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Mathematics Test - 13

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Mathematics Test - 13
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Weekly Quiz Competition
  • Question 1
    4 / -1

    The number of values of k for which the system of equations (k + 1) x + 8y = 4k and

    kx + (k + 3)y = 3k - 1 has infinitely many solutions is

    Solution
    For infinitely many solutions, the two equations become identical \(\Rightarrow \frac{k+1}{k}=\frac{8}{k+3}=\frac{4 k}{3 k-1}\)
    from first 2
    \(\frac{k+1}{k}=\frac{8}{k+3}\)
    \(\Rightarrow k=1,3\)
    From last 2
    \(\frac{8}{k+3}=\frac{4 k}{3 k-1}\)
    \(\Rightarrow k=1,2\)
    so value of \(k\) which satisfy both is 1
    so \(k=1\) so number of values of \(\mathrm{k}\) is one
  • Question 2
    4 / -1

    If \(a_{n}=\frac{\Sigma n}{n !}\), the sum of the infinite series \(\operatorname{da}_{n}\) is

    Solution
    \(a_{n}=\frac{\sum n}{n !}=\frac{n(n+1)}{2(n !)}=\frac{1}{2}\left[\frac{1}{(n-2) !}+\frac{2}{(n-1) !}\right]\)
    \(\sum a_{n}=\frac{1}{2}\left[\sum_{n=2}^{\infty} \frac{1}{(n-2) !}+2 \sum_{n=1}^{\infty} \frac{1}{(n-1) !}\right]=\frac{1}{2}(e+2 e)=\frac{3}{2} e\)
  • Question 3
    4 / -1

    \(\int x^{3} e^{x^{2}} d x\) is equal to

    Solution
    \(=\frac{1}{2}\left[t \mathrm{e}^{\mathrm{t}}-\mathrm{e}^{\mathrm{t}}\right]+\mathrm{C}=\frac{1}{2}\left[\mathrm{x}^{2} \mathrm{e}^{\mathrm{x}^{2}}-\mathrm{e}^{\mathrm{x}^{2}}\right]+\mathrm{C}=\frac{1}{2} \mathrm{e}^{\mathrm{x}^{2}}\left(\mathrm{x}^{2}-1\right)+\mathrm{C}\)
  • Question 4
    4 / -1

    Area of an equilateral triangle is √3 cm2. The length of each side of the triangle is

    Solution
    Area of equilateral triangle \(=\frac{\sqrt{3}}{4} a^{2}\) \(\frac{\sqrt{3}}{4} a^{2}=\sqrt{3}\)
    \(a^{2}=4\)
    \(a=2 \mathrm{cm}\)
  • Question 5
    4 / -1
    If \(|\vec{a}|=4,|\vec{b}|=4\) and \(|\vec{c}|=5\) such that \(\vec{a} \perp(\vec{b}+\vec{c})\)
    \(\vec{b} \perp \overrightarrow{(c+a)}\) and \(\vec{c} \perp(\vec{a}+\vec{b}),\) then \(|\vec{a}+\vec{b}+\vec{c}|=\)
    Solution
    \(\vec{a} \perp(\vec{b}+\vec{c}), \vec{b} \perp(\vec{c}+\vec{a}) \vec{c}-(\vec{a}+\vec{b})\)
    \(\vec{a} \cdot \vec{b}+\vec{c})=0, \vec{b} \cdot(\vec{c}+\vec{a})=0, \vec{c} \cdot(\vec{a}+\vec{b})=0\)
    \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=2 \sum \vec{a} \cdot \vec{b}=0\)
    \(|\vec{a}+\vec{b}+\vec{c}|^{2}=\sum a^{2}+2 \sum \vec{a} \cdot \vec{b}\)
    \(|\vec{a}+\vec{b}+\vec{c}|^{2}=\sum a^{2}+0=4^{2}+4^{2}+5^{2}=57\)
    \(|\vec{a}+\vec{b}+\vec{c}|=\sqrt{57}\)
  • Question 6
    4 / -1
    Let \(a_{1}, a_{2}, a_{3},\) be terms of an AP. If \(\frac{a_{1}+a_{2}+\ldots+a_{p}}{a_{1}+a_{2}+\ldots a_{q}}=\frac{p^{2}}{q^{2}} p \neq q,\) then \(\frac{a_{6}}{a_{21}}\) equals
    Solution
    \(\frac{a_{1}+a_{2}+\ldots+a_{p}}{a_{1}+a_{2}+\ldots+a_{q}}=\frac{p^{2}}{q^{2}}\)
    \(\Rightarrow \frac{\frac{p}{2}\left[2 a_{1}+(p-1) d\right]}{\frac{q}{2}\left[2 a_{1}+(q-1) d\right]}=\frac{p^{2}}{q^{2}}\)
    \(\Rightarrow \frac{\left(2 a_{1}-d\right)+p d}{\left(2 a_{1}-d\right)+q d}=\frac{p}{q}\)
    \(\Rightarrow\left(2 a_{1}-d\right)(p-q)=0\)
    As p is not equal to \(\Rightarrow a_{1}=\frac{d}{2}\)
    \(\mathrm{Now} \frac{\mathrm{a}_{6}}{\mathrm{a}_{21}}=\frac{\mathrm{a}_{1}+5 \mathrm{d}}{\mathrm{a}_{1}+20 \mathrm{d}}\)
    \(=\frac{\frac{d}{2}+5 d}{\frac{d}{2}+20 d}=\frac{11 d}{41 d}\)
    \(=\frac{11}{41}\)
  • Question 7
    4 / -1

    If the sum of two of the roots ofx3+px2+qx+r=0is zero, then pq =

    Solution

    Given that, α + β = 0

    α + β + γ = -p ⇒ γ = -p

    Substituting γ = -p in the given equation

    ⇒−p3+p3−pq+r=0⇒ pq =r

  • Question 8
    4 / -1

    The orthocenter of the triangle formed by the lines xy = 0 and x + y = 1 is

    Solution

    The triangle formed by the given lines is a right angle triangle right angled at (0, 0). Hence, it is the orthocenter.

  • Question 9
    4 / -1
    The value of the limit
    \(\lim _{x \rightarrow \frac{\pi}{2}}\left[1^{1 / \cos ^{2} x}+2^{1 / \cos ^{2} x}+3^{1 / \cos ^{2} x}+\ldots \ldots \ldots .+10^{1 / \cos ^{2} x}\right]^{\cos ^{2} x}\)
    Solution
    \(\mathrm{Let} \frac{1}{\mathrm{cos}^{2} \mathrm{x}}=\lambda \geq 1\)
    \(\left(a s 0 \leq \cos ^{2} x \leq 1\right)\)
    \(\therefore \lim _{x \rightarrow \frac{\pi}{2}}\left[1^{1 \cos ^{2} x}+2^{1 \cos ^{2} x}+3^{1 \cos ^{2} x}+\ldots \ldots \ldots+10^{1 \cos ^{2} x}\right]^{\cos ^{2} x}\)
    \(=\lim _{\lambda \rightarrow \infty}\left(1^{\lambda}+2^{\lambda}+\ldots+10^{\lambda}\right)^{\frac{1}{\lambda}}\)
    \(=\left(10^{\lambda}\right)^{\frac{1}{\lambda}} \lim _{\lambda \rightarrow \infty}\left[\left(\frac{1}{10}\right)^{\lambda}+\left(\frac{2}{10}\right)^{\lambda}+\ldots+1\right]^{\frac{1}{\lambda}}\)
    \(=10[0+0+\ldots+1]^{0}=10\)
  • Question 10
    4 / -1

    Let n(U) = 700, n(A) = 200, n(B) = 300 and n(A ∩ B) = 100,

    Then n(Ac∩Bc)=

    Solution

    n(Ac∩Bc) = n(U) - n(A ∪ B)

    = n(U) - [n(A) + n(B) - n(A ∩ B)]

    = 700 - [200 + 300 - 100] = 300.

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