Find the absolute maximum and minimum values of the function of given by
Question: Find the absolute maximum and minimum values of the function of given by $f(x)=\cos ^{2} x+\sin x, x \in[0, \pi]$ Solution: Given : $f(x)=\cos ^{2} x+\sin x$ $\Rightarrow f^{\prime}(x)=2 \cos x(-\sin x)+\cos x=-2 \sin x \cos x+\cos x$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow-2 \sin x \cos x+\cos x=0$ $\Rightarrow \cos x(2 \sin x-1)=0$ $\Rightarrow \sin x=\frac{1}{2}$ or $\cos x=0$ $\Rightarrow x=\frac{\pi}{6}$ or $\frac{\pi}{2}$ $[\because x \...
Read More →Find the maximum value
Question: Find the maximum value of $2 x^{3}-24 x+107$ in the interval $[1,3]$. Find the maximum value of the same function in $[-3,-1]$. Solution: Given : $f(x)=2 x^{3}-24 x+107$ $\Rightarrow f^{\prime}(x)=6 x^{2}-24$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow 6 x^{2}-24=0$ $\Rightarrow 6 x^{2}=24$ $\Rightarrow x^{2}=4$ $\Rightarrow x=\pm 2$ Thus, the critical points of $f$ in the interval $[1,3]$ are 1,2 and 3 . Now, $f(1)=2(1)^{3}-24(1)+107=85$ $f(2)=2...
Read More →Find the points on the line x + y = 4
Question: Find the points on the line x + y = 4 which lie at a unit distance from the line 4x + 3y = 10. Solution: Let (x1, y1) be any point lying in the equation x+ y = 4 x1+ y1= 4 ..1 Distance of the point (x1, y1) from the equation 4x + 3y = 10 Let $\left(x_{1}, y_{1}\right)$ be any point lying in the equation $x+y=4$ $\therefore \mathrm{x}_{1}+\mathrm{y}_{1}=4 \ldots .1$ Distance of the point $\left(x_{1}, y_{1}\right)$ from the equation $4 x+3 y=10$ $d=\frac{\left|A x_{0}+B y_{0}+C\right|}{...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : $x^{2}=-18 y$ Solution: Given equation : $x^{2}=-18 y$ Comparing given equation with parabola having equation, $x^{2}=-4 a y$ $4 a=18$ $\cdot a=\frac{9}{2}$ Focus: $F(0,-a)=F\left(0,-\frac{9}{2}\right)$ Vertex: $A(0,0)=A(0,0)$ Equation of the directrix: $y-a=0$ $y-\frac{9}{2}=0$ $y=\frac{9}{2}$ Lenth of latusrectum : $4 \mathrm{a}=18$...
Read More →If the solve the problem
Question: (i) $f(x)=4 x-\frac{x^{2}}{2}$ in $[-2,4,5]$ (ii) $f(x)=(x-1)^{2}+3$ in $[-3,1]$ (iii) $f(x)=3 x 4-8 x^{3}+12 x^{2}-48 x+25$ in $[0,3]$ (iv) $f(x)=(x-2) \sqrt{x-1}$ in $[1,9]$ Solution: (i) Given: $f(x)=4 x-\frac{x^{2}}{2}$ $\Rightarrow f^{\prime}(x)=4-x$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow 4-x=0$ $\Rightarrow x=4$ Thus, the critical points of $f$ are $-2,4$ and $4.5$. Now, $f(-2)=4(-2)-\frac{(-2)^{2}}{2}=-8-2=-10$ $f(4)=4(4)-\frac{(4)^{2...
Read More →If the solve the problem
Question: (i) $f(x)=4 x-\frac{x^{2}}{2}$ in $[-2,4,5]$ (ii) $f(x)=(x-1)^{2}+3$ in $[-3,1]$ (iii) $f(x)=3 x 4-8 x^{3}+12 x^{2}-48 x+25$ in $[0,3]$ (iv) $f(x)=(x-2) \sqrt{x-1}$ in $[1,9]$ Solution: (i) Given: $f(x)=4 x-\frac{x^{2}}{2}$ $\Rightarrow f^{\prime}(x)=4-x$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow 4-x=0$ $\Rightarrow x=4$ Thus, the critical points of $f$ are $-2,4$ and $4.5$. Now, $f(-2)=4(-2)-\frac{(-2)^{2}}{2}=-8-2=-10$ $f(4)=4(4)-\frac{(4)^{2...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : $x^{2}=-8 y$ Solution: Given equation : $x^{2}=-8 y$ Comparing given equation with parabola having equation, $x^{2}=-4 a y$ $4 a=8$ $\cdot a=2$ Focus: $F(0,-a)=F(0,-2)$ Vertex : $A(0,0)=A(0,0)$ Equation of the directrix : $y-a=0$ - $y-2=0$ - $y=2$ Lenth of latusrectum : $4 a=8$...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : $3 x^{2}=8 y$ Solution: Given equation : $3 x^{2}=8 y$ $x^{2}=\frac{8}{3} y$ Comparing the given equation with parabola having an equation, $x^{2}=4 a y$ $\cdot 4 a=\frac{8}{3}$ - $a=\frac{2}{3}$ Focus: $F(0, a)=F\left(0, \frac{2}{3}\right)$ Vertex : $A(0,0)=A(0,0)$ Equation of the directrix: $y+a=0$ $y+\frac{2}{3}=0$ $y=-\frac{2}{3}$ Lenth of ...
Read More →Find the equation of the lines
Question: Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14. Solution: The equation of line in intercept form is $\frac{x}{a}+\frac{y}{b}=1$ Where $\mathrm{a}$ and $\mathrm{b}$ are the intercepts on the axis. Given that $\mathrm{a}+\mathrm{b}=14$ The above equation can be written as $\Rightarrow \mathrm{b}=14-\mathrm{a}$ Substituting the value of $a$ and $b$ in equation 1 we get So, equation of line is ...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : $x^{2}=10 y$ Solution: Given equation : $x^{2}=10 y$ Comparing given equation with parabola having equation $x^{2}=4 a y$ $4 a=10$ $\cdot a=2.5$ Focus : $F(0, a)=F(0,2.5)$ Vertex : $A(0,0)=A(0,0)$ Equation of the directrix : $y+a=0$ $-y+2.5=0$ $\cdot y=-2.5$ Lenth of latusrectum : 4a = 10...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : $x^{2}=16 y$ Solution: Given equation : $x^{2}=16 y$ Comparing given equation with parabola having equation, $x^{2}=4 a y$ $4 a=16$ $\cdot a=4$ Focus: $\mathrm{F}(0, \mathrm{a})=\mathrm{F}(0,4)$ Vertex : $A(0,0)=A(0,0)$ Equation of the directrix : $y+a=0$ $-y+4=0$ $\cdot y=-4$ Lenth of latusrectum : 4a = 16...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : $5 y^{2}=-16 x$ Solution: Given equation : $5 y^{2}=-16 x$ $y^{2}=-\frac{16}{5} x$ Comparing the given equation with parabola having an equation, $y^{2}=-4 a x$ $\cdot 4 a=\frac{16}{5}$ $\cdot a=\frac{4}{5}$ Focus: $F(-a, 0)$ $=F\left(-\frac{4}{5}, 0\right)$ Vertex : $\mathrm{A}(0,0)=\mathrm{A}(0,0)$ Equation of the directrix : $x-a=0$ $x-\frac...
Read More →Find the angle between the lines
Question: Find the angle between the lines $\mathrm{y}=(2-\sqrt{3})(\mathrm{x}+5)$ and $\mathrm{y}=(2+\sqrt{3})(\mathrm{x}-7)$ Solution: Given equations are $y=(2-\sqrt{3})(x+5)$ and $(2+\sqrt{3})(x-7)$ The given equation can be written as $\Rightarrow y=(2-\sqrt{3}) x+(2-\sqrt{3}) 5 \ldots \ldots 1$ $\Rightarrow y=(2+\sqrt{3}) x-7(2+\sqrt{3}) \ldots \ldots .2$ Now, we have to find the slope of equation 1 Since, the equation 1 is in $y=m x+b$ form, we can easily see that the slope $\left(m_{1}\r...
Read More →Find the equation of the line passing
Question: Find the equation of the line passing through the point (5, 2) and perpendicular to the line joining the points (2, 3) and (3, 1). Solution: Given points are A (5, 2), B (2, 3) and C (3, -1) Firstly, we find the slope of the line joining the points $(2,3)$ and $(3,-1)$ Slope of the line joining two points $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ $\therefore \mathrm{m}_{\mathrm{BC}}=\frac{-1-3}{3-2}=-\frac{4}{1}=-4$ It is given that line passing through the point $(5,2)$ is perpendicular to $...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : $y^{2}=-6 x$ Solution: Given equation : $y^{2}=-6 x$ Comparing given equation with parabola having equation, $y^{2}=-4 a x$ $4 a=6$ $\cdot a=\frac{3}{2}$ Focus: $F(-a, 0)=F\left(-\frac{3}{2}, 0\right)$ Vertex: $A(0,0)=A(0,0)$ Equation of the directrix : $x-a=0$ $x-\frac{3}{2}=0$ $\mathrm{X}=\frac{3}{2}$ Lenth of latusrectum : $4 a=6$...
Read More →Find the equation of the straight line
Question: Find the equation of the straight line which passes through the point (1, 2) and cuts off equal intercepts from axes. Solution: The equation of line in intercept form is $\frac{x}{a}+\frac{y}{b}=1$ Where $a$ and $b$ are the intercepts on the axis. Given that $a=b$ $\Rightarrow \frac{x}{a}+\frac{y}{a}=1$ The above equation can be written as $\Rightarrow \frac{x+y}{a}=1$ On cross multiplication we get $\Rightarrow x+y=a \ldots .1$ If equation 1 passes through the point $(1,-2)$, we get $...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola : $y^{2}=-8 x$ Solution: Given equation : $y^{2}=-8 x$ Comparing given equation with parabola having equation, $y^{2}=-4 a x$ $4 a=8$ - $a=2$ Focus: $F(-a, 0)=F(-2,0)$ Vertex: $A(0,0)=A(0,0)$ Equation of the directrix: $x-a=0$ $\cdot x-2=0$ $\cdot x=2$ Lenth of latusrectum : $4 \mathrm{a}=8$...
Read More →If the solve the problem
Question: (i) $f(x)=4 x-\frac{x^{2}}{2}$ in $[-2,4,5]$ (ii) $f(x)=(x-1)^{2}+3$ in $[-3,1]$ (iii) $f(x)=3 x 4-8 x^{3}+12 x^{2}-48 x+25$ in $[0,3]$ (iv) $f(x)=(x-2) \sqrt{x-1}$ in $[1,9]$ Solution: (i) Given: $f(x)=4 x-\frac{x^{2}}{2}$ $\Rightarrow f^{\prime}(x)=4-x$ For a local maximum or a local minimum, we must have $f^{\prime}(x)=0$ $\Rightarrow 4-x=0$ $\Rightarrow x=4$ Thus, the critical points of $f$ are $-2,4$ and $4.5$. Now, $f(-2)=4(-2)-\frac{(-2)^{2}}{2}=-8-2=-10$ $f(4)=4(4)-\frac{(4)^{2...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: $3 y^{2}=8 x$ Solution: Given equation : $3 y^{2}=8 x$ $y^{2}=\frac{8}{3} x$ Comparing the given equation with parabola having equation, $y^{2}=4 a x$ $4 a=\frac{8}{3}$ - $a=\frac{2}{3}$ Focus: $F(a, 0)=F\left(\frac{2}{3}, 0\right)$ Vertex : $A(0,0)=A(0,0)$ Equation of the directrix : $x+a=0$ $x+\frac{2}{3}=0$ $\mathrm{x}=-\frac{2}{3}$ Lenth of ...
Read More →Prove that
Question: Prove that $f(x)=\sin x+\sqrt{3} \cos x$ has maximum value at $x=\frac{\pi}{6}$. Solution: We have, $f(x)=\sin x+\sqrt{3} \cos x$ $\Rightarrow f^{\prime}(x)=\cos x+\sqrt{3}(-\sin x)$ $\Rightarrow f^{\prime}(x)=\cos x-\sqrt{3} \sin x$ For $f(x)$ to have maximum or minimum value, we must have $f^{\prime}(x)=0$ $\Rightarrow \cos x-\sqrt{3} \sin x=0$ $\Rightarrow \cos x=\sqrt{3} \sin x$ $\Rightarrow \cot x=\sqrt{3}$ $\Rightarrow x=\frac{\pi}{6}$ Also, $f^{\prime \prime}(x)=-\sin x-\sqrt{3}...
Read More →If the solve the problem
Question: If $f(x)=x^{3}+a x^{2}+b x+c$ has a maximum at $x=-1$ and minimum at $x=3$. Determine $a, b$ and $c$. Solution: We have, $f(x)=x^{3}+a x^{2}+b x+c$ $\Rightarrow f^{\prime}(x)=3 x^{2}+2 a x+b$ As, $f(x)$ is maximum at $x=-1$ and minimum at $x=3$. So, $f(-1)=0$ and $f(3)=0$ $\Rightarrow 3(-1)^{2}+2 a(-1)+b=0$ and $3(3)^{2}+2 a(3)+b=0$ $\Rightarrow 3-2 a+b=0 \quad \ldots$ (i) and $27+6 a+b=0 \quad \ldots$ (ii) (ii) - (i), we get $27-3+6 a+2 a=0$ $\Rightarrow 8 a=-24$ $\Rightarrow a=-3$ Su...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: $y^{2}=10 x$ Solution: Given equation: $\mathrm{y}^{2}=10 \mathrm{x}$ Comparing given equation with parabola having equation, $y^{2}=4 a x$ $4 a=10$ $\cdot a=2.5$ Focus : $F(a, 0)=F(2.5,0)$ Vertex : $A(0,0)=A(0,0)$ Equation of the directrix : $x+a=0$ - $x+2.5=0$ - $x=-2.5$ Lenth of latusrectum : $4 \mathrm{a}=4 \cdot(2 \cdot 5)=10$...
Read More →The third term of G.P. is 4.
Question: The third term of G.P. is 4. The product of its first 5 terms is (A) 43 (B) 44 (C) 45 (D) None of these Solution: (C) 45 Explanation: Given the third term of G.P, T3= 4 To find the product of first five terms We know that, Tn= a rn 1 It is given that, T3= 4 ⇒ar3 1= 4 ⇒ar2= 4 (i) Product of first 5 terms = a ar ar2 ar3 ar4 = a5r1+2+3+4 = a5r10 = (ar2)5 = (4)5[from (i)] Hence, the correct option is (C)...
Read More →Find the coordinates of the focus and the vertex, the equations of the
Question: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola: $y^{2}=12 x$ Solution: Given equation : $y^{2}=12 x$ Comparing given equation with parabola having equation, $y^{2}=4 a x$ $4 a=12$ - $a=3$ Focus : $F(a, 0)=F(3,0)$ Vertex : $\mathrm{A}(0,0)=\mathrm{A}(0,0)$ Equation of the directrix : x+a=0 - $x+3=0$ $x=-3$ Lenth of latusrectum : $4 \mathrm{a}=4 .(3)=12$...
Read More →If the sum of n terms of an A.P.
Question: If the sum of n terms of an A.P. is given by Sn = 3n + 2n 2, then the common difference of the A.P. is (A) 3 (B) 2 (C) 6 (D) 4 Solution: (D) 4 Explanation: To find: Common Difference of A.P that is d Consider, Sn= 3n + 2n2 Putting n = 1, we get S1= 3(1) + 2(1)2 = 3 + 2 S1= 5 Putting n = 2, we get S2= 3(2) + 2(2)2 = 6 + 2(4) = 6 + 8 S2= 14 Now, we know that, S1= a1 ⇒a1= 5 And a2= S2 S1 = 14 5 = 9 Common Difference, d = a2 a1 = 9 5 = 4 Hence, the correct option is (D)...
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