}^{\pi / 4} \frac{4 \tan ^{2} x \sec ^{2} x}{\tan ^{6} x+2 \tan ^{3} x+1} d x\) Now, put \(\tan ^{3} x=t\), \(3 \tan ^{2} x \sec ^{2} x d x=d t\) So at \(\mathrm{x}=0, \mathrm{t}=0\) and at \(\mathrm{x}=\frac{\pi}{4}, \mathrm{t}=1\) So, \(I=\frac{4}{3} \int_{0}^{1} \frac{d t}{t^{2}+2 t+1}=\frac{4}{3} \int_{0}^{1} \frac{d t}{(t+1)^{2}}\) \(=\frac{4}{3}\left[\frac{-1}{(t+1)}\right]_{0}^{1}=\frac{-4}{3}\left[\frac{1}{2}-1\right]=\frac{-4}{3} \times \frac{-1}{2}=\frac{2}{3}\)
TS EAMCET-2020-11.09.2020
Integral Calculus
86663
Let \(\frac{d}{d x} F(x)=\left(\frac{e^{\sin x}}{x}\right), x>0\) if \(\int_{1}^{4} \frac{3}{x} e^{\sin x^{3}} d x=f(k)-f(1)\), then one of the possible value of \(k\), is
1 15
2 16
3 63
4 64
Explanation:
(D) : Given that, \(\frac{d}{d x} F(x)=\left(\frac{e^{\sin x}}{x}\right), x>0\) On integrating both sides we get - \(F(x)=\int \frac{e^{\sin x}}{x} d x\) \(\int_{1}^{4} \frac{3}{x} e^{\sin x^{3}} d x =\int_{1}^{4} \frac{3 x^{2}}{x^{3}} \cdot e^{\sin x^{3}} d x\) \(=F(k)-F(1)\) Let us consider, \(x^{3}=z\) \(3 d^{2} d x=d z\) \(\therefore \quad \int_{1}^{64} \frac{\mathrm{e}^{\sin z}}{\mathrm{z}} \mathrm{dz}=\mathrm{F}(\mathrm{k})-\mathrm{F}(1)\) \({[F(z)]_{1}^{64}=F(F)-F(1) \quad[\because \text { from equation }(i)]}\) \(F(64)-F(1)=F(k)-f(1)\) Comparing on both sides we get- \(\mathrm{k}=64\)
86617
The integral \(\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x\) is equal to
1 2
2 4
3 1
4 6
Explanation:
(C) : Let, \(I=\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x\) \(=\int_{2}^{4} \frac{2 \log x}{2 \log x+\log (6-x)^{2}} d x\) \(=\int_{2}^{4} \frac{2 \log x d x}{2[\log x+\log (6-x)]} d x\) \(I=\int_{2}^{4} \frac{\log x}{\log x+\log (6-x)} d x\) \(I=\int_{2}^{4} \frac{\log (6-x)}{\log (6-x)+\log x} d x \ldots . \text { (ii) } \tag{ii}\) On adding equation (i) and (ii), we get - \(2 I=\int_{2}^{4} \frac{\log x+\log (6-x)}{\log x+\log (6-x)} d x\) \(2 I=\int_{2}^{4} d x\) \(2 I=[x]_{2}^{4}\) \(2 I=2\) \(I=1\)
[JCECE-2018]
Integral Calculus
86625
Let \(I=\int_{a}^{b}\left(x^{4}-2 x^{2}\right) d x\). If \(l\) is minimum, then the ordered pair \((a, b)\) is
1 \((-\sqrt{2}, 0)\)
2 \((0, \sqrt{2})\)
3 \((\sqrt{2},-\sqrt{2})\)
4 \((-\sqrt{2}, \sqrt{2})\)
Explanation:
(D) : \(I=\int_{a}^{b}\left(x^{4}-2 x^{2}\right) d x\) \(y=x^{4}-2 x^{2}=0\) \(x=0, \pm \sqrt{2} \tag{i}\) Therefore, the graph of above curve Hence, from the graph we can conduce that the internal value I in minimum when the curve is founded between \((-\sqrt{2}, 0),(\sqrt{2} ., 0)\) \(\therefore\) The ordered pair \((a, b)\) is \((-\sqrt{2}, \sqrt{2})\)
JEE Main-2019-10.01.2019
Integral Calculus
86655
If \(f(x)=\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x\), then \(\int_{0}^{\pi / 4} \frac{\sin ^{2} 2 x}{f(x)} d x=\)
1 2
2 \(\frac{2}{3}\)
3 \(\frac{-2}{3}\)
4 \(\frac{1}{6}\)
Explanation:
(B) : Given, \(f(x)=\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x\) and \(I=\int_{0}^{\pi / 4} \frac{\sin ^{2} 2 x}{f(x)} d x\) \(=\int_{0}^{\pi / 4} \frac{4 \sin ^{2} x \cos ^{2} x}{\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x} d x\) \(=\int_{0}^{\pi / 4} \frac{4 \tan ^{2} x \sec ^{2} x}{\tan ^{6} x+2 \tan ^{3} x+1} d x\) Now, put \(\tan ^{3} x=t\), \(3 \tan ^{2} x \sec ^{2} x d x=d t\) So at \(\mathrm{x}=0, \mathrm{t}=0\) and at \(\mathrm{x}=\frac{\pi}{4}, \mathrm{t}=1\) So, \(I=\frac{4}{3} \int_{0}^{1} \frac{d t}{t^{2}+2 t+1}=\frac{4}{3} \int_{0}^{1} \frac{d t}{(t+1)^{2}}\) \(=\frac{4}{3}\left[\frac{-1}{(t+1)}\right]_{0}^{1}=\frac{-4}{3}\left[\frac{1}{2}-1\right]=\frac{-4}{3} \times \frac{-1}{2}=\frac{2}{3}\)
TS EAMCET-2020-11.09.2020
Integral Calculus
86663
Let \(\frac{d}{d x} F(x)=\left(\frac{e^{\sin x}}{x}\right), x>0\) if \(\int_{1}^{4} \frac{3}{x} e^{\sin x^{3}} d x=f(k)-f(1)\), then one of the possible value of \(k\), is
1 15
2 16
3 63
4 64
Explanation:
(D) : Given that, \(\frac{d}{d x} F(x)=\left(\frac{e^{\sin x}}{x}\right), x>0\) On integrating both sides we get - \(F(x)=\int \frac{e^{\sin x}}{x} d x\) \(\int_{1}^{4} \frac{3}{x} e^{\sin x^{3}} d x =\int_{1}^{4} \frac{3 x^{2}}{x^{3}} \cdot e^{\sin x^{3}} d x\) \(=F(k)-F(1)\) Let us consider, \(x^{3}=z\) \(3 d^{2} d x=d z\) \(\therefore \quad \int_{1}^{64} \frac{\mathrm{e}^{\sin z}}{\mathrm{z}} \mathrm{dz}=\mathrm{F}(\mathrm{k})-\mathrm{F}(1)\) \({[F(z)]_{1}^{64}=F(F)-F(1) \quad[\because \text { from equation }(i)]}\) \(F(64)-F(1)=F(k)-f(1)\) Comparing on both sides we get- \(\mathrm{k}=64\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Integral Calculus
86617
The integral \(\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x\) is equal to
1 2
2 4
3 1
4 6
Explanation:
(C) : Let, \(I=\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x\) \(=\int_{2}^{4} \frac{2 \log x}{2 \log x+\log (6-x)^{2}} d x\) \(=\int_{2}^{4} \frac{2 \log x d x}{2[\log x+\log (6-x)]} d x\) \(I=\int_{2}^{4} \frac{\log x}{\log x+\log (6-x)} d x\) \(I=\int_{2}^{4} \frac{\log (6-x)}{\log (6-x)+\log x} d x \ldots . \text { (ii) } \tag{ii}\) On adding equation (i) and (ii), we get - \(2 I=\int_{2}^{4} \frac{\log x+\log (6-x)}{\log x+\log (6-x)} d x\) \(2 I=\int_{2}^{4} d x\) \(2 I=[x]_{2}^{4}\) \(2 I=2\) \(I=1\)
[JCECE-2018]
Integral Calculus
86625
Let \(I=\int_{a}^{b}\left(x^{4}-2 x^{2}\right) d x\). If \(l\) is minimum, then the ordered pair \((a, b)\) is
1 \((-\sqrt{2}, 0)\)
2 \((0, \sqrt{2})\)
3 \((\sqrt{2},-\sqrt{2})\)
4 \((-\sqrt{2}, \sqrt{2})\)
Explanation:
(D) : \(I=\int_{a}^{b}\left(x^{4}-2 x^{2}\right) d x\) \(y=x^{4}-2 x^{2}=0\) \(x=0, \pm \sqrt{2} \tag{i}\) Therefore, the graph of above curve Hence, from the graph we can conduce that the internal value I in minimum when the curve is founded between \((-\sqrt{2}, 0),(\sqrt{2} ., 0)\) \(\therefore\) The ordered pair \((a, b)\) is \((-\sqrt{2}, \sqrt{2})\)
JEE Main-2019-10.01.2019
Integral Calculus
86655
If \(f(x)=\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x\), then \(\int_{0}^{\pi / 4} \frac{\sin ^{2} 2 x}{f(x)} d x=\)
1 2
2 \(\frac{2}{3}\)
3 \(\frac{-2}{3}\)
4 \(\frac{1}{6}\)
Explanation:
(B) : Given, \(f(x)=\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x\) and \(I=\int_{0}^{\pi / 4} \frac{\sin ^{2} 2 x}{f(x)} d x\) \(=\int_{0}^{\pi / 4} \frac{4 \sin ^{2} x \cos ^{2} x}{\sin ^{6} x+\cos ^{6} x+2 \sin ^{3} x \cos ^{3} x} d x\) \(=\int_{0}^{\pi / 4} \frac{4 \tan ^{2} x \sec ^{2} x}{\tan ^{6} x+2 \tan ^{3} x+1} d x\) Now, put \(\tan ^{3} x=t\), \(3 \tan ^{2} x \sec ^{2} x d x=d t\) So at \(\mathrm{x}=0, \mathrm{t}=0\) and at \(\mathrm{x}=\frac{\pi}{4}, \mathrm{t}=1\) So, \(I=\frac{4}{3} \int_{0}^{1} \frac{d t}{t^{2}+2 t+1}=\frac{4}{3} \int_{0}^{1} \frac{d t}{(t+1)^{2}}\) \(=\frac{4}{3}\left[\frac{-1}{(t+1)}\right]_{0}^{1}=\frac{-4}{3}\left[\frac{1}{2}-1\right]=\frac{-4}{3} \times \frac{-1}{2}=\frac{2}{3}\)
TS EAMCET-2020-11.09.2020
Integral Calculus
86663
Let \(\frac{d}{d x} F(x)=\left(\frac{e^{\sin x}}{x}\right), x>0\) if \(\int_{1}^{4} \frac{3}{x} e^{\sin x^{3}} d x=f(k)-f(1)\), then one of the possible value of \(k\), is
1 15
2 16
3 63
4 64
Explanation:
(D) : Given that, \(\frac{d}{d x} F(x)=\left(\frac{e^{\sin x}}{x}\right), x>0\) On integrating both sides we get - \(F(x)=\int \frac{e^{\sin x}}{x} d x\) \(\int_{1}^{4} \frac{3}{x} e^{\sin x^{3}} d x =\int_{1}^{4} \frac{3 x^{2}}{x^{3}} \cdot e^{\sin x^{3}} d x\) \(=F(k)-F(1)\) Let us consider, \(x^{3}=z\) \(3 d^{2} d x=d z\) \(\therefore \quad \int_{1}^{64} \frac{\mathrm{e}^{\sin z}}{\mathrm{z}} \mathrm{dz}=\mathrm{F}(\mathrm{k})-\mathrm{F}(1)\) \({[F(z)]_{1}^{64}=F(F)-F(1) \quad[\because \text { from equation }(i)]}\) \(F(64)-F(1)=F(k)-f(1)\) Comparing on both sides we get- \(\mathrm{k}=64\)
