What is X in the given reaction ?
Question: What is ' $X^{\prime}$ in the given reaction ? Correct Option: 1 Solution: $\begin{aligned}\mathrm{CH}_{2}-\mathrm{OH} \\\mathrm{CH}_{2}-\mathrm{OH}\end{aligned}+$ oxalic acid $\stackrel{210^{\circ} \mathrm{C}}{\rightarrow} \mathrm{CH}_{2}=\mathrm{CH}_{2}$...
Read More →Let f:
Question: Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $f(x)= \begin{cases}-55 x, \text { if } x-5 \\ 2 x^{3}-3 x^{2}-120 x, \text { if }-5 \leq x \leq 4 \\ 2 x^{3}-3 x^{2}-36 x-336, \text { if } x4\end{cases}$ Let $A=\{x \in R: f$ is increasing $\} .$ Then A is equal to :(1) $(-5,-4) \cup(4, \infty)$ (2) $(-5, \infty)$(3) $(-\infty,-5) \cup(4, \infty)$(4) $(-\infty,-5) \cup(-4, \infty)$Correct Option: 1, Solution: $f(x)=\left\{\begin{array}{ccc}-55 ; x-5 \\ 6\left(x^{2}-x-20\right) ...
Read More →Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m
Question: Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find (i) the area of the field and (ii) the length of the perpendicular from the opposite vertex on the side measuring 154 m. Solution: (i) Let: $a=85 \mathrm{~m}$ and $b=154 \mathrm{~m}$ Given : Perimeter $=324 \mathrm{~m}$ or, $a+b+c=324$ $\Rightarrow c=324-85-154=85 \mathrm{~m}$ $\therefore s=\frac{324}{2}=162 \mathrm{~m}$ By Heron's formula, we have: Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ ...
Read More →Match List-I with List-II
Question: (1) (a) $-($ i $),($ b $)-($ iii $),($ c $)-($ iv $),($ d $)-($ ii $)$(2) $(\mathrm{a})-(\mathrm{ii}),(\mathrm{b})-(\mathrm{iv}),(\mathrm{c})-(\mathrm{iii}),(\mathrm{d})-(\mathrm{i})$(3) $(\mathrm{a})-(\mathrm{ii}),(\mathrm{b})-(\mathrm{iii}),(\mathrm{c})-(\mathrm{iv}),(\mathrm{d})-(\mathrm{i})$(4) $(\mathrm{a})-(\mathrm{ii}),(\mathrm{b})-(\mathrm{iii}),(\mathrm{c})-(\mathrm{i}),(\mathrm{d})-($ iv $)$Correct Option: 1, Solution: (d) $\tan \phi=\frac{V_{\mathrm{L}}-\mathrm{V}_{\mathrm{C...
Read More →Identify A and B in the chemical reaction.
Question: Identify $\mathrm{A}$ and $\mathrm{B}$ in the chemical reaction. Correct Option: , 4 Solution:...
Read More →The minimum value of
Question: The minimum value of $\alpha$ for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha$ has at least one solution in $\left(0, \frac{\pi}{2}\right)$ is Solution: $f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$ Let $\sin x=t \quad \because x \in\left(0, \frac{\pi}{2}\right) \Rightarrow 0t1$ $f(t)=\frac{4}{t}+\frac{1}{1-t}$ $f^{\prime}(t)=\frac{-4}{t^{2}}+\frac{1}{(1-t)^{2}}$ $=\frac{t^{2}-4(1-t)^{2}}{t^{2}(1-t)^{2}}$ $=\frac{(t-2(1-t))(t+2(1-t))}{t^{2}(1-t)^{2}}$ $=\frac{(3 t-2)(2-...
Read More →The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field.
Question: The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the field. Also, find the cost of ploughing the field at₹ 5 per m2. Solution: Let the sides of the triangle be 25xm, 17xm and 12xm.We know:Perimeter = Sum of all sidesor, 540 = 25x+ 17x+ 12xor, 54x= 540or,x= 10Thus, we obtain the sides of the triangle.2510 = 250 m1710 = 170 m1210 = 120 m Now, Let: $a=250 \mathrm{~m}, b=170 \mathrm{~m}$ and $c=120 \mathrm{~m}$ $\therefore s=\frac{5...
Read More →The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m.
Question: The sides of a triangle are in the ratio 5 : 12 : 13 and its perimeter is 150 m. Find the area of the triangle. Solution: Let the sides of the triangle be 5xm, 12xm and 13xm.We know:Perimeter = Sum of all sidesor, 150 = 5x+ 12x+ 13xor, 30x= 150or,x= 5Thus, we obtain the sides of the triangle.55 = 25 m125 = 60 m135 = 65 m Now, Let: $a=25 \mathrm{~m}, b=60 \mathrm{~m}$ and $c=65 \mathrm{~m}$ $\therefore s=\frac{150}{2}=75 \mathrm{~m}$ By Heron's formula, we have : Area of triangle $=\sqr...
Read More →Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length.
Question: Find the area of a triangular field whose sides are 91 cm, 98 m and 105 m in length. Find the height corresponding to the longest side. Solution: Let : $a=91 \mathrm{~m}, b=98 \mathrm{~m}$ and $c=105 \mathrm{~m}$ $\therefore s=\frac{a+b+c}{2}=\frac{91+98+105}{2}=147 \mathrm{~m}$ By Heron's formula, we have : Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{147(147-91)(147-98)(147-105)}$ $=\sqrt{147 \times 56 \times 49 \times 42}$ $=\sqrt{7 \times 3 \times 7 \times 2 \times 2 \times ...
Read More →Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length.
Question: Calculate the area of the triangle whose sides are 18 cm, 24 cm and 30 cm in length. Also, find the length of the altitude corresponding to the smallest side. Solution: Let : $a=18 \mathrm{~cm}, b=24 \mathrm{~cm}$ and $c=30 \mathrm{~cm}$ $\therefore s=\frac{a+b+c}{2}=\frac{18+24+30}{2}=36 \mathrm{~cm}$ By Heron's formula, we have : Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{36(36-18)(36-24)(36-30)}$ $=\sqrt{36 \times 18 \times 12 \times 6}$ $=\sqrt{12 \times 3 \times 6 \times ...
Read More →Which of the following compound gives pink colour on reaction with phthalic anhydride in conc.
Question: Which of the following compound gives pink colour on reaction with phthalic anhydride in conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ followed by treatment with $\mathrm{NaOH}$ ? Correct Option: , 2 Solution:...
Read More →If the tangent to the curve $y=x^{3}$ at the point $Pleft(t, t^{3}ight)$ meets the curve again at $Q$, then the ordinate of the point which divides $mathrm{PQ}$ internally in the ratio $1: 2$ is :
Question: If the tangent to the curve $y=x^{3}$ at the point $P\left(t, t^{3}\right)$ meets the curve again at $Q$, then the ordinate of the point which divides $\mathrm{PQ}$ internally in the ratio $1: 2$ is :(1) $-2 t^{3}$ (2) $-\mathrm{t}^{3}$(3) 0(4) $2 t^{3}$Correct Option: 1, Solution: Equation of tangent at $\mathrm{P}\left(\mathrm{t}, \mathrm{t}^{3}\right)$ $\left(y-t^{3}\right)=3 t^{2}(x-t) \ldots$ (i) Now solve the above equation with $y=x^{3} \ldots$ (ii) By (i) and (ii), $x^{3}-t^{3}...
Read More →Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length.
Question: Find the area of the triangle whose sides are 42 cm, 34 cm and 20 cm in length. Hence, find the height corresponding to the longest side. Solution: Let : $a=42 \mathrm{~cm}, b=34 \mathrm{~cm}$ and $c=20 \mathrm{~cm}$ $\therefore s=\frac{a+b+c}{2}=\frac{42+34+20}{2}=48 \mathrm{~cm}$ By Heron's formula, we have : Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{48(48-42)(48-34)(48-20)}$ $=\sqrt{48 \times 6 \times 14 \times 28}$ $=\sqrt{4 \times 2 \times 6 \times 6 \times 7 \times 2 \t...
Read More →The base of a triangular field is three times its altitude.
Question: The base of a triangular field is three times its altitude. If the cost of sowing the field at Rs 58 per hectare is Rs 783, find its base and height. Solution: Let the height of the triangle behm. Base = 3hmNow, Area of the triangle $=\frac{\text { Total Cost }}{\text { Rate }}=\frac{783}{58}=13.5 \mathrm{ha}=135000 \mathrm{~m}^{2}$ We have: Area of triangle $=135000 \mathrm{~m}^{2}$ $\Rightarrow \frac{1}{2} \times$ Base $\times$ Heigh $t=135000$ $\Rightarrow \frac{1}{2} \times 3 h \ti...
Read More →The function
Question: The function $f(x)=\frac{4 x^{3}-3 x^{2}}{6}-2 \sin x+(2 x-1) \cos x:$(1) increases in $\left[\frac{1}{2}, \infty\right)$(2) decreases $\left(-\infty, \frac{1}{2}\right]$(3) increases in $\left(-\infty, \frac{1}{2}\right]$(4) decreases $\left[\frac{1}{2}, \infty\right)$Correct Option: 1, Solution: $f^{\prime}(x)=(2 x-1)(x-\sin x)$ $\Rightarrow f^{\prime}(x) \geq 0$ in $x \in\left[\frac{1}{2}, \infty\right)$...
Read More →Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm.
Question: Find the area of the triangle whose base measures 24 cm and the corresponding height measures 14.5 cm. Solution: We have:Base = 24 cmHeight = 14.5 cm Now, Area of triangle $=\frac{1}{2} \times$ Base $\times$ Height $=\frac{1}{2} \times 24 \times 14.5$ $=174 \mathrm{~cm}^{2}$...
Read More →Which is the final product (major) ' $A$ ' in the given reaction?
Question: Which is the final product (major) ' $A$ ' in the given reaction? Correct Option: , 3 Solution:...
Read More →be a real polynomial of degree 3 which vanishes at
Question: Let $\mathrm{P}(\mathrm{x})$ be a real polynomial of degree 3 which vanishes at $x=-3$. Let $\mathrm{P}(\mathrm{x})$ have local minima at $x=1$, local maxima at $x=-1$ and $\int_{-1}^{1} \mathrm{P}(\mathrm{x}) \mathrm{d} \mathrm{x}=18$, then the sum of all the coefficients of the polynomial $\mathrm{P}(\mathrm{x})$ is equal to Solution: Let $p^{\prime}(x)=a(x-1)(x+1)=a\left(x^{2}-1\right)$ $p(x)=a \int\left(x^{2}-1\right) d x+c$ $=a\left(\frac{x^{3}}{3}-x\right)+c$ Now $p(-3)=0$ $\Righ...
Read More →Let a tangent be drawn to the ellipse
Question: Let a tangent be drawn to the ellipse $\frac{x^{2}}{27}+y^{2}=1$ at $(3 \sqrt{3} \cos \theta, \sin \theta)$ where $\theta \in\left(0, \frac{\pi}{2}\right)$. Then the value of $\theta$ such that the sum of intercepts on axes made by this tangent is minimum is equal to: (1) $\frac{\pi}{8}$ (2) $\frac{\pi}{4}$(3) $\frac{\pi}{6}$(4) $\frac{\pi}{3}$Correct Option: 3, Solution: Equation of tangent be $\frac{x \cos \theta}{3 \sqrt{3}}+\frac{y \cdot \sin \theta}{1}=1, \quad \theta \in\left(0, ...
Read More →Solve this
Question: An $\mathrm{AC}$ current is given by $\mathrm{I}=\mathrm{I}_{1} \sin \omega \mathrm{t}+\mathrm{I}_{2} \cos \omega \mathrm{t}$. A hot wire ammeter will give a reading :(1) $\sqrt{\frac{\frac{I_{1}^{2}-I_{2}^{2}}{2}}{1}}$(2) $\sqrt{\frac{\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}}{2}}$(3) $\frac{\mathrm{I}_{1}+\mathrm{I}_{2}}{\sqrt{2}}$(4) $\frac{\mathrm{I}_{1}+\mathrm{I}_{2}}{2 \sqrt{2}}$Correct Option: 2, Solution: $\mathrm{I}=\mathrm{I}_{1} \sin \omega \mathrm{t}+\mathrm{I}_{2} \cos \omega...
Read More →Main Products formed during a reaction of 1-methoxy naphthalene with hydroiodic acid are:
Question: Main Products formed during a reaction of 1-methoxy naphthalene with hydroiodic acid are:Correct Option: , 2 Solution:...
Read More →Construct a ∆ABC in which ∠B = 45°, ∠C = 60° and the perpendicular from the vertex A to base BC is 4.5 cm.
Question: Construct a ∆ABCin which B= 45, C= 60 and the perpendicular from the vertexAto baseBCis 4.5 cm. Solution: Steps of construction:1. Draw a line segmentXY.2. Take a point Oon XYand draw POXY.3. Along PO, set off OA = 4.5 cm. 4. Draw a line LM $\| X Y$. 5. Draw $\angle L A B=45^{\circ}$ and $\angle M A C=60^{\circ}$, meeting $X Y$ at $B$ and $C$, respectively. Thus, ABC is the required triangle....
Read More →Construct a right triangle whose one side is 3.5 cm and the sum of the other side and the hypotenuse is 5.5 cm.
Question: Construct a right triangle whose one side is 3.5 cm and the sum of the other side and the hypotenuse is 5.5 cm. Solution: Steps of construction:1. Draw a line segment BC = 3.5 cm 2. Construct $\angle \mathrm{CBX}=90$ 3. With B as centre and 5.5 cm radius cut an arc on BX and name it as D.4. Join CD.5. Construct the perpendicular bisector of CD intersecting BD at point A. 6. Join AC. $\triangle \mathrm{ABC}$ is the required triangle....
Read More →Construct a square of side 4 cm.
Question: Construct a square of side 4 cm. Solution: Steps of construction:1. Draw a line segmentAB= 4 cm. 2. Construct $\angle B A X=90^{\circ}$ and $\angle A B Y=90^{\circ}$. 3. Set offAD= 4 cm andBC= 4 cm.4. JoinDC. Thus, $\square A B C D$ is the required square....
Read More →Considering the above chemical reaction,
Question: Considering the above chemical reaction, identify the product "X" :Correct Option: , 3 Solution:...
Read More →