Find the least number which must be added to 6203 to obtain a perfect square.
Question: Find the least number which must be added to 6203 to obtain a perfect square. Find this perfect square and its square root. Solution: Using the long division method:Thus, to get a perfect square greater than the given number, we take the square of the next natural number of the quotient, i.e. 78. $79^{2}=6241$ Number that should be added to the aiven number to make it a perfect square $=6241-6203$ = 38 The perfect square thus obtained is 6241 and its square root is 79....
Read More →In figure, 0 is the centre of a circle of radius 5 cm,
Question: In figure, 0 is the centre of a circle of radius 5 cm, T is a point such that OT = 13 and 0T intersects the circle at E, if AB is the tangent to the circle at E, find the length of AB. Solution: Given, OT = 13 cm and OP = 5 cm Since, if we drawn a line from the centre to the tangent of the circle. It is always perpendicular to the tangent i.e., OPPT. In right angled $\triangle O P T_{1} \quad O T^{2}=O P^{2}+P T^{2}$ [by Pythagoras theorem, (hypotenuse) $\left.^{2}=(\text { base })^{2}...
Read More →Find the least number which must be subtracted from 7581 to obtain a perfect square.
Question: Find the least number which must be subtracted from 7581 to obtain a perfect square. Find this perfect square and its square root. Solution: Using the long division method:Therefore, the number that should be subtracted from the given number to makeit aperfect square is 12. Perfect square = 7581-12 = 7569 Its square root is 87....
Read More →Find the least number which must be subtracted from
Question: Find the least number which must be subtracted from 2509 to make it a perfect square. Solution: Using the long division method:Therefore, the number that should be subtracted from the given number to make it a perfect square is 9....
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Question: Evaluate: $\sqrt{92416}$ Solution: Using the long division method: $\therefore \sqrt{92416}=304$...
Read More →An example of a function which is everywhere
Question: An example of a function which is everywhere continuous but fails to be differentiable exactly at two points is ____________. Solution: Consider the function $g(x)=|x-1|+|x+1|$. $|x-1|= \begin{cases}x-1, x \geq 1 \\ -(x-1), x1\end{cases}$ $|x+1|= \begin{cases}x+1, x \geq-1 \\ -(x+1), x-1\end{cases}$ $\therefore g(x)=|x-1|+|x+1|= \begin{cases}-(x-1)-(x+1), x-1 \\ -(x-1)+x+1, -1 \leq x1 \\ x-1+x+1, x \geq 1\end{cases}$ $\Rightarrow g(x)=|x-1|+|x+1|= \begin{cases}-2 x, x-1 \\ 2, -1 \leq x...
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Question: Evaluate: $\sqrt{19600}$ Solution: Using the long division method: $\therefore \sqrt{19600}=140$...
Read More →An example of a function which is everywhere
Question: An example of a function which is everywhere continuous but fails to be differentiable exactly at two points is ____________. Solution: Consider the function $g(x)=|x-1|+|x+1|$. $|x-1|= \begin{cases}x-1, x \geq 1 \\ -(x-1), x1\end{cases}$ $|x+1|= \begin{cases}x+1, x \geq-1 \\ -(x+1), x-1\end{cases}$ $\therefore g(x)=|x-1|+|x+1|= \begin{cases}-(x-1)-(x+1), x-1 \\ -(x-1)+x+1, -1 \leq x1 \\ x-1+x+1, x \geq 1\end{cases}$ $\Rightarrow g(x)=|x-1|+|x+1|= \begin{cases}-2 x, x-1 \\ 2, -1 \leq x...
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Question: Evaluate: $\sqrt{17956}$ Solution: Using the long division method: $\therefore \sqrt{17956}=134$...
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Question: Evaluate: $\sqrt{10404}$ Solution: Using the long division method: $\therefore \sqrt{10404}=102$...
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Question: Evaluate: $\sqrt{14161}$ Solution: Using the long division method: $\therefore \sqrt{14161}=119$...
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Question: Evaluate: $\sqrt{11449}$ Solution: Using the long division method: $\therefore \sqrt{11449}=107$...
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Question: Evaluate: $\sqrt{9025}$ Solution: Using the long division method: $\therefore \sqrt{9025}=95$...
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Question: Evaluate: $\sqrt{7056}$ Solution: Using the long division method: $\therefore \sqrt{7056}=84$...
Read More →In a figure the common tangents,
Question: In a figure the common tangents, AB and CD to two circles with centres 0 and O intersect at E. Prove that the points 0, E and O are collinear. Solution: Joint $A O, O C$ and $O^{\prime} D, O^{\prime} B$. Now, in $\triangle E O^{\prime} D$ and $\triangle E O^{\prime} B$, $O^{\prime} D=O^{\prime} B$ [radius] $O^{\prime} E=O^{\prime} E$ [common side] $E D=E B$ [since, tangents drawn from an external point to the circle are equal in length] $\therefore$ $\triangle E O^{\prime} D \cong \tri...
Read More →The set of points where the function
Question: The set of points where the function $f(x)=\left\{\begin{array}{cc}x+1, x2 \\ 2 x-1, x \geq 2\end{array}\right.$ is not differentiable, is_________________ Solution: The given function is $f(x)=\left\{\begin{array}{cl}x+1, x2 \\ 2 x-1, x \geq 2\end{array}\right.$. $(x+1)$ and $(2 x-1)$ are polynomial functions which are differentiable at each $x \in \mathrm{R} .$ So, $f(x)$ is differentiable for all $x2$ and for all $x2$. So, we need to check the differentiability off(x) atx= 2. We hav...
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Question: Evaluate: $\sqrt{6241}$ Solution: Using the long division method: $\therefore \sqrt{6241}=79$...
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Question: Evaluate: $\sqrt{4489}$ Solution: Using the long division method: $\therefore \sqrt{4489}=67$...
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Question: Evaluate: $\sqrt{1444}$ Solution: Using the long division method: $\therefore \sqrt{1444}=38$...
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Question: Evaluate: $\sqrt{576}$ Solution: Using the long division method: $\therefore \sqrt{576}=24$...
Read More →The set of point where the function
Question: The set of point where the function $f(x)=|2 x-1|$ is differentiable, is_______________ Solution: The given function is $f(x)=|2 x-1|$. $f(x)=|2 x-1|= \begin{cases}2 x-1, x \geq \frac{1}{2} \\ -(2 x-1), x\frac{1}{2}\end{cases}$ Now, $(2 x-1)$ and $-(2 x-1)$ are polynomial functions which are differentiable at each $x \in \mathrm{R} .$ So, $f(x)$ is differentiable for all $x\frac{1}{2}$ and for all $x\frac{1}{2}$. So, we need to check the differentiability of $f(x)$ at $x=\frac{1}{2}$. ...
Read More →Find the least square number which is exactly divisible by
Question: Find the least square number which is exactly divisible by each of the numbers 8, 12, 15 and 20. Solution: The smallest number divisible by each of these numbers is their L.C.M. L.C.M. of 8, 12, 15, 20 = 120 Resolving into prime factors: $120=2 \times 2 \times 2 \times 3 \times 5$ To make this into a perfect square, we need to multiply the number with $2 \times 3 \times 5=30$. Required number $=120 \times 30=3600$...
Read More →Prove that the tangent drawn at the mid-point
Question: Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. Solution: Let mid-point of an arc AMB be M and TMT be the tangent to the circle. Join AB, AM and MB. Since, $\operatorname{arc} A M=\operatorname{arc} M B$ Chord $A M=$ Chord $M B$ $\ln \triangle A M B$ $A M=M B$ $\Rightarrow \quad \angle M A B=\angle M B A$ [equal sides corresponding to the equal angle] ... (i) Since, $T M T^{\prime}$ is a tangent line. $\the...
Read More →Find the least square number which is exactly divisible by
Question: Find the least square number which is exactly divisible by each of the numbers 6, 9, 15 and 20. Solution: The smallest number divisible by each of these numbers is their L.C.M. L.C.M. of 6, 9, 15, 20 = 180 Resolving into prime factors: $180=2 \times 2 \times 3 \times 3 \times 5$ To make it a perfect square, we multiply it with 5. Required number $=180 \times 5=900$...
Read More →AB is a diameter and AC is a chord of a circle
Question: AB is a diameter and AC is a chord of a circle with centre 0 such that BAC = 30. The tangent at C intersects extended AB at a point D. Prove that BC = BD. Solution: A circle is drawn with centre O and AB is a diameter. AC is a chord such that BAC = 30. Given AB is a diameter and AC is a chord of circle with certre O, BAC = 30....
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