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Mathematics Test-25

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Mathematics Test-25
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  • Question 1
    4 / -1

    Evaluate \(\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x\)

    Solution

    Given:

    \(I=\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x\)

    Let

    \(I^{\prime}=\int e^x\left(f(x)+f^{\prime}(x)\right) d x=\int e^x f(x) d x+\int e^x f^{\prime}(x) d x\)

    Integration by Parts of first term:

    \(\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left[f^{\prime}(x) \int g(x) d x\right] d x\)

    Where \(\mathrm{f}\) is first function and \(\mathrm{g}\) is second function.

    Preference order generally adopted for the selection of the first function: ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)

    \(f=f(x) ; g=e^x\)

    \(I^{\prime}=\left\{f(x) \int e^x d x-\int f^{\prime}(x)\left[\int e^x d x\right] d x\right\}+\int e^x f^{\prime}(x) d x\)

    \(I^{\prime}=\left\{e^x f(x)-\int e^x f^{\prime}(x) d x\right\}+\int e^x f^{\prime}(x) d x=e^x f(x)\) 

    \(\therefore \int e^x\left(f(x)+f^{\prime}(x)\right) d x=e^x f(x)+c\)

    Here \(f(x)=\frac{1}{x} ; f^{\prime}(x)=-\frac{1}{x^2}\)

    \(I=\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x=\frac{e^x}{x}+c\)

  • Question 2
    4 / -1
    The angle between the vectors \(\hat{i}-\hat{j}\) and \(\hat{j}-\hat{k}\) is:
    Solution

    Given,

    vectors are:

    \(\vec{ a }=\hat{ i }-\hat{ j }\)

    \(\vec{ b }=\hat{ i }-\hat{ k }\)

    So, \(\vec{a} \cdot \vec{b}=(\hat{i}-\hat{j}) \cdot(\hat{i}-\hat{k})\)

    \(\vec{a} \cdot \vec{b}=(1 \times 0)+(-1 \times 1)+(0 \times(-1)) \)

    \(\Rightarrow \vec{a} \cdot \vec{b}=-1 \)

    Also, \(|\vec{a}|=\sqrt{1^{2}+(-1)^{2}+0}=\sqrt{2} \)

    \(|\vec{b}|=\sqrt{0+1^{2}}=\sqrt{2} \)

    We also know, \( \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cdot \cos \theta \)

    \(\Rightarrow-1=\sqrt{2} \cdot \sqrt{2} \cdot \cos \theta \)

    \(\Rightarrow \cos \theta=\frac{-1}{2} \)

    \(\Rightarrow \theta=\frac{2 \pi}{3}\)

    The angle between the vectors is \(\frac{2 \pi}{3}\).

  • Question 3
    4 / -1

    Find the coefficient of \(x^{4}\) in the expansion of \(\left(1+x+x^{2}+x^{3}\right)^{11}\).

    Solution

    Given:

    \(\left(1+x+x^{2}+x^{3}\right)^{11}\).

    By expanding given equation using expansion formula we can get the coefficient \(x^{4}\).

    \( 1+x+x^{2}+x^{3}\)

    \(\Rightarrow(1+x)+x^{2}(1+x)\)

    \(\Rightarrow(1+x)\left(1+x^{2}\right)\)

    So,

    \(\left(1+x+x^{2}+x^{3}\right)^{11}=(1+x)^{11}\left(1+x^{2}\right)^{11}\)

    \(=1+{ }^{11} C_{1} x^{2}+{ }^{11} C_{2} x^{2}+{ }^{11} C_{3} x^{3}+{ }^{11} C_{4} x^{4} \ldots \ldots\)

    \(=1+{ }^{11} C_{1} x^{2}+{ }^{11} C_{2} x^{4}+\ldots \ldots\)

    To find term in from the product of two brackets on the right-hand-side, consider the following products terms as,

    \(=1 \times{ }^{11} C_{2} x^{4}+{ }^{11} C_{2} x^{2} \times{ }^{11} C_{1} x^{2}+{ }^{11} C_{4} x^{4}\)

    \(\Rightarrow\left[{ }^{11} C_{2}+{ }^{11} C_{2} \times{ }^{11} C_{1}+{ }^{11} C_{4}\right] x^{4}\)

    \(\Rightarrow[55+605+330] x^{4}\)

    \(\Rightarrow 990 x^{4}\)

    So,

    The coefficient of \(x^{4}\) is 990.

  • Question 4
    4 / -1

    There are \(12\) points in a plane out of which \(5\) are collinear. The number of triangles formed by the points as vertices is:

    Solution

    Total number of triangles that can be formed with \(12\) points (if none of them are collinear).

    \(={ }^{12} \mathrm{C}_{3}\)

    (this is because we can select any three points and form the triangle if they are not collinear).

    With collinear points, we cannot make any triangle (as they are in straight line). Here \(5\) points are collinear. Therefore we need to subtract \({ }^{5} \mathrm{C}_{3}\) triangles from the above count.

    So, required number of triangles, \(={ }^{12} \mathrm{C}_{3}-{ }^{5} \mathrm{C}_{3}=\) \(\frac{12!}{{(12-3)!}{3!}}-\frac{5!}{(5-3)!×3!}=\)\(\frac{9!×10×11×12}{1×2×3×9!}-\frac{5×4×3!}{1×2×3!}\)

    \(=220-10\)

    \(=210\)

  • Question 5
    4 / -1

    The distance between two parallel tangents of a circle of radius \(4\) cm is:

    Solution

    As we know that, the distance between the two parallel tangent to a circle is equal to the diameter. As, according to the diagram given below,

    \(\mathrm{{PQ}={PO}+{OQ}}\)

    \(=2 \times \) radius

    \(=2 \times 4\) cm

    \(=8\) cm

  • Question 6
    4 / -1

    Solve \((2 y+x) \frac{d y}{d x}=1\).

    Solution

    As we know,

    In first order linear differential equation;

    \(\frac{d y}{d x}+P y=Q\), where \(P\) and \(Q\) are function of \(x\)

    Integrating factor \((IF)=\mathrm{e}^{\int Pdx}\)

    General solution: \(y \times(IF)=\int Q(IF) dx\)

    The given equation is,

    \((2 y+x) \frac{dy}{dx}=1\)

    \(\Rightarrow(2 y+x)=\frac{dx}{dy}\)

    \(\Rightarrow \frac{dx}{dy}-x=2 y\)

    \(\therefore\) It is linear differential equation is of first order.

    \(I F=e^{l-1 d y}\)

    \(\Rightarrow I F=e^{-y}\)

    Now, \(x \times(I F)=\int Q(I F) d y\)

    \(\Rightarrow x \times e^{-y}=\int 2 y \times e^{-y} d y\)

    \(\Rightarrow x e^{-y}=2\left[y \int e^{-y} d y-\int\left\{\frac{d y}{d y} \times \int e^{-y} d y\right\} d y\right]\)

    \(\Rightarrow x e^{-y}=2\left[-y e^{-y}+\int e^{-y} d y\right]+c\)

    \(\Rightarrow x e^{-y}=2\left[-y e^{-y}-e^{-y}\right]+c\)

    \(\Rightarrow x+2 y+2=c e^y\)

  • Question 7
    4 / -1

    Find the set of value of x for which f(x) = cos x − x is decreasing in

    Solution

    It is given that,

    f(x) = cos x – x

    Differentiating with respect to x, we get

    ⇒ f’(x) = -sin x – 1

    As we know that,

    -1 ≤ sin x ≤ 1, for x ∈ R

    ⇒ -1 ≤ -sin x ≤ 1, for x ∈ R

    ⇒ -1 - 1 ≤ -sin x - 1 ≤ 1 - 1, for x ∈ R

    ⇒ -2 ≤ f’(x) ≤ 0, for x ∈ R

    ∴ f’(x) ≤ 0, for x ∈ R

    Hence f(x) is decreasing in x ∈ R or x ∈ (-∞,∞).

  • Question 8
    4 / -1

    If \(a \cot \theta+b \operatorname{cosec} \theta=p\) and \(b \cot \theta+a\) \(\operatorname{cosec} \theta=q\), then \(p^2-q^2=\)

    Solution

    \(a \cot \theta+b \operatorname{cosec} \theta=p\)

    \(b \cot \theta+a \operatorname{cosec} \theta=q\)

    Squaring and subtracting,

    \(p^2-q^2=(a \cot \theta+b \operatorname{cosec} \theta)^2-(b \cot \theta+\) \(a \operatorname{cosec} \theta)^2\)

    \(=a^2 \cot ^2 \theta+b^2 \operatorname{cosec}^2 \theta+2 a b \cot \theta \operatorname{cosec} \theta-\left(b^2 \cot ^2 \theta+a^2 \operatorname{cosec}^2 \theta+2 a b \cot \theta \operatorname{cosec} \theta\right)\)

    \(=a^2 \cot ^2 \theta+b^2 \operatorname{cosec}^2 \theta+2 a b \cot \theta \operatorname{cosec} \theta-b^2 \cot ^2 \theta-a^2 \operatorname{cosec}^2 \theta-2 a b \cot \theta \operatorname{cosec} \theta\)

    \(=a^2\left(\cot ^2 \theta-\operatorname{cosec}^2 \theta\right)+b^2\left(\operatorname{cosec}^2 \theta-\cot ^2 \theta\right)\)

    \(=-a^2\left(\operatorname{cosec}^2 \theta-\cot ^2 \theta\right)+b^2\left(\operatorname{cosec}^2 \theta-\right.\) \(\left.\cot ^2 \theta\right)\)\(\left[\because \left(\operatorname{cosec}^2 \theta-\cot ^2 \theta =1 \right)\right]\)

    \(=-a^2 \times 1+b^2 \times 1=b^2-a^2\)

  • Question 9
    4 / -1

    \(\left|\begin{array}{lll}a+b & a & b \\ a & a+c & c \\ b & c & b+c\end{array}\right|=\)

    Solution

    \(\left|\begin{array}{ccc}a+b & a & b \\ a & a+c & c \\ b & c & b+c\end{array}\right|\)

    \(=(a+b)\left[(a+c)(b+c)-c^2\right]-a[a b+a c-b c]+b[a c-a b-b c]\)

    \(=(a+b)[a b+a c+c b]-a^2 b-a^2 c+a b c+a b c-a b^2-b^2 c\)

    \(=a^2 b+a^2 c+a b c+a b^2+a b c+b^2+a b c+c b^2-a^2 b-a^2 c+\)

    \(a b c+a b c-a b^2-b^2 c\)

    \(=4 a b c\)

  • Question 10
    4 / -1
    The vector equation of the plane passing through \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\), is \(\overrightarrow{\mathrm{r}}=\alpha \overrightarrow{\mathrm{a}}+\beta \overrightarrow{\mathrm{b}}+\gamma \overrightarrow{\mathrm{c}}\), provided that,
    Solution

    Given,

    A plane passing through \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\).

    Lines \(\vec{a}-\vec{b}\) and \(\vec{c}-\vec{a}\) lie on the plane.

    The parmetric equation of the plane can be written as:

    \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda_{1}(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})+\lambda_{2}(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}) \)

    \(\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}\left(1+\lambda_{1}+\lambda_{2}\right)-\lambda_{1} \overrightarrow{\mathrm{b}}+\lambda_{2} \overrightarrow{\mathrm{c}}\)

    Given that,

    \(\overrightarrow{\mathrm{r}}=\alpha \overrightarrow{\mathrm{a}}+\beta \overrightarrow{\mathrm{b}}+\gamma \overrightarrow{\mathrm{c}}\)

    \(\therefore \alpha+\beta+\gamma=1+\lambda_{1}-\lambda_{2}-\lambda_{1}+\lambda_{2}\)

    \(\alpha+\beta+\gamma=1\)

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