Mathematics Test - 29

Differentiating both sides, we get
\(\sqrt{1+\sin x} \quad f(x)=\frac{2}{3} \frac{3}{2}(1+\sin x)^{1 / 2} \cos x\)
\(\Rightarrow f(\mathrm{x})=\cos \mathrm{x}\)
Concepts:
Main Concept:
Antiderivative IntegralA function \(\varphi(x)\) is called a primitive or an anti - derivative of a function \(\$ f(x)\), \(i f \varphi\)
Let \(f(x)\) be a function. Then the collection of all its primitives is called the indefinite integral of \(f(x)\) and is denoted by \(\int f(x) \mathrm{dx}\)
Thus, \(\frac{d}{d x}(\phi(x)+c)=f(x) \Rightarrow \int f(x) d x=\phi(x)+c\)
Where \(\phi(x)\) is primitive of \(f(x)\) and \(c\) is an arbitrary constant known as the constant of integration.
Hence, the correct option is (A)

Let \(\quad S_{1} \equiv x^{2}+y^{2}-4 x-6 y-3=0\)
and \(S_{2} \equiv x^{2}+y^{2}+2 x+2 y+1=0\)
Centre and radius of \(S_{1}\) are \(C_{1}(2,3)\) and \(r_{1}=4\) and centre and radius of \(S_{2}\) are \(C_{2}(-1,-1)\) and \(r_{2}=1\)
\(\because \quad C_{1} C_{2}=\sqrt{(9+16)}=5=r_{1}+r_{2}\)
Hence the given circles are externally touching each other
Hence, number of common tangents = 3
Hence, the correct option is (C)

\(\mathrm{y}^{2}=12 \mathrm{x}\)
\(4 a=12 \quad\left(\because y^{2}=4 a x\right)\)
\(\Rightarrow \quad \mathrm{a}=3\)
Equation of tangent in slope form is
\(y=m x+\frac{3}{m}\)
It passes through (2,5)
\(\Rightarrow 5=2 m+\frac{3}{m}\)
\(\Rightarrow 5 m=2 m^{2}+3\)

Let \(\alpha\) be the common roots
\(\therefore \alpha^{2}+\alpha+a=0\)......(i)
and
\(\alpha^{2}+a \alpha+1=0\)......(ii)
\(\frac{\alpha^{2}}{1-a^{2}}=\frac{\alpha}{a-1}=\frac{1}{a-1}[a \neq 1]\)
Eliminating \(\alpha\), we get \((a-1)^{2}=\left(1-a^{2}\right)(a-1)\)
\(\Rightarrow(a-1)=1-a^{2}\)
\(\Rightarrow a^{2}+a-2=0\)
\(\Rightarrow a^{2}+2 a-a-2=0\)
\(\Rightarrow a(a+2)-1(a+2)=0\)
\(\Rightarrow(a+2)(a-1)=0\)
\(\Rightarrow a=-2 \quad[\because a \neq 1]\)
Hence, the correct option is (B)

The direction cosines of the line are
\(l^{2}+m^{2}+n^{2}=1\)
Now, \(\Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\)
\(\Rightarrow 1-\sin ^{2} \alpha+1-\sin ^{2} \beta+1-\sin ^{2} \gamma=1\)
\(\Rightarrow \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=2\)
Hence, the correct option is (C)

range of \(\sin ^{-1} x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
range of \(\cos ^{-1} x \in[0, \pi]\)
\(\therefore f(x)=\sin ^{-1} x-\cos ^{-1} x=\sin ^{-1} x+\cos ^{-1} x-2 \cos ^{-1} x\)
\(=\frac{\pi}{2}-2 \cos ^{-1} x \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]\) jdentity
range of \(f(x) \epsilon\left[\frac{\pi}{2}-2(\pi), \frac{\pi}{2}-2(0)\right]\)
\(\therefore f(x) \epsilon\left[-\frac{3 \pi}{2}, \frac{\pi}{2}\right]\)
Hence, the correct option is (A)

The equation of chord of contact at point \((h, k)\) is \(x h-y k=9\)
Comparing with \(x=9\), we have \(h=1, k=0\)
Hence equation of pair of tangents at point (1,0) is \(S S_{1}=T^{2}\)
\(\left(x^{2}-y^{2}-9\right)\left(1^{2}-0^{2}-9\right)=(x-9)^{2}\)
\(\Rightarrow-8 x^{2}+8 y^{2}+72=x^{2}-18 x+81\)
\(\Rightarrow 9 x^{2}-8 y^{2}-18 x+9=0\)
Hence , the correct option is B

\(\theta=\sin ^{-1} 2 x\)
\(\therefore \tan \left(\sin ^{-1}(2 x)\right)=\frac{2 x}{\sqrt{1-4 x^{2}}}\)
\(f^{\prime}(x)=\frac{2}{\sqrt{1-4 x^{2}}}+2 x\left(-\frac{1}{2}\right)\left(1-4 x^{2}\right)^{-\frac{3}{2}}(-8 x)\)
\(=\frac{2\left(1-4 x^{2}\right)+8 x^{2}}{\left(1-4 x^{2}\right)^{\frac{3}{2}}}\)
\(=\frac{2}{(1-4 x)^{\frac{3}{2}}}\)
\(f^{\prime}\left(\frac{1}{4}\right)=\frac{2}{\left(1-4 \times \frac{1}{16}\right)^{3 / 2}}=\frac{2}{\left(\frac{3}{4}\right)^{3 / 2}}=\frac{2}{\sqrt{\frac{27}{64}}}\)
\(=\frac{2}{3} \times \frac{8}{\sqrt{3}}=\frac{16 \sqrt{3}}{9}\)
\(\Rightarrow \mathrm{f}^{\prime}\left(\frac{1}{4}\right)=\frac{16 \sqrt{3}}{9}\)
Hence, the correct option is (A)

\(E_{1}=\) the event of drawing from the first box.
\(\mathrm{E}_{2}=\) the event of drawing form the second box.
\(B=\) the event of drawing a black ball
\(\mathrm{P}\left(\mathrm{E}_{1}\right)=\frac{3}{6}=\frac{1}{2}\) and \(\mathrm{P}\left(\mathrm{E}_{2}\right)=\frac{3}{6}=\frac{1}{2}\)
Clearly, \(E_{1}\) and \(E_{2}\) are mutually exclusive and exhaustive.
Now, \(\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{E}_{1}}\right)=\frac{3}{7}\) and \(\mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{E}_{2}}\right)=\frac{4}{7}\)
we have to find \(\mathrm{P}\left(\frac{\mathrm{E}_{1}}{\mathrm{B}}\right)\)
\(\mathrm{P}\left(\frac{\mathrm{E}_{1}}{\mathrm{B}}\right)=\frac{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{E}_{1}}\right)}{\mathrm{P}\left(\mathrm{E}_{1}\right) \cdot \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{E}_{1}}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right) \cdot \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{E}_{2}}\right)}=\frac{\frac{1}{2} \cdot \frac{3}{7}}{\frac{1}{2} \cdot \frac{3}{7}+\frac{1}{2} \cdot \frac{4}{7}}=\frac{3}{7}\)
Hence, the correct option is (D)