🔥 SSC 2025 highermath Suggestion.. Higher Mathematics Complete Practical Questions and Solutions

Описание: 🔬SSC Higher Math Practical Exam 2025.. ssc 2025 highermath practical.. ssc
) Draw the graph of y=2x+1y = 2^{x+1}y=2x+1 and find the inverse function.
1. Original function y=2x+1y = 2^{x+1}y=2x+1:
This function is an exponential function, whose graph will be an upward-sloping curve.
Finding some points:
xxx
y=2x+1y = 2^{x+1}y=2x+1
-2
2−1=0.52^{-1} = 0.52−1=0.5
-1
20=12^0 = 120=1
0
21=22^1 = 221=2
1
22=42^2 = 422=4
2
23=82^3 = 823=8
Now we need to draw the graph using these points.
2. Finding the inverse function:

y=2x+1y = 2^{x+1}y=2x+1

Let's say, y=2x+1y = 2^{x+1}y=2x+1
Take the log base of two:
log⁡2y=x+1\log_2 y = x+1log2​y=x+1

⇒ x=log⁡2y−1x = \log_2 y - 1x=log2​y−1

The inverse function will be:

f−1(x)=log⁡2x−1f^{-1}(x) = \log_2 x - 1f−1(x)=log2​x−1

(e) The points M(2,7),N(−4,3),P(−5,−4),Q(6,−8)M(2, 7), N(-4, 3), P(-5, -4), Q(6, -8)M(2,7),N(−4,3),P(−5,−4),Q(6,−8) are the four vertices of a quadrilateral. Draw the quadrilateral in the figure and find its area.

Here we will find the area of ​​the quadrilateral using the vertices. To find the area of ​​the quadrilateral, we will use the Shoelace Formula:
Area=12∣x1y2+x2y3+x3y4+x4y1−(y1x2+y2x3+y3x4+y4x1)∣\text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|Area=21​∣x1​y2​+x2​y3​+x3​y4​+x4​y1​−(y1​x2​+y2​x3​+y3​x4​+y4​x1​)∣
The values ​​of the points are:
M=(2,7)M = (2, 7)M=(2,7)
N=(−4,3)N = (-4, 3)N=(−4,3)
P=(−5,−4)P = (-5, -4)P=(−5,−4)
Q=(6,−8)Q = (6, -8)Q=(6,−8)
Apply In the formula:Area=12∣(2×3+(−4)(−4)+(−5)(−8)+6×7)−(7×(−4)+3×(−5)+(−4)×6+(−8)×2)∣\text{Area} = \frac{1}{2} \left| (2×3 + (-4)(-4) + (-5)(-8) + 6×7) - (7×(-4) + 3×(-5) + (-4)×6 + (-8)×2) \right|Area=21​∣(2×3+(−4)(−4)+(−5)(−8)+6×7)−(7×(−4)+3×(−5)+(−4)×6+(−8)×2)∣
=12∣(6+16+40+42)−(−28−15−24−16)∣= \frac{1}{2} \left| (6 + 16 + 40 + 42) - (-28 -15 -24 -16) \right|=21​∣(6+16+40+42)−(−28−15−24−16)∣
=12∣104−(−83)∣=12(104+83)=1872=93.5= \frac{1}{2} \left| 104 - (-83) \right| = \frac{1}{2} (104 + 83) = \frac{187}{2} = 93.5=21​∣104−(−83)∣=21​(104+83)=2187​=93.5
Answer: Area = 93.5 square units

SSC HigherMath Practical Exam 2025.. ssc highermath practical । ssc
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🔥 SSC 2025 highermath Suggestion.. Higher Math Complete Practical Questions and Solutions
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