Mathematics Test - 2

\(\mathrm{n}^{\text {th }}\) term of \(1^{\text {st }}\) series \(=n^{\text {th }}\) term of \(2^{\text {nd }}\) series

\(\Rightarrow 3+(n-1) 7=63+(n-1) 2\)

\(\Rightarrow(n-1) 5=60\)

\(\Rightarrow \mathrm{n}-1=12\)

\(\Rightarrow n=13\)

Hence, The correct answer is A.

\(\vec{a} \times(\vec{b} \times \vec{c})=\frac{\vec{b}}{\sqrt{2}}\)

\(\Rightarrow \overrightarrow{(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}}=\frac{\vec{b}}{\sqrt{2}}\)

\(\Rightarrow \vec{a} \cdot \vec{c}=\frac{1}{\sqrt{2}}\) and \(\vec{a} \cdot \vec{b}=0\)

\(\Rightarrow \vec{a}\) and \(\vec{c}\) are inclined at an angle of \(45^{\circ}\) and \(\vec{a} \perp \vec{b}\)

Hence option B is the correct answer.

Equation of tangent to the circle \(x^{2}+y^{2}=5\) at (1,-2) is \(x-2 y-5=0 \quad \ldots \ldots \ldots \ldots\)

Let this line touches the circle \(x^{2}+y^{2}-8 x+6 y+20=0\) at \(\left(x_{1}, y_{1}\right)\)

\(\therefore\) Equation of tangent at \(\left(x_{1}, y_{1}\right)\) is \(x x_{1}+y y_{1}+4\left(x+x_{1}\right)+3\left(y+y_{1}\right)+20=0\)

\(\Rightarrow x\left(x_{1}-4\right)+y\left(y_{1}+3\right)-4 x_{1}+3 y_{1}+20=0\)

Now, (1) and (2) represent the same line \(\therefore \frac{x_{1}-4}{1}=\frac{y_{1}+3}{-2}=\frac{-4 x_{1}+3 y_{1}+20}{-5}\)

\(\Rightarrow-2 x_{1}+8=y_{1}+3 \Rightarrow 2 x_{1}+y_{1}-5=0\)

Only the point (3,-1) satisfies it. Hence the point of contact is (3,-1)

Hence, The correct answer is B.

\(f(x)=x^{3}-3 x+[a]\)

Let \([a]=t\) (where \(t\) will be an integer \()\) \(f(x)=x^{3}-3 x+t\)

\(\Rightarrow f^{\prime}(x)=3 x^{2}-3\)

\(\Rightarrow f^{\prime}(x)=0\) has two real and distinct solution which are \(x=1\) and \(x\) \(=-1\)

so \(f(x)=0\) will have three distinct and real solution when \(f(1)\). \(f(-1)<0\)

Now, \(f(1)=(1)^{3}-3(1)+t=t-2\)

\(f(-1)=(-1)^{3}-3(-1)+t=t+2\)

From equation (ii)

\((t-2)(t+2)<0\)

\(\Rightarrow t \in(-2,2)\)

Now \(t=[a]\)

Hence \([a] \in(-2,2)\)

\(\Rightarrow a \in[-1,2)\)

Hence, The correct answer is D.

From cosine rule we have \(\Rightarrow c^{2}-2 b c \cos A+b^{2}-a^{2}=0\)

let \(c_{1}\) and \(c_{2}\) be the roots of the above equation. The sum of the roots \(=c_{1}+c_{2}=2 b \cos A\) Now, sum of area of two triangle

\(=\Delta=\Delta_{1}+\Delta_{2}=\frac{b c_{1} \sin A+b c_{2} \sin A}{2}=\frac{b\left(c_{1}+c_{2}\right) \sin A}{2}\)

\(\Rightarrow \Delta=b^{2} \cos A \sin A=\frac{b^{2} \sin 2 A}{2}\)

Hence option A is the correct answer.

Any plane through the given line is \(2 x-y+3 z+1+\lambda(x+y+z+3)=0\)

If this plane is parallel to the line \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) then the normal to the plane is also perpendicular to the above line. \(\therefore\left(I_{1} I_{2}+m_{1} m_{2}+n_{1} n_{2}=0\right)\)

\(\Rightarrow(2+\lambda) 1+(\lambda-1) 2+(3+\lambda) 3=0\)

\(\Rightarrow \lambda=-3 / 2\)

and the required plane is \(2 x-y+3 z+1+(-3 / 2)(x+y+z+3)=0\)

\(\Rightarrow x-5 y+3 z-7=0\)

or

\(x-5 y+3 z=7\)

Hence, The correct answer is A.

The total number of words that can be formed is \(10^{5}\) and number of these words in which no letters are repeated is \(10^{\mathrm{P}}\).

Hence the required number

\(=10^{5}-10 \mathrm{P}_{5}\)

\(=100000-10 \times 9 \times 8 \times 7 \times 6\)

\(=69760\)

Hence, The correct answer is A.

\(g(f(x))=|\sin x|=\sqrt{(\sin x)^{2}}\)

\(f(g(x))=(\sin \sqrt{x})^{2}=\sin ^{2} \sqrt{x}\)

\(\therefore f(x)=\sin ^{2} x\) and \(g(x)=\sqrt{x}\)

Hence, The correct answer is A.

\(f(\theta)=\cos ^{2} \theta+\sin ^{4} \theta\)

\(\Rightarrow f(\theta)=1-\sin ^{2} \theta+\sin ^{4} \theta\)

\(\Rightarrow f(\theta)=\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2}+\frac{3}{4}\)

Minimum value of \(f(\theta)=\frac{3}{4}\) And maximum value of

\(f(\theta) \frac{3}{4} \leq f(\theta) \leq 1\)

\(f(\theta)=\cos ^{2} \theta+\left(1-\cos ^{2} \theta\right)^{2}\)

\(=\frac{1+\cos 2 \theta}{2}+\left(1-\frac{1+\cos 2 \theta}{2}\right)^{2}\)

\(=\frac{1+\cos 2 \theta}{2}+\left(\frac{1-\cos 2 \theta}{2}\right)^{2}\)

So, period of \(f(\theta)=\frac{2 \pi}{2}=\pi\)

Hence option B is the correct answer.

Given, \(f(x-y), f(x), f(y)\) and \(f(x+y)\) are in AP \(\Rightarrow f(x-y)+f(x+y)=2 f(x) f(y) \quad \ldots(1)\)

Substitute \(x=0, y=0\) in (1) \(\Rightarrow 2 f(0)=2 f(0) f(0)\)

\(\Rightarrow f(0)(f(0)-1)=0\)

\(\Rightarrow f(0)=1(\because f(0) \neq 0\) given \()\)

Substitute \(x=0, y=x\) in \((1),\) we get \(f(-x)+f(x)=2 f(0) f(x)\)

\(\Rightarrow f(-x)=f(x) \ldots \ldots \ldots(2)\)

\(\Rightarrow f(-2)=f(2), f(-3)=f(3) \ldots \ldots .(3)\)

Differentiating ( 2 ) w.r.t \(x\) \(f^{\prime}(-x)(-1)=f^{\prime}(x)\)

\(f^{\prime}(-x)+f^{\prime}(x)=0\)

\(f^{\prime}(-2)+f^{\prime}(2)=0\) and \(f^{\prime}(-3)+f^{\prime}(3)=0\)

Hence option C is the correct answer.

The total number of red marbles is 6, numbered= {1,2,3,4,5,6}

The total number of white marbles is 4, numbered= {12,13,14,15}

The probability of a white ball is= 4/6 = 2/3

Among the white-numbered, we have 2 odd and 2 even-numbered marbles.

So, the probability of an odd-numbered white marble is= (2/3)×(1/2) =1/3

\(\int_{2}^{-4}(3-f(x)) d x=7\)

\(\Rightarrow \int_{2}^{4} 3 d x-\int_{2}^{-4}(f(x)) d x=7\)

\(\Rightarrow[3 x]_{2}^{-4}-\int_{2}^{-4}(f(x)) d x=7\)

\(\Rightarrow \int_{2}^{4}(f(x)) d x=-25 \ldots \ldots \ldots \ldots\)

\(\Rightarrow \int_{-4}^{-1}(f(x)) d x=-4 \ldots \ldots \ldots \ldots\)

(ii)

now we can write \(\int_{-4}^{2}(f(x)) d x=\int_{-4}^{-1}(f(x)) d x+\int_{-1}^{2}(f(x)) d x\)

\(\Rightarrow 25=-4+\int_{-1}^{2}(f(x)) d x\)

\(\Rightarrow \int_{-1}^{2}(f(x)) d x=29 \ldots \ldots \ldots \ldots \ldots(i i i)\)

Now, \(I=\int_{-2}^{1}(f(-x)) d x\)

let \(x=-t d x=-d t\)

when \(x=-2, t=2\)

and when \(x=1, t=-1\) so \(I=\int_{-2}^{1}(f(-x)) d x=\int_{2}^{-1}-(f(t)) d t\)

\(\Rightarrow I=\int_{-1}^{2}(f(t)) d t\)

\(\Rightarrow I=\int_{-1}^{2}(f(x)) d x\)

\(\Rightarrow I=29(\) from equation i i i\)

Hence, The correct answer is D.

\(I=\int e^{3 \log x}\left(x^{4}+1\right)^{-1} d x\)

\(=\int \frac{e^{\log z^{2}}}{1+x^{4}} d x\)

\(=\int \frac{x^{3}}{1+x^{4}} d x,\) put \(1+x^{4}=t\)

\(4 x^{3} d x=d t\)

\(I=\int_{t}^{1}\left(\frac{d t}{4}\right)=\frac{1}{4} \log t+C\)

\(=\frac{1}{4} \log \left(1+x^{4}\right)+C\)

Hence, The correct answer is A.

The projection of \(\frac{(\vec{x} \times \vec{y}) \cdot \vec{z}}{|\vec{z}|}\)

\(=\frac{|\vec{x} \vec{y} z|}{|\vec{z}|}\)

\(=\frac{1}{13}\left|\begin{array}{ccc}3 & -6 & -1 \\ 1 & 4 & -3 \\ 3 & -4 & -12\end{array}\right|\)

\(=\frac{1}{13}(3(-48-12)+6(-12+9)-(-4-12))\)

\(=\frac{1}{13}(-180-18+16)\)

\(=\frac{1}{13}(-182)\)

\(=-14\)

Hence, The correct answer is B.