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Mathematics Test - 53

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Mathematics Test - 53
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  • Question 1
    1 / -0

    tanθ + cotθ/tanθ − cotθ = 2, (0 ⩽ θ ⩽ 90∘), then the value of sinθ is?

    Solution
    \(\frac{\tan \theta+\cot \theta}{\tan \theta-\cot \theta}=2\)
    By componendo and dividendo \(\Rightarrow \frac{2 \tan \theta}{2 \cos \theta}=\frac{3}{1}\)
    \(\Rightarrow \frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\cos \theta}=3\)
    \(\Rightarrow \sin ^{2} \theta=3 \cos ^{2} \theta\)
    \(\Rightarrow \sin ^{2} \theta=3\left(1-\sin ^{2} \theta\right)\)
    \(\Rightarrow 4 \sin ^{2} \theta=3\)
    \(\Rightarrow \sin ^{2} \theta \Rightarrow \frac{3}{4}\)
    \(\Rightarrow \sin \theta=\frac{\sqrt{3}}{2}\)
    Alternate :
    \(\Rightarrow \frac{\tan \theta+\cot \theta}{\tan \theta-\cot \theta}=2\)
    By \(C\) and \(D\)
    \(\Rightarrow \frac{\tan \theta}{\cot \theta}=\frac{3}{1}\)
    \(\Rightarrow \tan ^{2} \theta=3\)
    \(\Rightarrow \tan \theta=\sqrt{3}\)
    \(\theta=60^{\circ}\)
    \(\Rightarrow \sin \theta\)
    \(\Rightarrow \sin 60^{\circ}\)
    \(\Rightarrow \frac{\sqrt{3}}{2}\)
  • Question 2
    1 / -0
    \(\frac{d}{d x}\left(\frac{1}{x^{4} \sec x}\right)=\)
    Solution
    \(\frac{d}{d x}\left(\frac{1}{x^{4} \sec x}\right)=\frac{d}{d x}\left(\frac{\cos x}{x^{4}}\right)=\frac{x^{4}(-\sin x)-\cos x\left(4 x^{3}\right)}{\left(x^{4}\right)^{2}}\)
    \(=\frac{-x^{3}(x \sin x+4 \cos x)}{x^{8}}=\frac{-(x \sin x+4 \cos x)}{x^{5}}\)
  • Question 3
    1 / -0
    \(\frac{d}{d x}\left(x^{2} \sin \frac{1}{x}\right)=\)
    Solution
    \(\frac{d}{d x}\left(x^{2} \sin \frac{1}{x}\right)=x^{2} \cos \left(\frac{1}{x}\right) \frac{d}{d x}\left(\frac{1}{x}\right)+2 x \sin \left(\frac{1}{x}\right)\)
    \(=-\frac{1}{x^{2}} \cdot x^{2} \cos \left(\frac{1}{x}\right)+2 x \sin \left(\frac{1}{x}\right)=2 x \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)\)
  • Question 4
    1 / -0

    The average of the first five multiples of 9 is:

    Solution
    Required average
    =(total sum of multiple of 9/5)
    =(9+18+27+36+45/5)=27
    Required average=(total sum of multiple of 95)=(9+18+27+36+455)=27
    Note that, average of 9 and 45 is also 27.
    And average of 18 and 36 is also 27.
  • Question 5
    1 / -0

    Find the value of \(\sqrt{(2 x-5)^{2}}+2 \sqrt{(x-1)^{2}}\), if \(1

    Solution

    Given,

    \(\sqrt{(2 x-5)^{2}}+2 \sqrt{(x-1)^{2}}\)

    \(=\sqrt{(5-2 x)^{2}}+2 \sqrt{(x-1)^{2}}(\) as \(1

    \(=(5-2 x)+2(x-1)\)

    \(=5-2 x+2 x-2\)

    \(=3\)

    \(\therefore \sqrt{(2 x-5)^{2}}+2 \sqrt{(x-1)^{2}}=3\)

  • Question 6
    1 / -0
    If \(y=x+\frac{1}{x}\), then
    Solution
    \(y=x+\frac{1}{x}\) \(\frac{d y}{d x}=1-\frac{1}{x^{2}} \quad\) Therefore
    \(x^{2} \cdot \frac{d y}{d x\left(1-\frac{1}{2}\right)}-x y+2=x^{2}\)
  • Question 7
    1 / -0

    Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.

    Solution
    \(1^{\text {st }}\) Method:
    \(8^{\text {th }}\) term \(=a+7 d=39\)......i
    \(12^{\text {th }}\) term \(=a+11 d=59\).......ii
    \((i)-(i i)\)
    Or, \(a+7 d-a-11 d=39-59\)
    Or, \(4 \mathrm{d}=20\)
    Or, \(d=5\)
    Hence, \(a+7 \times 5=39\)
    Thus, \(a=39-35=4\)
    \(2^{\text {nd }}\) Method (Thought Process):
    \(8^{\text {th }}\) term \(=39\) And, \(12^{\text {th }}\) term \(=59\) Here, we see that 20 is added to \(8^{\text {th }}\) term 39 to get \(12^{\text {th }}\) term 59 i.e. 4 times the common difference is added to 39 \(\mathrm{So}, \mathrm{CD}=\frac{20}{4}=5\)
    Hence, 7 times \(\mathrm{CD}\) is added to \(1^{\text {st }}\) term to get \(39 .\) That means 4 is the \(1^{\text {st }}\) term of the AP
  • Question 8
    1 / -0

    If cos20° = m and cos70° =n, then the value of m2 + n2 is?

    Solution
    \(\cos 20^{\circ}=m\)
    \(\cos 70^{\circ}=n\)
    \(\mathrm{So}\)
    \(\Leftrightarrow m^{2}+n^{2}=\cos ^{2} 20^{\circ}+\cos ^{2} 70^{\circ}\)
    \(\left[\text { If } \cos ^{2} A+\cos ^{2} B=1\right]\)
    (If, \(\left.A+B=90^{\circ}\right)\)
    \(\Leftrightarrow 1\)
  • Question 9
    1 / -0

    A boy agrees to work at the rate of one rupee on the first day, two rupees on the second day, and four rupees on third day and so on. How much will the boy get if he started working on the 1st of February and finishes on the 20th of February?

    Solution
    \(1^{\text {st}}\) term \(=1\)
    Common ration \(=2\)
    \(\begin{aligned} \operatorname{Sum}\left(S_{n}\right) &=a \times \frac{\left(r^{n}-1\right)}{(r-1)} \\ &=1 \times \frac{\left(2^{20}-1\right)}{(2-1)} \\ &=2^{20}-1 \end{aligned}\)
  • Question 10
    1 / -0

    In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?

    Solution
    No. of ways in which 10 paper can arranged is 10! Ways.
    When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
    These 9 papers can be arranged in 9! Ways.
    And two papers can be arranged themselves in 2! Ways.
    No. of arrangement when best and worst paper do not come together,
    = 10! - 9! × 2!
    = 9!(10 - 2)
    = 8 × 9!
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