The function
Question: The function $f(x)=2 x^{3}-3 x^{2}-12 x+4$, has (a) two points of local maximum (b) two points of local minimum (c) one maximum and one minimum (d) no maximum no minimum Solution: The given function is $f(x)=2 x^{3}-3 x^{2}-12 x+4$ $f(x)=2 x^{3}-3 x^{2}-12 x+4$ Differentiating both sides with respect tox, we get $f^{\prime}(x)=6 x^{2}-6 x-12$ $\Rightarrow f^{\prime}(x)=6\left(x^{2}-x-2\right)$ $\Rightarrow f^{\prime}(x)=6(x+1)(x-2)$ For maxima or minima, $f^{\prime}(x)=0$ $\Rightarrow ...
Read More →Which of the following sentences are statements?
Question: Which of the following sentences are statements? Justify (i) A triangle has three sides. (ii) 0 is a complex number. (iii) Sky is red. (iv) Every set is an infinite set. (v) 15 + 8 23. (vi) y + 9 = 7. (vii) Where is your bag? (viii) Every square is a rectangle. (ix) Sum of opposite angles of a cyclic quadrilateral is 180. (x) Sin2x + cos2 x = 0 Solution: (i) A triangle has three sides. A statement is a declarative sentence if it is either true or false but not both. So, the given sente...
Read More →Find the point on the y-axis which is equidistant from the points
Question: Find the point on the y-axis which is equidistant from the points A(3, 1, 2) and B(5, 5, 2). Solution: Consider, C(0,y,0) point which lies on y axis and is equidistant from points A(3, 1, 2) and B(5, 5, 2). AC = BC $\sqrt{(0-3)^{2}+(y-1)^{2}+(0-2)^{2}}=\sqrt{(0-5)^{2}+(y-5)^{2}+(0-2)^{2}}$ Squaring both sides, $(0-3)^{2}+(y-1)^{2}+(0-2)^{2}=(0-5)^{2}+(y-5)^{2}+(0-2)^{2}$ $9+y^{2}-2 y+1+4=25+y^{2}-10 y+25+4$ 8y = 40 Y = 5 The point C is (0,5,0)....
Read More →Find the equation of the curve formed by the set of all points which are
Question: Find the equation of the curve formed by the set of all points which are equidistant from the points A(-1, 2, 3) and B(3, 2, 1). Solution: Consider, $C(x, y, z)$ point equidistant from points $A(-1,2,3)$ and $B(3,2,1)$. AC = BC $\sqrt{(x+1)^{2}+(y-2)^{2}+(z-3)^{2}}=\sqrt{(x-3)^{2}+(y-2)^{2}+(z-1)^{2}}$ Squaring both sides, $(x+1)^{2}+(y-2)^{2}+(z-3)^{2}=(x-3)^{2}+(y-2)^{2}+(z-1)^{2}$ $x^{2}+2 x+1+y^{2}-4 y+4+z^{2}-6 z+9=x^{2}-6 x+9+y^{2}-4 y+4+z^{2}-2 z+1$ $8 x-4 z=0$ Equation of curve...
Read More →Maximum slope of the curve
Question: Maximum slope of the curve $y=-x^{3}+3 x^{2}+9 x-27$ is (a) 0 (b) 12 (c) 16 (d) 32 Solution: The given curve is $y=-x^{3}+3 x^{2}+9 x-27$. Slope of the curve, $m=\frac{d y}{d x}$ $\therefore m=\frac{d y}{d x}=-3 x^{2}+6 x+9$ $\Rightarrow \frac{d m}{d x}=-6 x+6$ For maxima or minima, $\frac{d m}{d x}=0$ $\Rightarrow-6 x+6=0$ $\Rightarrow x=1$ Now, $\frac{d^{2} m}{d x^{2}}=-60$ ⇒x = 1 is the point of local maximum So, the slope of the given curve is maximum whenx= 1. Maximum value of the...
Read More →Maximum slope of the curve
Question: Maximum slope of the curve $y=-x^{3}+3 x^{2}+9 x-27$ is (a) 0 (b) 12 (c) 16 (d) 32 Solution: The given curve is $y=-x^{3}+3 x^{2}+9 x-27$. Slope of the curve, $m=\frac{d y}{d x}$ $\therefore m=\frac{d y}{d x}=-3 x^{2}+6 x+9$ $\Rightarrow \frac{d m}{d x}=-6 x+6$ For maxima or minima, $\frac{d m}{d x}=0$ $\Rightarrow-6 x+6=0$ $\Rightarrow x=1$ Now, $\frac{d^{2} m}{d x^{2}}=-60$ ⇒x = 1 is the point of local maximum So, the slope of the given curve is maximum whenx= 1. Maximum value of the...
Read More →Show that the following points are collinear :
Question: Show that the following points are collinear : $P(3,-2,4), Q(1,1,1)$ and $R(-1,4,2)$ Solution: To prove: the 3 points are collinear Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$ Here, Vertices should be $R(-1,4,-2)$ The solution according is $\left(x_{1}, y_{1}, z_{1}\right)=(3,-2,4)$ $\left(x_{2}, y_...
Read More →x2 sin x + cos2x
Question: x2 sin x + cos2x Solution: Applying product rule of differentiation for given equation That is $\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{t} \cdot \mathrm{y})=\mathrm{y} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$ $\Rightarrow y=x^{2} \sin x+\cos 2 x$ $\Rightarrow \frac{d y}{d x}=\sin x \frac{d}{d x}\left(x^{2}\right)+x^{2} \frac{d}{d x}(\sin x)+\frac{d}{d x}(\cos 2 x)$ On differentiating we get $\Rightarrow \frac{d y}{d x}=\sin x(2 ...
Read More →Show that the following points are collinear :
Question: Show that the following points are collinear : $A(3,-5,1), B(-1,0,8)$ and $C(7,-10,-6)$ Solution: To prove: the 3 points are collinear. Formula: The distance between two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{Z}_{1}\right)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{Z}_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$ Here, $\left(x_{1}, y_{1}, z_{1}\right)=(3,-5,1)$ $\left(x_{2}, y_...
Read More →(2x – 7)2 (3x + 5)3
Question: (2x 7)2(3x + 5)3 Solution: Given $y=(2 x-7)^{2}(3 x+5)^{3}$ Applying product rule of differentiation that is $\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{t} \cdot \mathrm{y})=\mathrm{y} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$ $\Rightarrow y=(2 x-7)^{2}(3 x+5)^{3}$ $\Rightarrow \frac{d y}{d x}=(3 x+5)^{3} \frac{d}{d x}(2 x-7)^{2}+(2 x-7)^{2} \frac{d}{d x}(3 x+5)^{3}$ On differentiating we get $\Rightarrow \frac{d y}{d x}=(2)(3 x+5)^...
Read More →The function
Question: The function $f(x)=x^{x}$ has a stationary point at (a) $x=e$ (b) $x=\frac{1}{e}$ (c) $x=1$ (d) $x=\sqrt{e}$ Solution: The values of $x$ for which $f^{\prime}(x)=0$ are called stationary values. Let $f(x)=x^{x}$. $f(x)=x^{x}$ $\Rightarrow \log f(x)=\log x^{x}$ $\Rightarrow \log f(x)=x \log x$ Differentiating both sides with respect tox, we get $\frac{1}{f(x)} \times f^{\prime}(x)=x \times \frac{1}{x}+\log x \times 1$ $\Rightarrow f^{\prime}(x)=x^{x}(1+\log x)$ For stationary point, we ...
Read More →The function
Question: The function $f(x)=x^{x}$ has a stationary point at (a) $x=e$ (b) $x=\frac{1}{e}$ (c) $x=1$ (d) $x=\sqrt{e}$ Solution: The values of $x$ for which $f^{\prime}(x)=0$ are called stationary values. Let $f(x)=x^{x}$. $f(x)=x^{x}$ $\Rightarrow \log f(x)=\log x^{x}$ $\Rightarrow \log f(x)=x \log x$ Differentiating both sides with respect tox, we get $\frac{1}{f(x)} \times f^{\prime}(x)=x \times \frac{1}{x}+\log x \times 1$ $\Rightarrow f^{\prime}(x)=x^{x}(1+\log x)$ For stationary point, we ...
Read More →The function
Question: The function $f(x)=x^{x}$ has a stationary point at (a) $x=e$ (b) $x=\frac{1}{e}$ (c) $x=1$ (d) $x=\sqrt{e}$ Solution: The values of $x$ for which $f^{\prime}(x)=0$ are called stationary values. Let $f(x)=x^{x}$. $f(x)=x^{x}$ $\Rightarrow \log f(x)=\log x^{x}$ $\Rightarrow \log f(x)=x \log x$ Differentiating both sides with respect tox, we get $\frac{1}{f(x)} \times f^{\prime}(x)=x \times \frac{1}{x}+\log x \times 1$ $\Rightarrow f^{\prime}(x)=x^{x}(1+\log x)$ For stationary point, we ...
Read More →Show that the following points are collinear :
Question: Show that the following points are collinear : $A(-2,3,5), B(1,2,3)$ and $C(7,0,-1)$ Solution: To prove: the 3 points are collinear. Formula: The distance between two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{Z}_{1}\right)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$ Here, $\left(x_{1}, y_{1}, z_{1}\right)=(-2,3,5)$ $\left(x_{2}, y_{2}...
Read More →The maximum value
Question: The maximum value of $\left(\frac{1}{x}\right)^{x}$ is (a) $e$ (b) $e^{e}$ (c) $e^{1 / e}$ (d) $\left(\frac{1}{e}\right)^{1 / e}$ Solution: Let $f(x)=\left(\frac{1}{x}\right)^{x}$. $f(x)=\left(\frac{1}{x}\right)^{x}$ $\Rightarrow \log f(x)=\log \left(\frac{1}{x}\right)^{x}$ $\Rightarrow \log f(x)=x \log \left(\frac{1}{x}\right)$ $\Rightarrow \log f(x)=-x \log x$ Differentiating both sides with respect tox, we get $\frac{1}{f(x)} \times f^{\prime}(x)=-\left(x \times \frac{1}{x}+\log x \...
Read More →The maximum value
Question: The maximum value of $\left(\frac{1}{x}\right)^{x}$ is (a) $e$ (b) $e^{e}$ (c) $e^{1 / e}$ (d) $\left(\frac{1}{e}\right)^{1 / e}$ Solution: Let $f(x)=\left(\frac{1}{x}\right)^{x}$. $f(x)=\left(\frac{1}{x}\right)^{x}$ $\Rightarrow \log f(x)=\log \left(\frac{1}{x}\right)^{x}$ $\Rightarrow \log f(x)=x \log \left(\frac{1}{x}\right)$ $\Rightarrow \log f(x)=-x \log x$ Differentiating both sides with respect tox, we get $\frac{1}{f(x)} \times f^{\prime}(x)=-\left(x \times \frac{1}{x}+\log x \...
Read More →Find the value
Question: Show that D(-1, 4, -3) is the circumcentre of triangle ABC with vertices A(3, 2, -5), B(-3. 8, -5) and C(-3, 2, 1). Solution: To prove: D is circumcenter of triangle ABC Let us consider D as circumcenter of triangle ABC AD = BC = CD. Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_...
Read More →(ax2 + cot x) (p + q cos x)
Question: (ax2 + cot x) (p + q cos x) Solution: Given $y=\left(a x^{2}+\cot x\right)(p+q \cos x)$ Applying product rule of differentiation that $\Rightarrow \frac{d}{d x}(t . y)=y \cdot \frac{d t}{d x}+t \cdot \frac{d y}{d x}$ $\Rightarrow y=\left(a x^{2}+\cot x\right)(p+q \cos x)$ Now splitting the differentials, $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=(\mathrm{p}+\mathrm{q} \cos \mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\cot \mathrm{x}\right)+\left(\mathrm{ax}^{2}+\c...
Read More →Show that the points P
Question: Show that the points P(1, 3, 4), Q(-1, 6, 10), R(-7, 4, 7) and S(-5, 1, 1) are the vertices of a rhombus. Solution: To prove: Points P, Q, R, S forms rhombus. Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by $\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$ Here, $\left(x_{1}, y_{1}, z_{1}\right)=(1,...
Read More →The maximum value of sin x cos x is
Question: The maximum value of sinxcosxis Solution: Let $f(x)=\sin x \cos x$ $\Rightarrow f(x)=\frac{1}{2} \sin 2 x$ Differentiating both sides with respect to $x$, we get $f^{\prime}(x)=\frac{1}{2} \cos 2 x \times 2=\cos 2 x$ For maxima or minima, $f^{\prime}(x)=0$ $\Rightarrow \cos 2 x=0$ $\Rightarrow 2 x=\frac{\pi}{2}$ $\Rightarrow x=\frac{\pi}{4}$ Now, $f^{\prime \prime}(x)=-2 \sin 2 x$ $\Rightarrow f^{\prime \prime}\left(\frac{\pi}{4}\right)=-2 \sin \left(2 \times \frac{\pi}{4}\right)=-2 \s...
Read More →The maximum value of sin x cos x is
Question: The maximum value of sinxcosxis Solution: Let $f(x)=\sin x \cos x$ $\Rightarrow f(x)=\frac{1}{2} \sin 2 x$ Differentiating both sides with respect to $x$, we get $f^{\prime}(x)=\frac{1}{2} \cos 2 x \times 2=\cos 2 x$ For maxima or minima, $f^{\prime}(x)=0$ $\Rightarrow \cos 2 x=0$ $\Rightarrow 2 x=\frac{\pi}{2}$ $\Rightarrow x=\frac{\pi}{4}$ Now, $f^{\prime \prime}(x)=-2 \sin 2 x$ $\Rightarrow f^{\prime \prime}\left(\frac{\pi}{4}\right)=-2 \sin \left(2 \times \frac{\pi}{4}\right)=-2 \s...
Read More →If the solve the problem
Question: If $x$ is real, the minimum value $x^{2}-8 x+17$ is (a) $-1$ (b) 0 (c) 1 (d) 2 Solution: Let $f(x)=x^{2}-8 x+17$ Differentiating both sides with respect to $x$, we get $f^{\prime}(x)=2 x-8$ $f^{\prime}(x)=0$ $\Rightarrow 2 x-8=0$ $\Rightarrow x=4$ Now, $f^{\prime \prime}(x)=20$ So,x= 4 is the point of local minimum off(x). $\therefore$ Minimum value of $f(x)=f(4)=(4)^{2}-8 \times 4+17=16-32+17=1$ Thus, the minimum value of $x^{2}-8 x+17$ is 1 . Hence, the correct answer is option (c)....
Read More →Let the function
Question: Let the function $f: R \rightarrow R$ be defined by $f(x)=2 x+\cos x$, then $f(x)$ (a) has a minimum at $x=\pi$ (b) has a maximum at $x=0$ (c) is a decreasing function (d) is an increasing function Solution: The given function is $f(x)=2 x+\cos x$. $f(x)=2 x+\cos x$ Differentiating both sides with respect to $x$, we get $f^{\prime}(x)=2-\sin x$ We know $-1 \leq \sin x \leq 1$ $\therefore f^{\prime}(x)=2-\sin x0 \forall x \in \mathrm{R}$ $\Rightarrow f(x)$ is an increasing function for ...
Read More →The minimum value of the function
Question: The minimum value of the function $f(x)=2 x^{3}-21 x^{2}+36 x-20$ is (a) $-128$ (b) $-126$ (c) $-120$ (d) none of these Solution: (a) $-128$ Given : $f(x)=2 x^{3}-21 x^{2}+36 x-20$ $\Rightarrow f^{\prime}(x)=6 x^{2}-42 x+36$ For a local maxima or a local minima, we must have $f^{\prime}(x)=0$ $\Rightarrow 6 x^{2}-42 x+36=0$ $\Rightarrow x^{2}-7 x+6=0$ $\Rightarrow(x-1)(x-6)=0$ $\Rightarrow x=1,6$ Now, $f^{\prime \prime}(x)=12 x-42$ $\Rightarrow f^{\prime \prime}(1)=12-42=-300$ So, $x=1...
Read More →(sec x – 1) (sec x + 1)
Question: (sec x 1) (sec x + 1) Solution: Given $(\sec x-1)(\sec x+1)$ Let $y=(\operatorname{secx}-1)(\operatorname{secx}+1)$ The above equation can be written as $\Rightarrow y=(\sec x-1)(\sec x+1)=\sec ^{2} x-1=\tan ^{2} x$ $\Rightarrow y=\tan ^{2} x$ Now applying the chain rule we get $\Rightarrow \frac{d y}{d x}=\frac{d}{d(\tan x)}\left(\tan ^{2} x\right) \cdot \frac{d}{d x}(\tan x)$ $\Rightarrow \frac{d y}{d x}=2 \tan x \sec ^{2} x$...
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