86509
The value of \(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) where \(a\), \(b, c, k\) are constants, depends only on
1 a and \(\mathrm{k}\)
2 a and b
3 a, b and c
4 \(\mathrm{k}\)
Explanation:
(D) : Given, \(I=\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) \(=a \int_{-3}^{3} x^{5} d x+b \int_{-3}^{3} x^{3} d x+c \int_{-3}^{3} x d x+k \int d x\) If \(f(x)=x^{5}\) then \(f(-x)=(-x)^{5}=-x^{5}=-f(x)\) \(\therefore \mathrm{f}(\mathrm{x})\) is an odd function \(\therefore \int_{-3}^{3} \mathrm{x}^{5} \mathrm{~d} \mathrm{x}=0\) Similarly, \(\mathrm{x}^{3}\) and \(\mathrm{x}\) are odd function \(\therefore \int_{-3}^{3} \mathrm{x}^{3} \mathrm{dx}=\int_{-3}^{3} \mathrm{x} d \mathrm{x}=0\) So, The value of \(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) depend Only on k.
MHT CET-2019
Integral Calculus
86510
If \(\int_{0}^{k} \frac{d x}{2+18 x^{2}}=\frac{\pi}{24}\), then the value of \(k\) is
86509
The value of \(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) where \(a\), \(b, c, k\) are constants, depends only on
1 a and \(\mathrm{k}\)
2 a and b
3 a, b and c
4 \(\mathrm{k}\)
Explanation:
(D) : Given, \(I=\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) \(=a \int_{-3}^{3} x^{5} d x+b \int_{-3}^{3} x^{3} d x+c \int_{-3}^{3} x d x+k \int d x\) If \(f(x)=x^{5}\) then \(f(-x)=(-x)^{5}=-x^{5}=-f(x)\) \(\therefore \mathrm{f}(\mathrm{x})\) is an odd function \(\therefore \int_{-3}^{3} \mathrm{x}^{5} \mathrm{~d} \mathrm{x}=0\) Similarly, \(\mathrm{x}^{3}\) and \(\mathrm{x}\) are odd function \(\therefore \int_{-3}^{3} \mathrm{x}^{3} \mathrm{dx}=\int_{-3}^{3} \mathrm{x} d \mathrm{x}=0\) So, The value of \(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) depend Only on k.
MHT CET-2019
Integral Calculus
86510
If \(\int_{0}^{k} \frac{d x}{2+18 x^{2}}=\frac{\pi}{24}\), then the value of \(k\) is
86509
The value of \(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) where \(a\), \(b, c, k\) are constants, depends only on
1 a and \(\mathrm{k}\)
2 a and b
3 a, b and c
4 \(\mathrm{k}\)
Explanation:
(D) : Given, \(I=\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) \(=a \int_{-3}^{3} x^{5} d x+b \int_{-3}^{3} x^{3} d x+c \int_{-3}^{3} x d x+k \int d x\) If \(f(x)=x^{5}\) then \(f(-x)=(-x)^{5}=-x^{5}=-f(x)\) \(\therefore \mathrm{f}(\mathrm{x})\) is an odd function \(\therefore \int_{-3}^{3} \mathrm{x}^{5} \mathrm{~d} \mathrm{x}=0\) Similarly, \(\mathrm{x}^{3}\) and \(\mathrm{x}\) are odd function \(\therefore \int_{-3}^{3} \mathrm{x}^{3} \mathrm{dx}=\int_{-3}^{3} \mathrm{x} d \mathrm{x}=0\) So, The value of \(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) depend Only on k.
MHT CET-2019
Integral Calculus
86510
If \(\int_{0}^{k} \frac{d x}{2+18 x^{2}}=\frac{\pi}{24}\), then the value of \(k\) is
86509
The value of \(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) where \(a\), \(b, c, k\) are constants, depends only on
1 a and \(\mathrm{k}\)
2 a and b
3 a, b and c
4 \(\mathrm{k}\)
Explanation:
(D) : Given, \(I=\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) \(=a \int_{-3}^{3} x^{5} d x+b \int_{-3}^{3} x^{3} d x+c \int_{-3}^{3} x d x+k \int d x\) If \(f(x)=x^{5}\) then \(f(-x)=(-x)^{5}=-x^{5}=-f(x)\) \(\therefore \mathrm{f}(\mathrm{x})\) is an odd function \(\therefore \int_{-3}^{3} \mathrm{x}^{5} \mathrm{~d} \mathrm{x}=0\) Similarly, \(\mathrm{x}^{3}\) and \(\mathrm{x}\) are odd function \(\therefore \int_{-3}^{3} \mathrm{x}^{3} \mathrm{dx}=\int_{-3}^{3} \mathrm{x} d \mathrm{x}=0\) So, The value of \(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) depend Only on k.
MHT CET-2019
Integral Calculus
86510
If \(\int_{0}^{k} \frac{d x}{2+18 x^{2}}=\frac{\pi}{24}\), then the value of \(k\) is
86509
The value of \(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) where \(a\), \(b, c, k\) are constants, depends only on
1 a and \(\mathrm{k}\)
2 a and b
3 a, b and c
4 \(\mathrm{k}\)
Explanation:
(D) : Given, \(I=\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) \(=a \int_{-3}^{3} x^{5} d x+b \int_{-3}^{3} x^{3} d x+c \int_{-3}^{3} x d x+k \int d x\) If \(f(x)=x^{5}\) then \(f(-x)=(-x)^{5}=-x^{5}=-f(x)\) \(\therefore \mathrm{f}(\mathrm{x})\) is an odd function \(\therefore \int_{-3}^{3} \mathrm{x}^{5} \mathrm{~d} \mathrm{x}=0\) Similarly, \(\mathrm{x}^{3}\) and \(\mathrm{x}\) are odd function \(\therefore \int_{-3}^{3} \mathrm{x}^{3} \mathrm{dx}=\int_{-3}^{3} \mathrm{x} d \mathrm{x}=0\) So, The value of \(\int_{-3}^{3}\left(a x^{5}+b x^{3}+c x+k\right) d x\) depend Only on k.
MHT CET-2019
Integral Calculus
86510
If \(\int_{0}^{k} \frac{d x}{2+18 x^{2}}=\frac{\pi}{24}\), then the value of \(k\) is
