86391
If \(f(x)\) is continuous at \(x=0\) and \(f(0)=2\), then what is the value of \(\lim _{x \rightarrow 0}\left[\frac{\int_{0}^{x} f(u) d u}{x}\right]\) ?
1 0
2 1
3 2
4 \(\mathrm{f}(2)\)
Explanation:
(C): Given, \(\lim _{x \rightarrow 0} \int_{0}^{x} \frac{f(u) d u}{x}=\frac{0}{0}\) Then, differentiate \(=\lim _{x \rightarrow 0} \frac{f(x)}{1}=\frac{f(0)}{1}=\frac{2}{1}=2\) So, \(\lim _{x \rightarrow 0}\left[\frac{\int_{0}^{x} f(u) d u}{x}\right]=2\)
SCRA-2009
Integral Calculus
86392
Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be a differentiable function such that \(f(3)=3, f^{\prime}(3)=\frac{1}{2}\). Then the value of \(\lim _{x \rightarrow 3} \int_{3}^{f(x)}\left[\frac{2 t^{3}}{x-3}\right] \text { is }\)
86393
If \(f: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are continuous functions, then the value of the integral \(\int_{-\pi / 2}^{\pi / 2}(f(x)+f(-x))(g(x)-g(-x)) d x\) is
1 \(\pi\)
2 1
3 -1
4 0
Explanation:
(D) : Let, \(I=\int_{-\pi / 2}^{\pi / 2}(f(x)+f(-x))(g(x)-g(-x)) d x\) \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{-\pi / 2}^{\pi / 2}(f(-x)+f(x))(g(-x)-g(x)) d x \tag{ii}\) On adding equation (i) and (ii), we get - \(\left.2 \mathrm{I}=\int_{-\pi / 2}^{\pi / 2}(\mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x}))(\mathrm{g}(\mathrm{x})-\mathrm{g}(\mathrm{x})) \mathrm{d}(-\mathrm{x})\right)+(\mathrm{f}(-\mathrm{x})+\mathrm{f}(\mathrm{x}))\) \(\quad 2 \mathrm{I}=0\) \(\quad \mathrm{I}=0\)
AMU-2007
Integral Calculus
86394
If \(\int_{0}^{\pi n} f\left(\cos ^{2} x\right) \mathrm{d} x=k \int_{0}^{\pi} f\left(\cos ^{2} x\right) \mathrm{d} x\), then the value of \(k\) is
1 1
2 \(n\)
3 \(\frac{\mathrm{n}}{2}\)
4 none of these
Explanation:
(B) :Given,\(\int_{0}^{\pi n} f\left(\cos ^{2} x\right) d x\) Using \(\int_{a}^{b} f(x) d x=\int_{a}^{b}(a+b-x) d x\) \(=\int_{0}^{\pi n} f\left(\cos ^{2}(\pi n-x) d x\right.\) \(=\mathrm{n} \int_{0}^{\pi} \mathrm{f}\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}\) Comparing \(\mathrm{k} \int_{0}^{\pi} \mathrm{f}\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}\) So, \(\mathrm{k}=\mathrm{n}\)
AMU-2007
Integral Calculus
86395
If \([t]\) denotes the greatest integer \(\leq t\), then the value of \(\frac{3(e-1)}{e} \int_{1}^{2} x^{2} e^{[x]+\left[x^{3}\right]} d x\) is :
86391
If \(f(x)\) is continuous at \(x=0\) and \(f(0)=2\), then what is the value of \(\lim _{x \rightarrow 0}\left[\frac{\int_{0}^{x} f(u) d u}{x}\right]\) ?
1 0
2 1
3 2
4 \(\mathrm{f}(2)\)
Explanation:
(C): Given, \(\lim _{x \rightarrow 0} \int_{0}^{x} \frac{f(u) d u}{x}=\frac{0}{0}\) Then, differentiate \(=\lim _{x \rightarrow 0} \frac{f(x)}{1}=\frac{f(0)}{1}=\frac{2}{1}=2\) So, \(\lim _{x \rightarrow 0}\left[\frac{\int_{0}^{x} f(u) d u}{x}\right]=2\)
SCRA-2009
Integral Calculus
86392
Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be a differentiable function such that \(f(3)=3, f^{\prime}(3)=\frac{1}{2}\). Then the value of \(\lim _{x \rightarrow 3} \int_{3}^{f(x)}\left[\frac{2 t^{3}}{x-3}\right] \text { is }\)
86393
If \(f: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are continuous functions, then the value of the integral \(\int_{-\pi / 2}^{\pi / 2}(f(x)+f(-x))(g(x)-g(-x)) d x\) is
1 \(\pi\)
2 1
3 -1
4 0
Explanation:
(D) : Let, \(I=\int_{-\pi / 2}^{\pi / 2}(f(x)+f(-x))(g(x)-g(-x)) d x\) \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{-\pi / 2}^{\pi / 2}(f(-x)+f(x))(g(-x)-g(x)) d x \tag{ii}\) On adding equation (i) and (ii), we get - \(\left.2 \mathrm{I}=\int_{-\pi / 2}^{\pi / 2}(\mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x}))(\mathrm{g}(\mathrm{x})-\mathrm{g}(\mathrm{x})) \mathrm{d}(-\mathrm{x})\right)+(\mathrm{f}(-\mathrm{x})+\mathrm{f}(\mathrm{x}))\) \(\quad 2 \mathrm{I}=0\) \(\quad \mathrm{I}=0\)
AMU-2007
Integral Calculus
86394
If \(\int_{0}^{\pi n} f\left(\cos ^{2} x\right) \mathrm{d} x=k \int_{0}^{\pi} f\left(\cos ^{2} x\right) \mathrm{d} x\), then the value of \(k\) is
1 1
2 \(n\)
3 \(\frac{\mathrm{n}}{2}\)
4 none of these
Explanation:
(B) :Given,\(\int_{0}^{\pi n} f\left(\cos ^{2} x\right) d x\) Using \(\int_{a}^{b} f(x) d x=\int_{a}^{b}(a+b-x) d x\) \(=\int_{0}^{\pi n} f\left(\cos ^{2}(\pi n-x) d x\right.\) \(=\mathrm{n} \int_{0}^{\pi} \mathrm{f}\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}\) Comparing \(\mathrm{k} \int_{0}^{\pi} \mathrm{f}\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}\) So, \(\mathrm{k}=\mathrm{n}\)
AMU-2007
Integral Calculus
86395
If \([t]\) denotes the greatest integer \(\leq t\), then the value of \(\frac{3(e-1)}{e} \int_{1}^{2} x^{2} e^{[x]+\left[x^{3}\right]} d x\) is :
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Integral Calculus
86391
If \(f(x)\) is continuous at \(x=0\) and \(f(0)=2\), then what is the value of \(\lim _{x \rightarrow 0}\left[\frac{\int_{0}^{x} f(u) d u}{x}\right]\) ?
1 0
2 1
3 2
4 \(\mathrm{f}(2)\)
Explanation:
(C): Given, \(\lim _{x \rightarrow 0} \int_{0}^{x} \frac{f(u) d u}{x}=\frac{0}{0}\) Then, differentiate \(=\lim _{x \rightarrow 0} \frac{f(x)}{1}=\frac{f(0)}{1}=\frac{2}{1}=2\) So, \(\lim _{x \rightarrow 0}\left[\frac{\int_{0}^{x} f(u) d u}{x}\right]=2\)
SCRA-2009
Integral Calculus
86392
Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be a differentiable function such that \(f(3)=3, f^{\prime}(3)=\frac{1}{2}\). Then the value of \(\lim _{x \rightarrow 3} \int_{3}^{f(x)}\left[\frac{2 t^{3}}{x-3}\right] \text { is }\)
86393
If \(f: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are continuous functions, then the value of the integral \(\int_{-\pi / 2}^{\pi / 2}(f(x)+f(-x))(g(x)-g(-x)) d x\) is
1 \(\pi\)
2 1
3 -1
4 0
Explanation:
(D) : Let, \(I=\int_{-\pi / 2}^{\pi / 2}(f(x)+f(-x))(g(x)-g(-x)) d x\) \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{-\pi / 2}^{\pi / 2}(f(-x)+f(x))(g(-x)-g(x)) d x \tag{ii}\) On adding equation (i) and (ii), we get - \(\left.2 \mathrm{I}=\int_{-\pi / 2}^{\pi / 2}(\mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x}))(\mathrm{g}(\mathrm{x})-\mathrm{g}(\mathrm{x})) \mathrm{d}(-\mathrm{x})\right)+(\mathrm{f}(-\mathrm{x})+\mathrm{f}(\mathrm{x}))\) \(\quad 2 \mathrm{I}=0\) \(\quad \mathrm{I}=0\)
AMU-2007
Integral Calculus
86394
If \(\int_{0}^{\pi n} f\left(\cos ^{2} x\right) \mathrm{d} x=k \int_{0}^{\pi} f\left(\cos ^{2} x\right) \mathrm{d} x\), then the value of \(k\) is
1 1
2 \(n\)
3 \(\frac{\mathrm{n}}{2}\)
4 none of these
Explanation:
(B) :Given,\(\int_{0}^{\pi n} f\left(\cos ^{2} x\right) d x\) Using \(\int_{a}^{b} f(x) d x=\int_{a}^{b}(a+b-x) d x\) \(=\int_{0}^{\pi n} f\left(\cos ^{2}(\pi n-x) d x\right.\) \(=\mathrm{n} \int_{0}^{\pi} \mathrm{f}\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}\) Comparing \(\mathrm{k} \int_{0}^{\pi} \mathrm{f}\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}\) So, \(\mathrm{k}=\mathrm{n}\)
AMU-2007
Integral Calculus
86395
If \([t]\) denotes the greatest integer \(\leq t\), then the value of \(\frac{3(e-1)}{e} \int_{1}^{2} x^{2} e^{[x]+\left[x^{3}\right]} d x\) is :
86391
If \(f(x)\) is continuous at \(x=0\) and \(f(0)=2\), then what is the value of \(\lim _{x \rightarrow 0}\left[\frac{\int_{0}^{x} f(u) d u}{x}\right]\) ?
1 0
2 1
3 2
4 \(\mathrm{f}(2)\)
Explanation:
(C): Given, \(\lim _{x \rightarrow 0} \int_{0}^{x} \frac{f(u) d u}{x}=\frac{0}{0}\) Then, differentiate \(=\lim _{x \rightarrow 0} \frac{f(x)}{1}=\frac{f(0)}{1}=\frac{2}{1}=2\) So, \(\lim _{x \rightarrow 0}\left[\frac{\int_{0}^{x} f(u) d u}{x}\right]=2\)
SCRA-2009
Integral Calculus
86392
Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be a differentiable function such that \(f(3)=3, f^{\prime}(3)=\frac{1}{2}\). Then the value of \(\lim _{x \rightarrow 3} \int_{3}^{f(x)}\left[\frac{2 t^{3}}{x-3}\right] \text { is }\)
86393
If \(f: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are continuous functions, then the value of the integral \(\int_{-\pi / 2}^{\pi / 2}(f(x)+f(-x))(g(x)-g(-x)) d x\) is
1 \(\pi\)
2 1
3 -1
4 0
Explanation:
(D) : Let, \(I=\int_{-\pi / 2}^{\pi / 2}(f(x)+f(-x))(g(x)-g(-x)) d x\) \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{-\pi / 2}^{\pi / 2}(f(-x)+f(x))(g(-x)-g(x)) d x \tag{ii}\) On adding equation (i) and (ii), we get - \(\left.2 \mathrm{I}=\int_{-\pi / 2}^{\pi / 2}(\mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x}))(\mathrm{g}(\mathrm{x})-\mathrm{g}(\mathrm{x})) \mathrm{d}(-\mathrm{x})\right)+(\mathrm{f}(-\mathrm{x})+\mathrm{f}(\mathrm{x}))\) \(\quad 2 \mathrm{I}=0\) \(\quad \mathrm{I}=0\)
AMU-2007
Integral Calculus
86394
If \(\int_{0}^{\pi n} f\left(\cos ^{2} x\right) \mathrm{d} x=k \int_{0}^{\pi} f\left(\cos ^{2} x\right) \mathrm{d} x\), then the value of \(k\) is
1 1
2 \(n\)
3 \(\frac{\mathrm{n}}{2}\)
4 none of these
Explanation:
(B) :Given,\(\int_{0}^{\pi n} f\left(\cos ^{2} x\right) d x\) Using \(\int_{a}^{b} f(x) d x=\int_{a}^{b}(a+b-x) d x\) \(=\int_{0}^{\pi n} f\left(\cos ^{2}(\pi n-x) d x\right.\) \(=\mathrm{n} \int_{0}^{\pi} \mathrm{f}\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}\) Comparing \(\mathrm{k} \int_{0}^{\pi} \mathrm{f}\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}\) So, \(\mathrm{k}=\mathrm{n}\)
AMU-2007
Integral Calculus
86395
If \([t]\) denotes the greatest integer \(\leq t\), then the value of \(\frac{3(e-1)}{e} \int_{1}^{2} x^{2} e^{[x]+\left[x^{3}\right]} d x\) is :
86391
If \(f(x)\) is continuous at \(x=0\) and \(f(0)=2\), then what is the value of \(\lim _{x \rightarrow 0}\left[\frac{\int_{0}^{x} f(u) d u}{x}\right]\) ?
1 0
2 1
3 2
4 \(\mathrm{f}(2)\)
Explanation:
(C): Given, \(\lim _{x \rightarrow 0} \int_{0}^{x} \frac{f(u) d u}{x}=\frac{0}{0}\) Then, differentiate \(=\lim _{x \rightarrow 0} \frac{f(x)}{1}=\frac{f(0)}{1}=\frac{2}{1}=2\) So, \(\lim _{x \rightarrow 0}\left[\frac{\int_{0}^{x} f(u) d u}{x}\right]=2\)
SCRA-2009
Integral Calculus
86392
Let \(f: \mathrm{R} \rightarrow \mathrm{R}\) be a differentiable function such that \(f(3)=3, f^{\prime}(3)=\frac{1}{2}\). Then the value of \(\lim _{x \rightarrow 3} \int_{3}^{f(x)}\left[\frac{2 t^{3}}{x-3}\right] \text { is }\)
86393
If \(f: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}\) are continuous functions, then the value of the integral \(\int_{-\pi / 2}^{\pi / 2}(f(x)+f(-x))(g(x)-g(-x)) d x\) is
1 \(\pi\)
2 1
3 -1
4 0
Explanation:
(D) : Let, \(I=\int_{-\pi / 2}^{\pi / 2}(f(x)+f(-x))(g(x)-g(-x)) d x\) \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\) \(I=\int_{-\pi / 2}^{\pi / 2}(f(-x)+f(x))(g(-x)-g(x)) d x \tag{ii}\) On adding equation (i) and (ii), we get - \(\left.2 \mathrm{I}=\int_{-\pi / 2}^{\pi / 2}(\mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x}))(\mathrm{g}(\mathrm{x})-\mathrm{g}(\mathrm{x})) \mathrm{d}(-\mathrm{x})\right)+(\mathrm{f}(-\mathrm{x})+\mathrm{f}(\mathrm{x}))\) \(\quad 2 \mathrm{I}=0\) \(\quad \mathrm{I}=0\)
AMU-2007
Integral Calculus
86394
If \(\int_{0}^{\pi n} f\left(\cos ^{2} x\right) \mathrm{d} x=k \int_{0}^{\pi} f\left(\cos ^{2} x\right) \mathrm{d} x\), then the value of \(k\) is
1 1
2 \(n\)
3 \(\frac{\mathrm{n}}{2}\)
4 none of these
Explanation:
(B) :Given,\(\int_{0}^{\pi n} f\left(\cos ^{2} x\right) d x\) Using \(\int_{a}^{b} f(x) d x=\int_{a}^{b}(a+b-x) d x\) \(=\int_{0}^{\pi n} f\left(\cos ^{2}(\pi n-x) d x\right.\) \(=\mathrm{n} \int_{0}^{\pi} \mathrm{f}\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}\) Comparing \(\mathrm{k} \int_{0}^{\pi} \mathrm{f}\left(\cos ^{2} \mathrm{x}\right) \mathrm{dx}\) So, \(\mathrm{k}=\mathrm{n}\)
AMU-2007
Integral Calculus
86395
If \([t]\) denotes the greatest integer \(\leq t\), then the value of \(\frac{3(e-1)}{e} \int_{1}^{2} x^{2} e^{[x]+\left[x^{3}\right]} d x\) is :